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Forces and Kinematics Problem - Am I right?

  1. Jan 17, 2006 #1
    Can someone tell me if I'm right? 0.98 seems a bit high for the coefficient of friction?

    A 25kg sled is pulled at a steady 1.3 m/s by a rope inclined 40 degrees above horizontal. If the pulling force is 26.7 N, what is the coefficient of friction between the sled runners and the snow?

    m = 25 kg, v = 1.3 m/s, Fp = 26.7 N, mgcos40 = Fn = 187.7 N,

    Ftotal = ma (Ftotal = 0 because of no acceleration).
    Fp + mgsin40 - Ff = 0,
    26.7 + 157.5 = Ff, Ff = 184.2.

    u = Ff/Fn = 184.2/187.7 = 0.98.
     
  2. jcsd
  3. Jan 17, 2006 #2

    Hootenanny

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    Using g as 9.8 I got 0.697(3sf)
     
  4. Jan 17, 2006 #3
    Can you please post your work? :rofl:
     
  5. Jan 17, 2006 #4

    Hootenanny

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    [tex]F= \mu R
    \\
    \mu R + 25gcos50=26.7
    \\
    \mu 25gsin50=26.7 - 25gcos50
    \\
    \mu = \frac{26.7-25gcos50}{25gsin50}
    \\
    \mu = 0.697[/tex]
    Sorry it took my a while to sought out the LaTex. lol :blushing: I just wanted you to know I was workin on it.
     
    Last edited: Jan 17, 2006
  6. Jan 17, 2006 #5
    I don't understand that very much. what's muRmuR? like...lol
     
  7. Jan 17, 2006 #6
    Sorry, I'm having a lot of trouble and that whole long equation just confused me more.
     
  8. Jan 17, 2006 #7

    Hootenanny

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    it shouldn't come out like that. I'm jus correcting it. I Hate Latex
     
  9. Jan 17, 2006 #8
    ok cause i dunno where you got all the numbers from
     
  10. Jan 17, 2006 #9

    Hootenanny

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    [tex]F= \mu R[/tex]
    [tex]\mu R + 25gcos50=26.7[/tex]
    [tex]\mu 25gsin50=26.7 - 25gcos50[/tex]
    [tex]\mu = \frac{26.7-25gcos50}{25gsin50}[/tex]
    [tex]\mu = 0.697[/tex]
    I've tried to lay it out better
     
  11. Jan 17, 2006 #10
    R is the normal force?

    by the way, thank you for helping me along so much so far :)
     
  12. Jan 17, 2006 #11

    Hootenanny

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    No worries. Yeah, R is the normal reaction. mu is the co-effiecent of friction.
     
  13. Jan 17, 2006 #12
    you used 50 degrees as the angle. it's 40 though? :\
     
  14. Jan 17, 2006 #13

    Hootenanny

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    cos50=sin40.
     
  15. Jan 17, 2006 #14

    Hootenanny

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    [tex]F= \mu R[/tex] is the equation for maximal frictional force.
     
  16. Jan 17, 2006 #15
    numerically, that's correct. but i don't understand that concept at all. can you demonstrate it with 40 degrees? i apologize for being so lost :(
     
  17. Jan 17, 2006 #16
    F = uR is, yes.
     
  18. Jan 17, 2006 #17

    Hootenanny

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    Okay. Have you drawn a diagram with the slope and all the forces acting?
     
  19. Jan 17, 2006 #18
    F(total) = ma

    F(total) = 0

    F(normal) = (m)(g)(cos40) = (25)(9.8)(cos40) = 187.7

    F(pulling) - F(friction) + 25(9.8)(sin40) = 0

    26.7 + 25(9.8)(sin40) = F(friction)

    26.7 + 157.5 = F(friction)

    184.2 = F(friction)

    F(friction) = F(normal) * mu

    mu = F(friction)/F(normal) = 184.2/187.7 = 0.98

    The coefficient of friction equals 0.98.
     
  20. Jan 17, 2006 #19
    Yes, I drew a free body diagram.
     
  21. Jan 17, 2006 #20

    Hootenanny

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    Third line down. it should read: F(pulling)-F(Friction)-25(9.8)(sin40)=0 because the component of the weight it acting in the opposite direction to the pulling force.
     
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