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Coefficient of Friction problem!

  1. Oct 22, 2006 #1
    The question states: "What is the coefficient of friction between a sled and a plane inclined at 30 degrees from the horizontal if the sled just slides without accelerating when given an initial push?"

    I have drawn a free-body diagram and labeled all of the forces (I think i labeled all of them) but now I don't know what to do.. What equations will I need to write and what numbers will I need to use? If somebody could guide me through this it would be most appreciated..
     
  2. jcsd
  3. Oct 22, 2006 #2

    Hootenanny

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    Welcome to the forums,

    Could you describe your FBD, what forces do you have acting in which directions?
     
  4. Oct 22, 2006 #3
    I whipped up a little diagram of what I thought the forces were

    [​IMG]
     
  5. Oct 22, 2006 #4

    Hootenanny

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    There is one error there, would the weight really be acting in that direction?
     
  6. Oct 22, 2006 #5
    Whoops, shouldn't it be going from the sled STRAIGHT down, perpendicular to the bottom of the triangle?
     
  7. Oct 22, 2006 #6

    Hootenanny

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    Indeed it should.
     
  8. Oct 22, 2006 #7
    With that being said, I am going to need to write my equations.. I believe that frictional force can be modeled with the equation (Ff = μ Fn) and Fn in this context is equal to mgcos30. Also, mgsin30 minus the frictional force would be equal to ma. However, if we don't know M how can we solve for the coefficient of friction in the first equation?
     
  9. Oct 22, 2006 #8

    arildno

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    I assume you with M means the mass of the sled.

    Why don't you set up Newton's 2.law and see what can be done with that pernicious M? :smile:
     
  10. Oct 22, 2006 #9
    I still don't understand what to substitute into Newton's 2.Law; Acceleration must be zero if the sled slides without accelerating with an initial push, and we don't know the value for m, as well as the value for the net force.. By playing around with the forces I labeled, I got the equation (mgsin30 - μk*mgcos30) = ma) but this may not be correct.

    Ahhh, I can't seem to understand this =(
     
  11. Oct 22, 2006 #10

    arildno

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    "Also, mgsin30 minus the frictional force would be equal to ma."

    Correct!
    Hence, with a=0, do you agree that we have:
    [tex]mg\sin(30)=F_{fric}[/tex]
    where [itex]F_{fric}[/itex] is the frictional force?
     
  12. Oct 22, 2006 #11

    arildno

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    This IS correct.

    So, what does that equation say when you know that a=0?
     
  13. Oct 22, 2006 #12
    It will say after simplifying:

    4.9m - μk * 8.48m = 0
     
  14. Oct 22, 2006 #13

    arildno

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    To set in numbers is NOT simplification!!
    Do as follows:
    [tex]mg\sin(30)=\mu{m}g\cos(30)[/tex]
    Now, divide with [itex]mg\cos(30)[/itex]:
    [tex]\frac{mg\sin(30)}{mg\cos(30)}=\frac{\mu{mg}\cos(30)}{mg\cos(30)}[/tex]

    Now, what factors cancels in each of the fractions?
     
  15. Oct 22, 2006 #14
    ahh i got it now.. Thanks so much for your help everyone!
     
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