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Another friction problem!

  1. Sep 17, 2004 #1
    Ok here is another problem that I am stuck on!
    A 20.0kg sled is being pulled across a horizontal surface at a constant velocity. The pulling force has a magnitude of 80.0N and is directed at an angle 30.0 degrees above the horizontal. Determine the coefficient of friction.
    I think that Fn=(20.0)(9.80)(cos 30) and I know that uk=Fk/Fn and a=0m/s(sqr) but I am stuck on how to go about finding Fk?
    Am I on the right track?
     
  2. jcsd
  3. Sep 17, 2004 #2

    Pyrrhus

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    You're using Newton's First Law [tex] \sum^n_{i=0} \vec{F_{i}} = 0 \Rightarrow \vec{V} = constant [/tex] that's good :smile: .

    Now, Remember you got a weight pointing down, the normal force pointing up, friction that always points against movement, so in this case is pointing left, and you got a force pointing at angle 30 degrees, so put the force in its components Fy and Fx.

    [Edits: I was just checking LaTex :biggrin:]
     
    Last edited: Sep 17, 2004
  4. Sep 17, 2004 #3

    Pyrrhus

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    If Fn is normal force then you are not calculating it right.
     
  5. Sep 17, 2004 #4
    So is it W=20(9.80)=196N and then 80(cos 30)-Ff=0 and Ff=69N
    I know that something is still wrong because I am still getting the wrong answer!
     
  6. Sep 17, 2004 #5

    Pyrrhus

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    Y-axis:

    [tex] F_{y} + N = mg [/tex]

    [tex] N = mg - F_{y}[/tex]

    X-axis:

    [tex] F_{f} = F_{x} [/tex]

    [tex] \mu N = F_{x} [/tex]
     
  7. Sep 17, 2004 #6

    Pyrrhus

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    By the way your problem is asking:
    not the force of friction!

    If you notice Coefficient of Friction has no units and its represented as [tex] \mu [/tex]
     
  8. Sep 17, 2004 #7
    what is N, is that the normal force?
     
  9. Sep 17, 2004 #8
    From what I did, your Ff is right, but you need to solve for Fn too.

    Sum forces in each direction, set it equal to Ma

    Sum in the Y = -m*g + Fn + 80sin(30) = ma = 0 // NOT ACCELERATING

    Sum in the X = 90cos(30) - Ff = 0 // Not accelerating again.

    So solve top for Fn, solve bottom for Ff, then use Ff = mu*Fn

    I get 0.44 = mu_k
     
  10. Sep 17, 2004 #9

    Pyrrhus

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    N is the normal force, yes, i represented it as such
     
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