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Inclined force problem

  1. Jun 14, 2008 #1
    Hi it's been a while since I posted here so I will avoid the latex on this question and hope you understand.

    The coefficient of kinetic friction for a 22kg bobsled on a track is .10. What force is required to push it down a 6.0 degree incilne and achieve a speed of 60km/h at the end of 75m?

    So this is what I did:

    I found the acceleration of the sled to be 1.85 m/s^2
    Then I found the normal force= mg times cos 6= 214N
    I then found the force of friction to be 21N
    I then set Net Force equal to ma= 22 times 1.85 which equaled 40.7
    Then I set 40.7 equal to forcex minus force of friction
    and solved for forcex

    Can anyone point out what I did wrong?
    Thanks
     
  2. jcsd
  3. Jun 14, 2008 #2

    Doc Al

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    What forces acting on the sled parallel to the track?
     
  4. Jun 14, 2008 #3
    You mean like force of friction and Sin (6) times mg which is the horizontal component of Force of gravity
     
    Last edited: Jun 14, 2008
  5. Jun 14, 2008 #4

    Doc Al

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    That's the one you missed. (Three forces act, including the applied force which you are trying to find.)

    So set up an equation for the net force.

    Yes.
     
  6. Jun 14, 2008 #5
    O I was confused by that. So wouldn't then net force just equal the applied force?
     
  7. Jun 14, 2008 #6

    Doc Al

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    Why would you say that? The net force is the sum of all the forces acting on the sled--the applied force is just one of those forces. (Pay attention to direction--sign--when you add the force components.)
     
  8. Jun 14, 2008 #7
    Well I figured the ohter forces cancel out since the object was at rest in the beginning and needed an applied force to move.
     
  9. Jun 14, 2008 #8

    Doc Al

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    No reason to assume so. Especially when you can calculate the forces and know for sure.
     
  10. Jun 14, 2008 #9
    Yea I forgot it could be moving but just not accelerating. Thanks for your help.
     
  11. Jun 14, 2008 #10

    Doc Al

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    We must assume (lacking information to the contrary) that it starts from rest. But that doesn't mean it was just sitting there waiting to be pushed.
     
  12. Jul 1, 2009 #11
    I'm solving the same problem. I don't understand how to find the friction force. Can someone explain in this particular problem?
     
  13. Jul 1, 2009 #12
    There is a formula for finding the force of friction. It's the coefficient of friction multiplied by the normal force. If you can find the normal force then I don't see your problem.
     
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