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Friction and Constant velocity

  1. Sep 24, 2008 #1
    1. The problem statement, all variables and given/known data
    A car of mas m = 1000kg is sliding down a hill. The coefficients of friction between the car's tires and the ground are us=0.89 and uk=0.61. For what inclination angle will the car slide down the hill with a constant velocity?


    2. Relevant equations



    3. The attempt at a solution

    Now I realize that because it is constant velocity, the acceleration is 0, so the summation of the forces needs to be 0, but i don't know how to set this up. Any help?

    Thanks. :)
     
  2. jcsd
  3. Sep 25, 2008 #2

    tiny-tim

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    Hi CaptainSFS! :smile:

    Hint: there are only three forces … the weight, the normal force, and the friction force … so set up a force triangle :smile:
     
  4. Sep 25, 2008 #3
    hello, yes, so i have my two frictional forces pointing upwards on the slope. I set it up like this... (1000)(9.81cos(theta))(.89)+(1000)(9.81cos(theta))(.61)-?=0 ...If this is correct, which I am not so sure b/c it just seems like I wouldn't be able to create an opposite force to equal it, so that the acceleration is 0. These frictional forces really confuse me, and I feel like I keep setting it up wrong.
     
  5. Sep 25, 2008 #4

    tiny-tim

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    Nooo …

    i] you can't have both kinetic friction and static friction … it's one or the other

    ii] what about the weight force?
     
  6. Sep 25, 2008 #5
    okay, so I'm assuming them I will be using the static coefficient because that's the kind tires use on pavement and whatnot. For the weight force... would that be the force pulling the car down the slope? m*g*sin(theta)?

    or do u mean the weight force as 1000*9.81?
     
  7. Sep 25, 2008 #6

    tiny-tim

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    Whoa!

    that would be for wheels that are rolling

    the car is sliding (in other words, the wheels are locked)
    the weight force is mg, and you use its component in the required direction.
     
  8. Sep 25, 2008 #7
    okay, so if I'm using it in the required direction, don't I have to use m*g*sin(theta) because gravity isn't pulling it from the side, it's pulling straight down on it?
     
  9. Sep 25, 2008 #8

    tiny-tim

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    Yes! Get on with it! :rolleyes:
     
  10. Sep 25, 2008 #9
    thanks!
     
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