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Determine acceleration on slope using coefficient of static friction

  1. May 14, 2008 #1
    A sports car is accelerating up a hill that rises 21.2 ° above the horizontal. The coefficient of static friction between the wheels and the road is μs = 0.880. It is the static frictional force that propels the car forward. (a) What is the magnitude of the maximum acceleration that the car can have? (b) What is the magnitude of the maximum acceleration if the car is being driven down the hill?

    I set up my fbd with the static frictional force acting in the -x direction, the normal force acting perpedicular to the inclined plane, and the weight acting in the -y direction. Without knowing the weight of the car or if it is at a constant velocity, I'm not sure how to go about setting up this problem.
  2. jcsd
  3. May 14, 2008 #2


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    Welcome to PF linzeluv,

    I think you'll find the in the end you don't need the mass of the car, it'll just cancel out in your calculations.

    Secondly, you already know that it isn't travelling at a constant velocity because it's accelerating!
  4. May 14, 2008 #3
    Setting up the equation

    I thought this would be on the right track, but my answer for part A still isn't right:

    Maximum Acceleration = Net Force in X-direction/ mass = -mg sin 21.2deg + maximum static frictional force/ mass

    Maximum static frictional force = coefficient of static friction * Normal Force

    Net Force in y-direction = ma = 0

    Net Force in y-direction = Normal Force - mg cos 21.2deg = 0
    Normal Force = mg cos 21.2deg

    Final Equation

    Maximum acceleration = -mg sin 21.2deg + .880 (mg cos 21.2deg)/ mass

    Mass cancels out and answer is 11.6 m/s^2

    Not sure where I'm going wrong.
  5. May 14, 2008 #4


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    All your working and your final equation is correct, but your final answer is wrong. You must have punched the numbers into your calculator incorrectly.
  6. Jun 6, 2009 #5
    make sure your calculator is in degrees mode not radians, I fell foul of that several times :-)
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