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Coefficient of Static Friction with Acceleration

  1. Mar 1, 2012 #1
    1. The problem statement, all variables and given/known data

    A car can decelerate at -4.90 m/s^2 without skidding when coming to rest on a level road.
    What would its deceleration be if the road were inclined at 11° uphill? Assume the same static friction coefficient.

    2. Relevant equations

    Force of Static Friction / Normal Force = coefficient of static friction
    F (normal) - mg *sin 11° = 0

    3. The attempt at a solution

    (4.90m/s^2)/(9.8 m/s^2) = 0.5

    I didn't know the mass so I just left it out.
    F - mg *sin 11° = 0
    F (normal) = mg * sin 11°
    F = (9.8 m/s^2) * sin 11°
    Normal force = 1.86 m/s^2

    I know those are acceleration units so something more than likely went wrong there.

    Force of static friction = (0.5)(1.86 m/s^2)
    Force of Static Friction = 0.93 m/s^2
     
  2. jcsd
  3. Mar 1, 2012 #2

    Doc Al

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    Staff: Mentor

    I don't understand this equation.
    Good.

    To find the acceleration, use ƩF = ma. What's the net force when going uphill?

    (Hint: What are the parallel and perpendicular components of the weight?)
     
  4. Mar 1, 2012 #3
    Doc Al - Sorry, the first one was a little unclear. I was attempting to find the normal force by saying the normal force equals the y component of the weight because there is no acceleration in the "y" direction. I think I should have used cosine instead of sine. Thank you for your advice.
     
  5. Mar 1, 2012 #4

    Doc Al

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    Staff: Mentor

    Exactly.
     
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