1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Net Forces on a Car on a Turn

  1. Dec 10, 2008 #1
    1. The problem statement, all variables and given/known data
    At a certain instant of time, a 1200 kg car traveling along a curve 250 m in radius is moving at a speed of 10 m/s but is slowing down at a rate of 2 m/s2. Ignoring air friction, what is the total static friction force on the car as a fraction of its weight at that instant?

    2. Relevant equations
    Fc = mv2/R
    Ff = [tex]\mu[/tex]FN = [tex]\mu[/tex]mg

    3. The attempt at a solution
    I have an idea at how to attempt this...
    mv2/R = Ff
    (since the centripetal force is provided by the frictional force)
    but I need clarification: I see it that the frictional force and the force due to the acceleration (deceleration) of the car act perpendicular to each other and therefore on different axes.
    How would you account for this force due to deceleration in calculating the centripetal force? Is this just a matter of calculating the speed of the car at that instant, in which case the deceleration doesn't matter?
  2. jcsd
  3. Dec 10, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Careful. Friction provides all the force accelerating the car, not just the centripetal force. There's both a tangential and a centripetal component.
  4. Dec 10, 2008 #3
    Ah, okay. So the total frictional force is equal to mv^2/R - 2m = -1.6m?
  5. Dec 10, 2008 #4


    User Avatar
    Homework Helper

    Unfortunately it's not that simple.

    The accelerations are vectors and they are not acting in the same direction.

    Figure the sum of the vectors first because the slowing is tangential and the centripetal is radial.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Net Forces on a Car on a Turn
  1. Car turning (Replies: 1)

  2. Net Force and Net Work (Replies: 1)