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The force of a mazda RX-8

  1. Jan 2, 2010 #1
    1. The problem statement, all variables and given/known data

    Pick a car, find its acceleration. calculate the force necessary to move the car with this acceleration and the minimum coefficient of friction to accomplish this. repeat the calculation of the coefficient of friction on a 20 degree incline.

    2. Relevant equations

    How do i calculate the coefficient of friction? on a 20 degree incline as well?

    3. The attempt at a solution
    0-60= 5.8s

    acceleration= change in velocity/ change in time
    acceleration= 26.82m/s / 5.8s = 4.62 m/s2

    weight= massx acceleration of gravity
    1392.7= m(9.81 m/s2)
    mass= 141.97

    force= mass x acceleration
    F= 141.97x 4.62 m/s2
    F= 646.49 N

    How do i find coefficient of friction and the coefficient of friction on a 20 degree incline?
    force= mass x acceleration
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jan 2, 2010 #2
    Since you have the amount of force on the plane, you can take that and set it equal to the force of friction.

    646.49 = m* g * Cf (Coefficient of Friction, since keyboards don't have mew), divide the mass and g off and you have the coefficient of friction
  4. Jan 2, 2010 #3


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    Homework Helper

    Hi emonroe14, welcome to PF.
    Post the complete text of the problem.
  5. Jan 3, 2010 #4
    Friction is always parallel to a surface and is figured by the component of force normal to the surface.

    The CF is a ratio between the normal force and the force of friction it generates. It varies with surfaces (including the tires).

    On a flat surface the normal force is typically the mass times the acceleration of gravity. But when the surface is tilted, the force of gravity is still straight down, while the force causing the friction is perpendicular to the surface. (Think of using some trig here.)

    The friction force must be greater than the force of acceleration, so that sets the lower limit.
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