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Newton's Law Friction Problem

  1. Feb 20, 2009 #1
    Ok, I was watching a physics lecture and there was an interesting example given

    Heres the Example

    "A rubber tire has a static coefficient of one, so at an angle of 45 degrees, the car will start to slide, which is independent of the area of the tires and mass of the car."

    So i did the calculations real quick to see if this is true.

    For the F (Friction Max) = Coefficient Friction x Normal Force

    On an incline:
    In the X direction, the Force can be measured in Mass(Gravity)Sin(Theta)
    In the Y Direction, the force can be measured in Mass(Gravity)Cos(Theta)

    (Correct me if i get any of this stuff wrong)

    According to Newton's Law, The force that the car exerts on the ground, the ground as to exert an equal amount of force if the car has no acceleration in the y direction.

    So: Mass(Gravity)Cos(Theta) Equals (=) Normal Force (Nf)

    And at the Point of Breaking off toe Accelerate, the equation would be:

    Mass(gravity)sin(Theta) - F(friction Max) = 0

    Substitute F(friction Max) with (static friction = Ms)(Mass(gravity)Cos(Theta))

    So you would get

    Mass(gravity)sin(Theta) - Ms(M)gCos(theta) = 0

    Deriving it from the equation, Than

    Tan(theta) = Ms (Static Friction)


    So I plugged Theta = 45 degrees
    and 100 kg to those equations

    F(Friction Max) = Mu x Normal Force

    Tan(45)(100kg)(9.81m/s2)(Cos45) = 693.6717523

    The Force in the X direction = (100kg)(9.81m/s2)Sin45 = 693.6717523 also

    Since the forces are the same, how can the car move? The forces cancel each other out, granted a tiny force will able to make it slide down but if nothing touches is, the car will still be stationary. So, I can't quite figure out why the professor is saying the car will start to slide down at an angle of 45 degrees since both the Friction Max Force is equal to the X direction Force.

    Anyone want to explain this for me?
  2. jcsd
  3. Feb 20, 2009 #2


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    Homework Helper

    Seems to me you explained it yourself! Nice job, too.
    Zero force holding it in place marks the point between sliding and not sliding, which you might call the beginning of the slide.
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