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Coefficent of friction

  1. Apr 18, 2008 #1
    Hello, I'm towards the end of my first semester of Physics 1 (cal based) and I've had this homework problem that has been making want to pull my hair out. I know that the coeffeicent of friction is between 0 and 1 and I have worked out the problem like my professor did but keep getting the wrong answer. Thank you for any help.

    Michael

    1. The problem statement, all variables and given/known data
    A banked circular highway curve is designed for traffic moving at 70 km/h. The radius of the curve is 215 m. Traffic is moving along the highway at 40 km/h on a rainy day. What is the minimum coefficient of friction between tires and road that will allow cars to negotiate the turn without sliding off the road?



    2. Relevant equations
    f-mg*sin(x)=ma*cos(x)
    N-mg*cos(x)=ma*sin(x)

    tan(x)=v^2/Rg

    where:
    m= mass = cancels out
    g= gravity = 9.8 m/s^2
    a= acceleration = v^2/R = 11.11^2/215 = .574 m/s^2
    x= angle
    v= velocity = 40 km/h = 11.11 m/s
    R= radius = 215m

    3. The attempt at a solution

    coefficent of friction = (g*sin(x)-a*cos(x))/(g*cos(x)+a*sin(x))

    x= tan^-1(11.11^2/(215*9.8))
    x= 3.353

    f = (9.8*sin(3.353)-.574*cos(3.353))
    (9.8*cos(3.353)-.574*sin(3.353))

    f = -1.01*10^-15
     
  2. jcsd
  3. Apr 18, 2008 #2

    alphysicist

    User Avatar
    Homework Helper

    Hi 72_Cutlass_S,

    I think there is a problem with your angle. When the problem says that the highway is designed for a speed of 70km/h, that means that at that speed friction is not needed to carry the car around the curve. The formula [itex]\tan\theta = v^2/(rg)[/itex] applies to that special speed. So I think you put in the wrong speed.

    There is also a typo in your first equation (f-mg*sin(x)=ma*cos(x)); the right hand side needs a negative sign to be consistent with the left hand side. However, the minus sign appears later on, which is why I think it you just accidently left it out.

    The next to the last line also has a minus sign in the denominator which I don't think should be there.
     
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