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Coefficient of Sliding Friction

  1. May 21, 2007 #1
    Golf Cart Questions

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

    Ms. Boylan on her 1750-kg cart is traveling at a constant speed of 8.5-m/s on a wet, straight, and slippery road. A mindless student steps out in from of the cart. The wheels lock and she begins skidding and slides to a halt in 12.0-m. What is the coefficient of sliding friction between the tires and the slippery road?

    2. Relevant equations

    vf^2 = vi^2 + 2ad

    a = (Fapplied - Fopposed) / mass

    u = Ffriction / weight

    3. The attempt at a solution

    Her acceleration is found with:
    (0)^2 = (8.5)^2 + 2(a)(12)
    a = -3.01 m/s^2

    So, -3.01 equals applied force minus Ffriction over 1750.

    Then I get stuck. If I could just figure that out, I could divide it by (1750 * 9.80) to get the coefficient. Thanks for your help.
     
    Last edited: May 21, 2007
  2. jcsd
  3. May 21, 2007 #2
    There is no force applied.
    Initially, she is going at constant speed so the forces are balanced. Force net = 0.
    Afterwards, she is skidding so the only force acting when she is slowing down is the force of friction.
    Ff=ma
    And you know from here on. :)
     
    Last edited: May 21, 2007
  4. May 21, 2007 #3
    Thanks a bunch! Just one other golf cart question (not friction related).

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

    Two campus supervisors in golf carts approach each other, one from the east (Cart A) and the other from the south (Cart B). Cart A has a mass of 1750-kg and cart B has a mass of 1275-kg and is traveling at 6.7056 meters per second. They collide, become locked, and move together at an angle of 30.0 degrees. What was the speed of cart A?

    2. Relevant equations

    (m1)(v1) + (m2)(v2) = (m1+m2)(v)

    3. The attempt at a solution

    (1750)(v1) + (1275)(6.7056) = (3025)(v)

    I'm stuck here trying to solve for v1. I obviously need to find the velocity of the combined mass, but I'm not sure how to use 30 degrees.
     
    Last edited: May 21, 2007
  5. May 22, 2007 #4
    Momentum is a vector, not a scalar. It also has direction.
    P1 + p2 = pf
    f=final
    This only works as a vector sum and if the collision is in one dimension. Your collision is in two dimensions. So in two dimensions, if you draw a diagram, you have that p1 + p2 = pf (as vectors).
    You then have a right angle triangle with the hypotenuse pf and the other two sides are p1 and p2.
    p1^2 + p2^2 = pf^2
    You also know the angle, which can help you determine another relationship.
     
    Last edited: May 22, 2007
  6. May 22, 2007 #5
    Oh, so it's just 6.7 tan 30 = 3.87 ?
     
  7. May 22, 2007 #6
    Well actually I don't know. It doesn't say if the angle of 30 is formed with the horizontal or with the vertical. But that's how you would do it. I just don't know where the 30 degree angle is.
     
  8. May 23, 2007 #7
    Thanks for your help, Husky.
     
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