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Finding coefficient of static friction with velocity and radius

  1. Aug 28, 2011 #1
    1. The problem statement, all variables and given/known data

    A bus passenger has her laptop sitting on the flat seat beside her as the bus, traveling at 10.0 m/s, goes around a turn with a radius of 25.0 m. What minimum coefficient of static friction is necessary to keep the laptop from sliding?

    Given:

    V = 10 m/s
    r = 25.0 m

    2. Relevant equations

    Fc = mv ^ 2 / r

    f = Us x Fn

    3. The attempt at a solution

    Fs = Force of static friction


    So far I have:

    Fc = mv^2 / r

    Fs + mg = mv^2 / r

    Fs = v^2 / r - g

    Fs = 10 ^ 2 / 25 - 9.8

    Fs = -5.8


    I don't think I am on the right track with this as my Fs is a negative. Any hint to the proper direction would be much appreciated. Thank you
     
  2. jcsd
  3. Aug 28, 2011 #2
    I do not understand what Fs is and why do you add the weight to it. Since your object does not slide you can find a relation involving only the centripetal force and the friction force. From that you solve for the mu_s_min
     
  4. Aug 28, 2011 #3
    Fs = Force of Static friction ?

    Do I not need to find Fs?
     
  5. Aug 28, 2011 #4

    PeterO

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    Homework Helper

    It is the adding of mg that is wrong. gravity acts vertically down. Fs acts horizontally. The centripetal force required is horizontal.

    The normal force would be useful on two fronts. It balances the weight force and along with the coefficient of friction, enables you to calculate that coefficient.

    Fs + mg + Fn = mv^2 / r is what you needed - and that is a vector sum formula so don't just go numerically adding and subtracting bits unless they are in the same direction
     
  6. Aug 30, 2011 #5
    So how would I go about finding Fn and Fs ?

    This is where im totally lost, the only givens I have are the radius = 25 m and V = 10 m/s
     
  7. Aug 30, 2011 #6

    PeterO

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    Your 3rd equation should have been

    Fs + mg + Fn = mv^2 / r

    If we assume the seat is horizontal [a reasonable assumption] then the normal force will be the same size as the weight force, so they will balance, and could have been left out all together.

    Fs = mju * Fn

    where mju is the coefficient of friction.

    Ultimately the mass will cancel out - which is useful as it would be unfortunate if only the light/heavy passengers in a bus could go around the corner with the bus.
     
  8. Jun 20, 2012 #7
    I am also working on this question and stumbled upon the answer of 0.4. Without plugging in all of my work is anyone able to tell me if I am correct or completely wrong?
     
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