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Forces on a bike

  1. Oct 21, 2003 #1
    A biology student rides her bike around a corner of radius 30 meter at a steady speed of 8.1 m/sec. The combined mass of the student and the bike is 89 kg. The coefficent of static friction between the bike and the road is ìs = 0.32.

    a) If she is not skidding, what is the magnitude of the force of friction on her bike from the road?
    b) What is the minimum value the coefficient of static friction can have before the bike tire will skid?
    c) What is the magnitude of the total force between the bike tire and the road?
    I'm stuck on part c, but maybe doing part a and b can help out also.

    a) A=v^2/R
    so A=8.1^2/30=2.187m/s/s
    Then I used F=m*a => 89kg*2.187m/sec/sec => 194.643 newtons

    b) Well, I used Newton's second law for the x axis, and figured out that
    -fs= -m(v^2/R)
    where fs is static friction force, m is mass, v is velocity, R is radius
    and I know that fs(max)=Us*N
    where Us=static coefficient of friction
    so I plug in for fs and solve for Us.
    since N=m*g
    Us=(m*v^2)/(m*g*R) or Us= (v^2)/(g*R)
    plug in my numbers, and I get minimum Us to be .223. That is when the bike is on the verge of skidding.

    c) Here is where I get stuck. I want the total force between bike tire and the road. I did a force diagram of the bike, with a mg force pointing down, a Normal force pointing up, and a frictional force to the left.
    So I tried adding weight of the bike/girl to the frictional force, but that doesn't work. I'm pretty sure I need to add the normal force of the girl into there, but to what other force, if any, I'm stuck on.
  2. jcsd
  3. Oct 21, 2003 #2

    Chi Meson

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    Science Advisor
    Homework Helper

    You are correct so far.

    Consider the forces from the ground on the bike:

    THe normal force, pointing up, and the static friction, pointing to the left. THat's it. Find the resultant of these two perpendicular components.
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