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Coefficient of static friction

  1. Feb 25, 2010 #1
    1. The problem statement, all variables and given/known data

    If a curve with a radius of 77.0 m is perfectly banked for a car traveling 65.8 km/hr, what must be the coefficient of static friction for a car not to skid when traveling at 91.2 km/hr?

    2. Relevant equations

    [tex]\mu[/tex]s = v^2/(g*r)

    3. The attempt at a solution

    1 Incorrect. (Try 1) 0.609
    2 Incorrect. (Try 2) 0.61
    3 Incorrect. (Try 3) 0.616
    4 Incorrect. (Try 4) 0.443
    5 Incorrect. (Try 5) 0.173
    6 Incorrect. (Try 6) 0.0335
    7 Incorrect. (Try 7) 0.0336
    8 Incorrect. (Try 8) 0.287
    9 Incorrect. (Try 9) 0.847

    I still have my last try. Please help!

    Thanks!
     
  2. jcsd
  3. Feb 26, 2010 #2

    rl.bhat

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    Using the formula for angle of banking, tan(theta) = V^2/g*r, find theta. At the perfect banking, the car can round the curve without any help of the frictional force. When you increase speed addition force towards the center is needed to keep the car on the track. this force is obtained by the friction.
    The additional force is (m*v2^2/r - m*v1^2/r). And the frictional force is mu*m*g*sin(theta). Now solver for mu.
     
  4. Feb 26, 2010 #3
    From your explaination, I got as follow:

    tan [tex]\theta[/tex] = v^2 / (g * r) = 18.28^2 / (9.81 * 77) => [tex]\theta[/tex] = 23.86 degree.

    I dont understand "The additional force is (m*v2^2/r - m*v1^2/r). And the frictional force is mu*m*g*sin(theta). Now solver for mu." ??? What is the equation here? Thanks!
     
  5. Feb 26, 2010 #4

    tiny-tim

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    Hi huybinhs! :smile:

    (have a theta: θ and a degree: º and try using the X2 tag just above the Reply box :wink:)
    It's good ol' Newton's second law … Ftotal = ma, in the horizontal direction, and Ftotal = 0, in the vertical direction.

    You know from the given measurements that a = mv2/r, horizontally,

    and you know that the only two forces are the normal force, N, and the down-slope friction force of (maximum) µN.

    Now take horizontal and vertical components of all this …

    what do you get? :smile:
     
  6. Feb 26, 2010 #5
    N = F = ma; but a = mv^2/r => N=F = (M2 * v2) /r.

    And Fs = µs * N.

    Is this correct?

    I dont understand "Now take horizontal and vertical components of all this" ???
     
  7. Feb 26, 2010 #6

    tiny-tim

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    This is hopless. :redface:

    We'd better do it in stages …

    what are the three forces on the car, and what are their directions? ​
     
  8. Feb 26, 2010 #7
    1. mg-down vertical direction

    2. F(N)*cos(theta)- up vertical direction.

    3. F(N)*sin(theta)- horizontal direction.

    Correct? What then? ;)
     
  9. Feb 26, 2010 #8
    With vertical direction, we have:

    F(N)*cos(theta) - mg = 0 <=> F(N) = mg/cos(theta) (i)

    With horizontal direction, we have:

    F(N)*sin(theta) = m (v^2/r) (ii)

    From i and ii, we have:

    tan(theta) = v^2 (r*g).
     
  10. Feb 26, 2010 #9
    then I calculate:

    tan(theta) = 18.28^2 / (9.81 * 77) => = 23.86 degree.

    What's next, man?
     
  11. Feb 26, 2010 #10

    tiny-tim

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    (I've been away for an hour)

    No, not correct …

    what about the friction force??

    Start again.​
     
  12. Feb 26, 2010 #11
    The friction force is:

    F(fr) = [tex]\mu[/tex]* N.

    What's then?
     
  13. Feb 26, 2010 #12

    tiny-tim

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    ok, so what are the three forces on the car, and what are their directions?
     
  14. Feb 26, 2010 #13
    1. F(N) = mg - vertical up

    2. F(G) = mg - vertical down

    3. F(fr) = [tex]\mu[/tex]s * F(N).

    Correct? What's next?
     
  15. Feb 26, 2010 #14

    tiny-tim

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    No, not correct …

    are we doing the same question? :confused:

    the track is banked, at an angle θ.

    Try again.​
     
  16. Feb 26, 2010 #15
    1. F(N) = F(N)* cos (theta)- vertical up

    2. F(G) = mg - vertical down

    3. F(fr) = [tex]\mu[/tex]s * F(N) = [tex]\mu[/tex]s * F(N)* cos (theta) - horizontal right.

    ????
     
  17. Feb 26, 2010 #16

    tiny-tim

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    this isn't making any sense

    i'm going to bed now :zzz: … i'll look at it again in the morning
     
  18. Feb 26, 2010 #17
    Oh.....come on... I still dont get anything from you yet :(
     
  19. Feb 26, 2010 #18

    rl.bhat

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    When the angle of banking is [tex]\theta[/tex], the car will go round the curve with the given velocity without the help of frictional force. The necessary centripetal force is provided by the component of N. When the velocity of the car increases, centripetal force provided by the component N is not sufficient to keep the car on the road. The additional force is needed to keep the on the road. And this force can be provided by the frictional force between the tire and the road. The frictional force is mu*N. On the flat road, normal force in mg. On the banked road what is the normal force? The additional force is (m*v2^2/r - m*v1^2/r).
     
    Last edited: Feb 26, 2010
  20. Feb 26, 2010 #19
    You mean:

    mu*m*g*sin(theta) + (m*v2^2/r - m*v1^2/r) = 0 ????

    And solve for mu???
     
  21. Feb 26, 2010 #20
    Ok, I calculated as follow:

    tan(theta) = v^2 / (g * r) = 18.28^2 / (9.81 * 77) => theta = 23.9 degree

    then:

    mu*m*g*cos(theta) = (m*v2^2/r - m*v1^2/r)

    => mu = 0.442

    Here is my forth try and it was incorrect:

    4 Incorrect. (Try 4) 0.443 ???

    Why?
     
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