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Friction in a banked turn

  1. Sep 30, 2008 #1
    A car moving at 41 km/h negotiates a 140 m-radius banked turn designed for 60 km/h. What coefficient of friction is needed to keep the car on the road?


    Ok can some one tell me what i did wrong?

    Fc= (m*V2)/r = m µ g

    Vmax= square root of ((µ) g r)

    60 = square root (µ*9.8*140)

    for mu i got 2.62391

    i used the 60 km/h because that was the maximum velocity of the curve

    how ever when i checked it its wrong ... my book does not say anything about the friction involved with a banked curve. help:cry:
     
  2. jcsd
  3. Sep 30, 2008 #2

    LowlyPion

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    Isn't there an angle of bank for the curve?

    Edit: I see it's designed for 60 km/h.

    As for your speed you will need to convert them to m/s from km/h.
     
  4. Sep 30, 2008 #3

    LowlyPion

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    This equation is not right because you haven't accounted for the angle of the bank.
     
  5. Sep 30, 2008 #4
    i changed the speed from km/h to m/s 60 km/h = 16.67 m/s

    there is no given curve however:

    if Fc cos (theta) = Weight sin (theta)

    (V2/r) cos (theta) = g sin (theta)

    then (V2/ g r ) = sin theta / cos theta = tan theta

    theta = tan inverse (V2/ g r)

    theta = 11.446

    but how does that help me find mu?

    btw after adjusting the km/h to meters/ s for mu after plugging it in the equation i got .202 but it was also wrong
     
  6. Sep 30, 2008 #5

    LowlyPion

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    Consider first the 60km/h case.
    Draw a force diagram. There is the centripetal force. There is the gravitational force down the incline. Since they say it is designed for 60km/h I think they must mean that there is no slipping for any μ.

    Mass drops out. Leaving θ, because you know g and V and R

    Now develop your equation for the slower speed. There your MV2/R plus your μ*g component must be sufficient to balance the downward force of the m*g*Sinθ component.
     
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