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Circular motion- banked curve, show that friction is necessary

  1. Jun 2, 2012 #1
    1. The problem statement, all variables and given/known data

    A 1200 kg car rounds a banked curve of radius 70 m. If the banked angle is 12° and the car is travelling at 90 km/h, show that friction is necessary in order for the car to safely make the turn.

    2. Relevant equations

    Force of gravity: Fg=m*g
    to calculate the normal force: opposite side of angle: Fncos(12°)=mg
    centripetal force: Fc=mv2/R
    Fc=Ff
    μ=Ff/Fn

    3. The attempt at a solution
    So these are the things we know:
    m=1200kg
    R=70m
    θ=12°
    v=90km/h=25m/s
    car is travelling on a banked curve

    With this information we can start off by calculating the Fg=mg=(1200kg)*(9.8m/s2)=11760N

    The normal force can be calculated from:
    Fncos(12°)=mg
    Fn=11760N/cos(12°)= 12022.725N

    From here, I calculated the Fc:
    Fc=(1200kg)*(25m/s)/(70m)= 10714.286 N

    I think we are to assume that Fc=Ff, because its a banked curve? Im not entirely sure on this reasoning... some clarification would be appreciated :)

    so from here I calculated μ:
    μ=Ff/Fn=10714.286N/12022.725N= 0.89

    END OF MY FIRST ATTEMPT
    _________________________________________
    BEGINNING OF SECOND ATTEMPT (for this attempt I used another problem as a guide that I found online... not entirely sure of the explanation given at the end, I have a hard time visualizing it)

    I calculated the Force trying to pull the car down the bank:
    from the adjacent side: F=mgsin(12°)= (1200kg)*(9.8m/s2)sin(12)= 2445.041N

    The needed centripetal force:
    Fc=mv2/R= (1200kg)*(25m/s)2/(70m)= 10714.286N

    The component of Fc that is parallel to the road surface is:
    opposing side: Fc*cos(12°)=10480.153 N

    The difference between component of centripetal force parallel to the roadway and the force due to gravity component parallel to the roadway is:
    10480.153 N- 2445.041 N= 8035.111N

    This force must be made up by friction, toward the center of the circle but parallel to the roadway surface.
    _____________________________

    SO, if anyone could lend me a hand and let me know where I went wrong and which attempt better suites this type of problem that would be greatly appreciated! Thank you so much in advance!
     
  2. jcsd
  3. Jun 2, 2012 #2
    The question asks you to prove that a frictional force is necessary, not to find the friction coefficient :wink: To do this, all you need to do is show that the force that pulls the car downwards the banked road(gravitational force) is less than the force that is 'pushing' it away(due to circular motion). Hence, your method of second attempt would be much more appropriate.

    -------------------
    First attempt error(s)!

    You seem to be missing a v2 term here.

    Nope. Draw out a diagram, see which forces balance the frictional force. You're missing one essential component i.e gravitational force.
     
  4. Jun 2, 2012 #3
    There is no force 'pushing the car away' !!!!!! Just show that the force towards the centre due to the banking is too small for the circular motion. The rest of the force must be provided by friction.
     
  5. Jun 2, 2012 #4
    Yes, there isn't. Thats why my the ''. But considering pseudo forces in your reference frame....
     
  6. Jun 2, 2012 #5
    The needed centripetal force:
    Fc=mv2/R= (1200kg)*(25m/s)2/(70m)= 10714.286N

    The normal force can be calculated from:
    Fncos(12°)=mg
    Fn=11760N/cos(12°)= 12022.725N
    ------------------
    Fc supply by Fn=FnSin(12°)=2499.665

    Should be added by friction =FfCos(12°)=10714.286-2499.665

    You have Fn calculated without resolving mg to its components, Thus it is orthogonal to Fc.

    I guess it is static friction.
     
    Last edited: Jun 3, 2012
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