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Ffr=ma=m(v^2/r) is this right?

  1. Nov 27, 2005 #1
    radius=76.3
    angle=30
    v=14.3
    mass=2600
    frictional force=?

    ffr=ma=m(v^2/r) is this right?
     
  2. jcsd
  3. Nov 27, 2005 #2

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    Hi there,

    could you clarify the question that you've been asked to solve ? A few more details please.
     
  4. Nov 27, 2005 #3
    a curve is banked. the car is on the curve. the car weighs 2600kg. is at an angle of 35. u=0.12 ideal speed for this curve is 22.88. i used V=sqr root tan(angle)*R*9.8

    The next curve that the car approches also has a radius of curvature 76.3m. it is banked at an angle of 30. the ideal speed for this curve is Vc(banked so that the car experiences no frictional force) the speed of the car Vs as it rounds this curve is Vs=0.625Vc
    what is the magnitude of the frictional force needed to keep it from sliding sideways?
     
  5. Nov 27, 2005 #4

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    Thanks for posting the question details.
    In your first post, your answer
    is wrong, I'm afraid. You have the friction force equal to the centripetal force and that only happens when the road is unbanked.

    Anyway, we are told that Vs is less than Vc, so what does that tell you about the direction that Fr acts in ?
    You have three forces acting on the car, Its own weight vertically downwards, the friction force from the road and the normal force from the road surface.
    The vertical component of the road's normal force, Fn, supports the weight of the car. Once you know in which direction Fr acts, you can add it or subtract it from Fn to give the centripetal force providing the velocity of Vs.
     
  6. Nov 27, 2005 #5
    ?

    u add them. so does that mean that Fny+Ffr=ma?
     
  7. Nov 27, 2005 #6

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    You don't add them :frown:

    Vs is less than Vc so the car will tend to slide down the slope of the banked road. So the friction must act up the slope. If the car is on the left hand side and the centre of rotation is on the right, then the (resolved) friction force acts to the left and the (resolved) normal force, Fnsin@, acts to the right. The difference is the centriptal force, Fc, acting to the right.
     
  8. Nov 27, 2005 #7
    :cry: so now Fc=Fny-Ffr=m(V^2/r)
     
  9. Nov 27, 2005 #8

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    Hang a mo. I'll draw a diagram.
     
  10. Nov 27, 2005 #9

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    In Fig1, I've shown the three forces acting on the car. The weight of the car itself, mg, acting vertically downwards, the friction force, Fr, acting up the slope, and the normal force, or support, from the road surface, Fn, acting normal to the road.

    In Fig2, I have resolved the forces into horizontal and vertical components.
    The horizontal forces provide the centripetal force, so

    Fc = Fnsin@ - Frcos@ = mv²/R

    [​IMG]
     
  11. Nov 27, 2005 #10
    gracias

    thanks for your help
     
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