1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Banking a Curve without Friction

  1. Oct 26, 2014 #1
    1. The problem statement, all variables and given/known data
    A highway curve of radius 80 m is banked at 45°. Suppose that an ice storm hits, and the curve
    is effectively frictionless. What is the speed with which to take the curve without tending to
    slide either up or down the surface of the road?

    [itex]r = 80m[/itex]
    [itex]\theta = 45^o[/itex]
    [itex]g = 9.8m/s^2[/itex]

    2. Relevant equations
    [itex]F_c = F_N sin \theta = mv^2/r[/itex]
    [itex]F_N = mgcos \theta[/itex]

    3. The attempt at a solution
    [itex]F_N sin \theta = \frac {mv^2}{r} \Rightarrow (mg cos \theta)sin \theta = \frac {mv^2}{r} \Rightarrow v = \sqrt{rg cos \theta sin \theta}[/itex]
    [itex]\Rightarrow \sqrt{(80m)(9.8m/s^2)cos 45^o sin 45^o} = \sqrt{392m^2/s^2} = 19.8m/s [/itex]

    I know the answer's supposed to be 28m/s and I'm missing a tangent somewhere to make it [itex]v = \sqrt{rgtan \theta}[/itex] instead, but I can't figure out where.

    Thank-you
     
  2. jcsd
  3. Oct 26, 2014 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The last line is wrong. The normal force here it is not the same as the normal force when sliding on a slope.
    In that case, the normal force cancelled the normal component of gravity. Here the normal force has to provide the horizontal centripetal force, so the vertical force components have to cancel. Make a drawing! The resultant of the normal force and gravity must be horizontal.
     
  4. Oct 26, 2014 #3
    I thought the normal force always acted perpendicular to a surface?
     
  5. Oct 26, 2014 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Yes, it is normal to the surface. But it is not mgcos(theta) always.
     
  6. Oct 26, 2014 #5
    If the forces have to cancel out to keep the object stationary, then the normal force needs to be equal to the sum of the centripetal force and mg, right? Or is it that since the centripetal force acts inwards, then it must be counteracted by the sum of mg and normal force?
     
    Last edited: Oct 26, 2014
  7. Oct 26, 2014 #6

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The car is not in equilibrium, but travels along a horizontal circle. Circular motion requires that the forces applied on the car supply the centripetal force. Draw the free-body diagram for the car.
     
  8. Oct 26, 2014 #7
    Sorry, this problem just doesn't make sense to me. BTW, I've drawn the FBD many times and even have a reference picture, even before I made this thread. Thanks anyways.
     
  9. Oct 26, 2014 #8

    ehild

    User Avatar
    Homework Helper
    Gold Member

    can you show your picture?
     
  10. Oct 26, 2014 #9

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    It's best not to think of the object as being stationary. After all, it's going around in a circle. It's also better to think in terms of centripetal acceleration instead of a centripetal force, which is not a third force acting on the car. There are only two forces on the car, the normal force and its weight. They show up on the lefthand side of ##\sum \vec{F} = m\vec{a}## while the centripetal acceleration appears on the righthand side. Can you write down the equations for the horizontal and vertical directions now?
     
  11. Oct 26, 2014 #10

    NTW

    User Avatar

    I get the same result as logan3...

    The side forces, parallel to the road, are m * g * sin θ and m * v2/r * cos θ

    For equilibrium, they must be equal => m * g * sin θ = m * v2/r * cos θ

    Solving for v => v = 19,8 m/s
     
  12. Oct 26, 2014 #11
    I had to step away and think about it for awhile. Actually, I pretty much thought about it non-stop for the past two hours. It helped thinking about how the normal force changed in other problems, such as a box on a horizontal flat surface and a box sliding down an incline.

    The components of the normal force are the horizontal and vertical forces -- the horizontal is the centripetal force and the vertical is the force acting opposite the force of weight (mg). Since the only net force is the centripetal force, then the vertical normal force component must cancel out with the force of weight acting down.
    [itex]F_N cos \theta = mg \Rightarrow F_N = \frac{mg}{cos \theta}[/itex]
     
    Last edited: Oct 26, 2014
  13. Oct 26, 2014 #12

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You got it at last!!!!
    Here is the free-body diagram:

    banking.JPG

    From the yellow triangle you see that the vector sum of N and mg is equal to the centripetal force Fcp.
     
  14. Oct 26, 2014 #13

    NTW

    User Avatar

    The weight mg has two components, one normal to the surface, m * g * cosθ, and the other one parallel to the surface, m * g* sinθ...
     
  15. Oct 26, 2014 #14

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Yes, but you do not want the car sliding downward, but moving horizontally around a circle.
     
  16. Oct 26, 2014 #15

    NTW

    User Avatar

    Well, I'm contemplating an instantaneous situation, 'a frame of the movie', and the forces parallel to the road are, for the component of the weight, m * g * sin θ, and the component of the centrifugal, 'inertial force', parallel to the road, that may be fictitious, but is valid from my frame of reference is m * (v2/r) * cosθ.

    Equating the two, and solving for v, it results v = 19,8 m/s...
     
  17. Oct 26, 2014 #16

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You're making an algebra mistake solving for ##v##. If you move the ##\cos \theta## over, you get
    $$\left(\frac{mg}{\cos\theta}\right)\sin\theta = m\frac{v^2}{r}$$ which is exactly the same equation logan3 finally got.
     
  18. Oct 26, 2014 #17

    NTW

    User Avatar

    Yes, sorry, you are right. My mistake... Now everything is clear...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Banking a Curve without Friction
Loading...