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Radial Net Force/Circular Motion Valley Question

  1. Oct 31, 2011 #1
    1. The problem statement, all variables and given/known data

    A car is traveling through a valley at a constant speed, though not a constant velocity and (I believe) it is at the bottom. If the car's speed is 25 m/s, its mass is 1200 kg and the radius of the valley (r) is 25 meters, Use your force diagram to create a Fnet=ma equation for the Y-Direction. Then determine the magnitude of the force of the road acting on the car. (i.e. the Force Normal)

    2. Relevant equations

    m = mass; v = velocity; r = radius
    Fnet=(mv2)/r

    3. The attempt at a solution

    So I know the force body diagram. Fn (Normal Force) is greater than Fg (gravity).
    Then when I continue to work with the equation I can come up with
    Fg - Fn = (mv2)/r which then becomes
    Fn = Fg - (mv2)/r
    However after this step I don't know what to plug in or even how to solve for the numbers that need to be plugged in. Well gravity will be 11760 N but after that I am lost.
    Thanks!
     
  2. jcsd
  3. Oct 31, 2011 #2

    Doc Al

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    Staff: Mentor

    You have your signs a bit mixed up:
    ƩF = ma
    Fn - Fg = mv2/r
    The normal force and the acceleration are both upward.
    You are given the mass, speed, and radius. Just plug them in.
     
  4. Oct 31, 2011 #3
    Wow that was really simple...sorry sometimes even the simplest stuff in physics just goes over my head. Thank you, especially for my equation corrections!

    So Fn = 41,760 N
     
  5. Oct 31, 2011 #4

    Doc Al

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    Staff: Mentor

    You are most welcome.
     
  6. Oct 31, 2011 #5
    Ah wait. There is a second question that is really similar. Instead of being in a valley the car is on top of a hill.

    m = 1200 kg
    r = 25 meters
    speed = 43 km/hr = 11.94 m/s

    Does that original formula change? The first question was Fn - Fg = (mv2)/r
    However, when I used this equation my Fn was 18608.148. I thought that the Fn was less than the Fg because the car is at the top of a hill
     
  7. Oct 31, 2011 #6

    Doc Al

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    Think it through. What's the direction of the acceleration when going over a hilltop?
     
  8. Oct 31, 2011 #7
    Well the acceleration is straight down since the car is at the top of the hill. So that would then add to Fg and make it larger (not the actual force of gravity, it would just be Fg + a).
    I feel like I should simply make Fg negative but I don't understand why.
     
  9. Oct 31, 2011 #8
    Oh a = v2/r
    I think I get it now.
    So there is no upward force at all? Can you explain how that works? My teacher was talking about it in class but since it's new I don't understand it all yet
     
  10. Oct 31, 2011 #9

    Doc Al

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    The normal force Fn acts upward. (The road always pushes up, right?) And gravity always acts downward.

    The difference between the two situations--the valley versus the hill--is the direction of the acceleration (and thus the direction of the net force).

    For the valley:
    ƩF = ma
    Fn - Fg = mv2/r
    Fn = Fg + mv2/r

    The acceleration is up and thus positive in the equation. (I choose up as positive as my sign convention.)

    For the hill:
    ƩF = ma
    Fn - Fg = -mv2/r (the acceleration is now down, thus negative)
    Fn = Fg - mv2/r

    This should make some intuitive sense. Think of a roller coaster. When going over the top of a hill, you feel light--a smaller than usual normal force acts. But when going through a valley, you are pressed firmly against the seat: the normal force is greater than usual.

    Make sense?
     
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