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Force Normal Question

  1. Oct 16, 2013 #1
    1. The problem statement, all variables and given/known data

    A photo of this problem is available at: http://imgur.com/l5lnI56
    (the visual is needed to understand the question)

    Problem text: A ski jump training hill can be thought of as featuring a near vertical plunge followed by a gradually decreasing (in magnitude) slope until it completely levels as shown in Fig. 1. A circle has been drawn to indicate the radius of curvature of the curve part way down the hill. Take Δh = -100m and R = 40m. Neglecting air resistance and friction, please answer the following question:

    Write down Newton's equations and find the normal force at each position. After falling through a height Δh the speed of the skier (squared) is v2=-2gΔh


    2. Relevant equations

    Fnet = ma
    ω = v/r
    α = at/r
    ar = v2/r = ω2r
    Fc = mv2/r


    3. The attempt at a solution

    ƩFx, i = 0
    ƩFy, i = -Fg
    --> since any contact between the object and the surface is negligible at i, there is no normal force.

    ƩFx, ii = Fg(sinθ)
    ƩFy, ii = N + Fc - Fgcosθ = maii

    ƩFx, iii = 0
    ƩFy, iii = N - Fg = 0
    N = Fg

    So at this point, there is a bit of a problem. When simplifying forces at points ii and iii, it becomes clear that mass is needed.
    e.g.
    N = Fg
    N = mg -> what is m?

    I can only assume that the answer comes from use of angular motion equations.

    At this point, I have the following information:
    radius (40m)
    height (-100m) --> used to find velocity using given formula --> velocity
    angle is 45 degrees.

    I calculated v to be 44.27m/s (using given velocity formula and height)
    I calculated ω to be 1.10677rad/s (using the given velocity formula and ω=v/r)
    I calculated ar to be 49m/s2 (using v2/r)

    I know that (defining x as the tangent to the circle, and y as toward the middle of the circle) in the y direction, the forces acting are (N + Fc) - 9.8mcos45. In the x direction, only 9.8msin45 is acting (force due to gravity down the incline at the point of contact).

    Other pieces of important that seem important to the solution can't be calculated without m.
    at = 9.8sin45
    α = 9.8msin45/40 = 0.245msin45

    At this point, I'm not sure what other steps to take to find mass, or if mass can be cancelled out in some sort of a proportional reasoning setup. I would appreciate a suggestion to help me figure out the rest. I'm not looking for a step-by-step solution, or even an answer.

    If any part of my post is unclear/inappropriately stated, let me know.
     
  2. jcsd
  3. Oct 17, 2013 #2

    CWatters

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    Homework Helper

    Perhaps I'm wrong but I can't see a way to calculate a value for the normal force without knowing the mass either. I haven't checked your working.
     
  4. Oct 17, 2013 #3
    This is resolved. Thanks to CWatters.

    Professor forgot to mention that he wasn't expecting a numerical answer, but instead, he wanted us to express the normal force in terms of mass (no need to find the magnitude of the mass). That made this a 5-minute question instead of a many-hour question.
     
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