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Determining length of highway deceleration ramp, with friction

  1. Dec 5, 2007 #1
    1. The problem statement, all variables and given/known data
    "At the base of a hill (blah blah) there is a ramp inclined with an angle of 30 degrees and has a surface with a coefficient of friction of 1.5. How long should this ramp be to be able to stop a 80,000lb truck traveling at 80 mph"

    2. Relevant equations
    (delta)K + (delta)U = -fk * d (K is kinetic energy, U is potential energy, fk is friction force)

    fk = u*n (u is frictional coefficient)

    3. The attempt at a solution

    set Ki. = Uf + fk*d (having fk be positive, since you'd be subracting a negative anyway, right?) Ki = initial K.E., Uf = final P.E.)

    (1/2) m v^2 = m g h + u n d (u is frict. coefficient, n is normal force, d is distance or hypotenuse of triangle, h is height.)

    for the normal force I got n = m g cos(30)
    and h = d sin(30)
    {both of the above just using trig}

    so (1/2) m v^2 = m g d sin(30) + u m g cos(30) d
    m's cancel out
    (1/2) v^2 = g d sin(30) + u g d cos(30)
    solve for d
    d = (v^2) / [g(sin(30) + u cos(30))]

    plugging in v = 35.8 m/s (80 mph)
    g = 9.8 m/s^2
    u = 1.5

    gives 72.7 meters. I don't have much faith in this answer. Can I do Ki. = Uf + fk*d for this problem?
  2. jcsd
  3. Dec 5, 2007 #2
    Since it's giving you the coefficient of friction, I think that you'd want to use this equation to solve the problem:


    Solve for [tex]x-x_0[/tex]

    Figure out [tex]a[/tex]

    [tex]a=-g(\sin{\theta}+\mu \cos{\theta})[/tex]


    [tex]x=\frac{v^2-v_0^2}{2a}=\frac{v_0^2}{2 g(\sin{\theta}+\mu \cos{\theta})}[/tex]
    Last edited: Dec 5, 2007
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