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Homework Help: Truck Ramps

  1. Apr 1, 2009 #1
    1. The problem statement, all variables and given/known data

    In Fig. 8-36, a runaway truck with failed brakes is moving downgrade at 130 km/h just before the driver steers the truck up a frictionless emergency escape ramp with an inclination of θ = 15°. The truck's mass is 1.2 x 104 kg. What minimum length L must the ramp have if the truck is to stop (momentarily) along it?


    2. Relevant equations

    Mechanical Energy
    E,mec = K + U
    K = 1/2mV^2
    U = mgy

    3. The attempt at a solution

    Going through the problem I got K2 + U2 = K1 + U1 [2 being at the top of the inclined ramp and 1 being the bottom of the inclined ramp]. With substitution, I get:

    1/2mv2^2 + mgy2 = 1/2mv1^2 + mgy1

    Since velocity on the top of the ramp will equal zero and the length at the bottom of the ramp equals zero. the new formula will be:

    gy2 = 1/2v1^2

    subbing y2 with Lcos(theta):

    Lcos(theta) = V1^2 / 2g

    L = V1^2 / (2gcos(theta) )

    I get an incorrect answer (247.97 m) The correct answer should be 260 m. Must be an misinterpretation somewhere. Any help would be nice :]
  2. jcsd
  3. Apr 1, 2009 #2
    You don't know how to solve a right triangle by trig. Try again on that part. Most of your set up is fine.
  4. Apr 1, 2009 #3
    eff me. sin(theta) no cos(theta) thanks.
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