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Engineering Dynamics - Car Striking a Barrier

  1. Sep 12, 2014 #1
    1. The problem statement, all variables and given/known data
    A car, assumed to be rigid and having a mass of 800 kg, strikes a barrel-barrier installation without the driver applying the brakes. From experiments the magnitude of the force of resistance F_r, created by deforming the barrels successive, is shown as a function of the vehicle penetration, s. If the car strikes the barrier traveling at a velocity v_c = 60 km/h, determine approximately the distance s to which the car penetrates the barrier.

    http://imgur.com/LYc87dW - here is a picture of my question. This has the graph of how the barrels resistance force increases with s.

    2. Relevant equations

    v_c = 60 km/h = 16.67 m/s
    m = 800 kg

    T (Kinetic energy) = 1/2 * mv^2

    U_1->2 = ∫F ds

    ∑U_1->2 = T_2 - T_1

    3. The attempt at a solution

    Starting off I assumed that once the car had reached the distance s it will have zero kinetic energy. So my equation became

    ∑U_1->2 = - T_1,

    ∫F ds = - 1/2 *mv^2, would my bounds of integration be 0 -> s ?

    I don't really know what to do, am I using the right principles here?
     
  2. jcsd
  3. Sep 12, 2014 #2
    You got it somewhat figured out. Keep in mind the signs when simplifying this:
    $$E_{\text{Final}} - E_{\text{Initial}} = \int \vec{F}(\vec{s}) \cdot \text{d}\vec{s}$$
     
  4. Sep 12, 2014 #3
    OK so then I would have,

    ∫F ds 0 -> s = - 1/2 *mv2

    ∫F ds 0 -> s = 111,155.56 J

    So from the graph, 40 kN x 0.75m = 30 J

    Then I need 111,155.56 J - 30,000 J = 81,155.56 J,

    so 60 kN x s = 81,155.56 J

    s = 81,155.56 J / 60 kN

    s = 1.35 m, then 1.35m + 0.75m = 2.10m
     
  5. Sep 13, 2014 #4
    I did something similar, but I don't get how you end up with 111,155.56 J. -1/2 (800)(60)^2 = 1,440,000. I wanted to go about it the same way as you but as I added up the separate distances, they went past the graph and still didn't get close to the amount of energy needed. Can anyone please clarify this?

    ... 5 seconds later I realize we use m/s not km/h in SI, after spending two hours trying to figure it out.... sorry thanks anyways
     
    Last edited: Sep 13, 2014
  6. Sep 14, 2014 #5

    I think I'm taking the same dynamics course as you, I finished my CAP assignment yesterday so I might be able to help.

    Remember Fr is opposite to the displacement therefore negative such that:

    T1 + ∑U1→2 = T2
    1/2mv^2 - Area = 0

    Also you are measuring area so your units will be according to the graph "kN m" not "J" such that:

    Area = 111155.56 N ft = 111.16 kN ft


    I also got "s = 2.10 m" as my final answer, so I'm assuming it's right.
     
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