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Accident Reconstruction using Friction, Energy, Momentum, and Work

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  1. Feb 15, 2013 #1
    (I believe I got this right, but it's necessary context to the next problem.)
    PROBLEM 1
    1. The problem statement, all variables and given/known data
    "The police department determined that the force required to drag a 130 N (29 lb) car tire across the pavement at a constant velocity is 100 N (23 lb). Specifications from the truck’s manufacturer claim that (for technical reasons) the effective coefficient of friction for truck tires is only 70% that of car tires. What was the coefficient of kinetic friction between the tires and the road for both the car and truck?"

    since F= ma and there's a constant velocity || ƩFx = 0 N
    Fa = 100 N

    Fg = 130 N

    I'm also given that the angles traveled after the accident are 7° for the truck and 33° for the car.


    2. Relevant equations
    μ = Ff/Fn

    3. The attempt at a solution

    ƩFy = Fg - Fn = 0
    Fn = 130 N

    ƩFx = Fa - Ff = Fa - μcFn
    0 = 100 N - μ(130 N)
    (of car) μc = 10/13 = 0.769

    7μc/10 = μt
    (of truck) μt = 70/130 = 0.538

    PROBLEM 2
    1. The problem statement, all variables and given/known data
    After collision, the truck and car skidded at the angles shown in the diagram. The car skidded a distance of 8.2 m (27 ft) before stopping while the truck skidded 11 m (37 ft) before stopping. The weight of the car is 13,600 N (3050 lb) and the weight of the truck is 69,700 N (15,695 lb). What was the speed of each vehicle just after the collision?

    μc = 0.769
    μt = 0.538
    ΔDc = 8.2 m
    ΔDt = 11 m
    Fgc = 13,600 N
    Fgt = 69,700 N

    2. Relevant equations
    μ = Ff/Fn
    Wf = FfΔD
    Wf = μFgΔD = ΔK = 1/2mv2
    μFgΔD = 1/2mv2
    μgΔD = 1/2v2
    v = sqrt(2μgΔD)

    3. The attempt at a solution

    (for car) v = sqrt(2μcgΔDc)
    vc = sqrt(2(0.769)(9.81)(8.2)) = 11.1 m/s

    (for truck) v = sqrt(2μtgΔDt)
    vt = sqrt(2(0.538)(9.81)(11)) = 10.8 m/s

    I don't think these are the correct answers, but I'm not sure where the flaw in my solving lies. Help, please?
    Thanks in advance. :)

    Link to actual page:
    physics.info/momentum-two-three/accident-reconstruction-1.pdf
     
    Last edited: Feb 15, 2013
  2. jcsd
  3. Feb 15, 2013 #2

    mfb

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    2016 Award

    Staff: Mentor

    Why? I didn't check all numbers (a computer can do this), but the formulas and the order of magnitude look correct.
     
  4. Feb 15, 2013 #3
    Perhaps I've just gotten too used to second guessing myself on my calculations. Sorry! Thanks for letting me know though. The fact that he gave me weight and I didn't use it threw me off.
     
    Last edited: Feb 15, 2013
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