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Force exerted by a seatbelt, on driver, after collision

  1. Sep 7, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose that the car is stopped from 60mph within right after the collision and the driver (75kg) didn’t push the brake.

    Seatbelt is non-stretchable and produces a constant force to stop the driver in 10cm. Calculate the acceleration of the driver and the force that the seatbelt acts on the driver for above two conditions.


    2. Relevant equations

    Time for the driver to hit the seatbelt: t=d/v
    Force that the seatbelt exerts on the driver: F=m(vo-vt)/t
    Acceleration of the driver: a=F/m


    3. The attempt at a solution

    60mph=27m/s

    Time for the driver to hit the seatbelt:
    t=d/v = 0.1m/(27m/s) = 3.7ms

    Force that the seatbelt exerts on the driver:
    F=m(vo-vt)/t = 75kg(27m/s - 0m/s)/3.7ms = 547,297N

    Acceleration of the driver:
    a=F/m = 547,297N/75kg = 7,297.3 m/s^2

    Is this right, the force seems so high?
     
  2. jcsd
  3. Sep 7, 2012 #2

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    You appear to be out by a factor of 2...

    I would use the equation of motion..

    V2= U2+2as

    where
    V2 = 0
    U = 60mph = 26.8m/s
    s = 0.1m

    Solve for "a"

    a = -U2/2s
    a = - 26.82/(2*0.1)
    = -3591m/s2

    Thats about 360 "g"

    Then..

    F=m * a
    = 75 * 3591
    = 269,325 N

    The high force is why cars have crumple zones at the front so you don't stop in 10cm.
     
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