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FORD F150 Centrifuge

  1. Feb 6, 2008 #1
    In the Ford F150 commerical that debuted during the Superbowl (), Ford says that it subejcted the tow hooks on the truck to 6g's of force.

    I timed the period of the truck during the portion in which the narrator is visible on the screen and got 1.80 seconds roughly.

    I read online that the truck was swung in a 50 foot circle (http://www.the-signal.com/?module=displaystory&story_id=53067&format=html)

    If I use a radius of 25 feet (7.62 meters) for the circle and a period of 1.80 seconds and use v=2(pi)(r)/T I find that the truck goes approximately 27 m/s (60 mph as the article claims).

    If I then use a = v^2/r to get the centripetal accleration, I find that the acceleration is 96 m/s^2.

    96 m/s^s is more like 10g's, not 6 g's. Am I doing something wrong?
    Last edited by a moderator: Sep 25, 2014
  2. jcsd
  3. Feb 6, 2008 #2
    Yes you are trying to relate real life physics to TV. Most likely it was not actually spun around as in the commercial. That is just a good editing team. Lets face it that would be very expensive to build, run, and dismantle. So no your math and physics are not wrong. The commercial is.
  4. Feb 6, 2008 #3
    Doh! You saw it on TV, it must be true!
  5. Feb 10, 2008 #4
    F-150 centrifuge

    check it out
    It looks like the vehicle was moving faster in the commercial than it appeared in the video of the commercial being taped.
    It was an actual demonstration, however. It beats last year's big commercial, which was an F-150 crashing through a giant birthday cake...
  6. Feb 10, 2008 #5
    Maybe because the force was spread out over the 2 tow hooks?
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