An Application of Cross Products: Torque

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  • Опубликовано: 27 дек 2010
  • This video explains how to solve a problem involving torque using cross products as well as another method.
    mathispower4u.yolasite.com/

Комментарии • 7

  • @everythingispossible5431
    @everythingispossible5431 3 года назад

    i wanted this

  • @maskd3515
    @maskd3515 Год назад +1

    🎉🎉🎉🎉🎉🎉🎉🎉🎉🎉🎉🎉🎉😊

  • @luigianchondo7241
    @luigianchondo7241 10 месяцев назад +1

    Or I did way simpler than what he did
    180°-110°=70°
    So 40lbs*1ft*sin(70°) = 37.6 lb-ft

  • @alexandgarciacalle
    @alexandgarciacalle 2 года назад

    In min 5:21, isn't it supposed to be negative torque? Since the determinate signs go ---> T= (0)-(0)+(-37.6k)=-37.6k. Thank you

    • @powerllesss2672
      @powerllesss2672 2 месяца назад

      He said take the sum of the blue products subtract the green products which results in T= (0) - (-37.6k)= 37.6k

  • @theBasilDestiny
    @theBasilDestiny 8 лет назад +2

    megnitude