Dynamics 17.4a Fixed Axis

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  • čas přidán 7. 08. 2024

Komentáře • 2

  • @Mr02071961
    @Mr02071961 Před 4 měsíci

    14:28 - you can treat the problem as a static one. If you calculate sum of moments around point A, than: 0.4C - 0.4mg + 0.3Ff = 0
    Pushing of the entire system (including AB member) by friction force Ff, happens at the wheel bearing (point B). Hence, friction force Ff acts directly on AB member at the point B. Therefore the distance for the moment around point A, caused by force Ff, should be 0.3 instead of 0.5
    Such calculation yields correct result.

  • @otnielechagarruga3955
    @otnielechagarruga3955 Před 4 lety

    For the last example, the question mentions the coefficient of kinetic friction but you did not use it. I assume you simplified the problem to "neglect friction" so its more straight forward to explain. Thank you!