FE Exam Review - Dynamics - Kinetics of Particles

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  • Опубликовано: 11 янв 2025

Комментарии • 7

  • @lancea673
    @lancea673 2 года назад

    M*(v^2/r) is equal to 12.25. If you want to calculate W*cos x, then it will equal to 0.85. Final answer should be closer to A, 12.25 - 0.85 = 11.4

    • @directhubfeexam
      @directhubfeexam  2 года назад +2

      Just checked this. You would need to add the Wcos(x) since it's on the left side of the equal sign. Therefore:
      M*(v^2/r) + W*cos x
      12.25 + 0.85 = 13.1

    • @lancea673
      @lancea673 2 года назад +1

      @@directhubfeexam Thanks my mistake. I see it now: F = Normal - Weight*cos x
      Normal = F + Weight*cos x

    • @directhubfeexam
      @directhubfeexam  2 года назад

      @@lancea673 I’m glad you caught that 👍

    • @mikefaraday1720
      @mikefaraday1720 2 года назад

      @@directhubfeexam Thanks for this. I tried this before looking at your solution. my initial solution did not include the effect of weight at all. I then watched your solution and my initial thinking was the same as Lance was thinking. Thanks for this clarifying. 💯

  • @SophiaTheKittycat
    @SophiaTheKittycat 3 года назад

    Where can I find the formula of a=v2/r ?

    • @directhubfeexam
      @directhubfeexam  3 года назад +2

      It will be under the "Particle Curvilinear Motion" topic on page. 115 in FE handbook 10.1.
      Note that ρ will be the radius of curvature (r) in this case.