Conceptual Dynamics Example Problem 4.3-5: Rigid-Body Kinematics (mechanisms)

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  • Опубліковано 4 жов 2024
  • This example problem is from the Undergraduate Mechanics text: Conceptual Dynamics. This problem analyzes the velocities of a 4-bar mechanism and is an example of the general planar motion of rigid bodies. The approach applied is relevant to the analysis of mechanisms in general.

КОМЕНТАРІ • 20

  • @cagdastemel3729
    @cagdastemel3729 3 роки тому +1

    Thank you for all your work.
    This was amazing explanation for people who were struggling with this kind of questions.

  • @lamrex18
    @lamrex18 9 років тому +15

    Very helpful video, Thanks!
    P.S. there is a calculation error in the last step while calculating Vp. 0.3xWbc = 1.191 not 11.91.

    • @hillrickc
      @hillrickc  9 років тому

      Iamrex18 Thank you for the keen eye!

  • @serderoglukf8222
    @serderoglukf8222 8 років тому +3

    woww!! it was amazing helpful video!
    thanks for it sir, i totally got it!

  • @gita_JU
    @gita_JU 4 роки тому +1

    Thank you very much. I understood very well

  • @isagumus1
    @isagumus1 7 років тому +1

    Excellent explanation thank you !

  • @tonyjaahu6928
    @tonyjaahu6928 9 років тому +3

    This video was helpful

  • @olivermcclelland3997
    @olivermcclelland3997 5 років тому +1

    Very helpful video! Completely understand the concept now! Do you have a video regarding the kinematics in rigid bodies including acceleration and relative acceleration? I like the way you explained this one, now need to understand relative acceleration! Thanks!

    • @hillrickc
      @hillrickc  5 років тому

      You can try this one: www.conceptualdynamics.com/files/krb/krb_gen_ep1.htm
      Or this one that my colleague did:
      www.conceptualdynamics.com/files/krb/krb_ic_ep2.htm

  • @aashishsys
    @aashishsys 2 роки тому +1

    Life saver

  • @nuraizaelisyabintihasim5960
    @nuraizaelisyabintihasim5960 4 роки тому

    The last value is 1.191 when 0.3x3.97

    • @hillrickc
      @hillrickc  4 роки тому

      Yes, it is. Thank you for the keen eye!

  • @ziddidunya883
    @ziddidunya883 2 роки тому

    Very helpful.
    Sir please can you tell me the name of book and writer name.

    • @hillrickc
      @hillrickc  2 роки тому

      The book is Conceptual Dynamics and I am a co-author with my colleague Dr. Kirstie Plantenberg. You can find additional information and resources at: www.conceptualdynamics.com/

    • @ziddidunya883
      @ziddidunya883 2 роки тому

      @@hillrickc ok Sir thanks . Sir is there is only one chapter (4) and how can i get solution pdf.

    • @hillrickc
      @hillrickc  2 роки тому

      @@ziddidunya883 The full book is available for purchase, but we only provide the solution manual to instructors that have adopted the book for their course.

    • @ziddidunya883
      @ziddidunya883 2 роки тому

      @@hillrickc Sir i have need this chapter solution. And i want help you.
      SAMPLE 14.3 A cheerleader throws her baton up in the air in the vertical xy-plane. At an instant when the baton axis is at theta = 60 degrees * fr the horizontal, the velocity of end A of the baton is overline v A =2m/si+ sqrt 3 m/sj . At the same instant, end B of the baton has velocity in the negative x-direction (but | vec v B | t = 1/2 * m is not known). If the length of the baton is and the center-of-mass is in the middle of the baton, find the velocity of the center-of-mass.
      You have any idea from which book this numerical related

  • @daborshy4089
    @daborshy4089 6 років тому +1

    Wow, that's a lot of work to find the velocity of a random point lol

  • @abudyalrehaili
    @abudyalrehaili 7 років тому

    how is Wbc is clockwise

    • @hillrickc
      @hillrickc  7 років тому +1

      When you start the solution of this problem, note that you DON'T need to know the direction of wbc. You can assume a direction and proceed with your solution. If your calculated value for wbc is negative, then you know the true direction of wbc is opposite what you assumed.
      In this case, we can understand the direction of rotation of link BC based on the motions of point B and point C. Since wcd rotates counterclockwise, point C moves strictly to the left at the instant shown (perpendicular to link CD). Further, link BC pushes on link AB causing it to also rotate counterclockwise. Therefore, at the instant shown, point B will move up and to the left (perpendicular to link AB). From the motion of point B and C, it is implied that link BC rotates clockwise (at this instant).