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Design a automatic locking mechanism for two concentric cylinders

  1. Apr 24, 2016 #1
    How much force required to pull the actuator pin from the solid cylinder
    Weight of the solid cylinder is 15kg
    And angle of inclination 30 ,
    Coefficient of friction 1.5 WP_20160424_001.jpg
     
  2. jcsd
  3. Apr 24, 2016 #2

    billy_joule

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    Is this homework?

    Do you know Newtons second law? Can you draw a free body diagram?
    Is friction acting on the pin and/or the cylinder bore?
     
  4. Apr 24, 2016 #3
    Yes Friction acting on solid cylinder due to slides on a hallow cylinder ...

    Actually I m already calculate the force of solid cylinder acting on a actuator pin
    F=wsin30-sliding friction force
     

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  5. Apr 24, 2016 #4

    billy_joule

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    If there's no friction on the pin then it doesn't require any force at all to retract it. In fact it'll require an upward force to prevent it retracting under it's own weight.
     
  6. Apr 24, 2016 #5
    Thank you ...
    But I need how much force required to pull the pin from the solid cylinder ....
    Here shear force acting on the actuator pin
     
  7. Apr 24, 2016 #6

    billy_joule

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    Shear force doesn't matter. Was there something in post #4 you didn't understand?
    Is there friction acting on the pin or not?
     
  8. Apr 24, 2016 #7
    No there is no friction acting on an actuator pin the friction acting only on solid cylinder
     
  9. Apr 24, 2016 #8

    billy_joule

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    OK, then post #4 stands.

    Obviously real mechanical systems do have friction so the pin will probably require some pull force to disengage.
     
  10. Apr 24, 2016 #9
    Ya right but that was I asking how much force required to disengage the pull
    I
     
  11. Apr 25, 2016 #10

    rbelli1

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    So less than zero as billy joule said in #4 and #6.

    BoB
     
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