Design a automatic locking mechanism for two concentric cylinders

  • #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
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?
 
  • #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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  • #4
Yes Friction acting on solid cylinder due to slides on a hallow cylinder ...
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.
 
  • #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
 
  • #6
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
Shear force doesn't matter. Was there something in post #4 you didn't understand?
Is there friction acting on the pin or not?
 
  • #7
Shear force doesn't matter. Was there something in post #4 you didn't understand?
Is there friction acting on the pin or not?
No there is no friction acting on an actuator pin the friction acting only on solid cylinder
 
  • #8
No there is no friction acting on an actuator pin the friction acting only on solid cylinder
OK, then post #4 stands.

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

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