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Force to hold a cylinder about its axis with Center of Gravity offset from the center

  1. Nov 4, 2012 #1
    This is a work related problem. I was calculating the amount of force required to hold a cylinder in a position when its Center of gravity is not at the center.

    Problem Background

    • Assume a 60' long cylinder.
    • Circumferential it is made of 8 sections.
    • For maintenance purpose, one section is completely removed (Side View shown in Picture3 & front view shown in Picture2). Hence the shift in center of gravity. (Shown in Picture2)
    • This entire arrangement is rotated with the help of gear and pinion arrangement. (Picture1.)



    To find :
    Find Pt in Picture1 (Pt is the counter force)

    Given Data:
    • Bigger Circle : Gear
    • Smaller Circle : pinion
    • Center of Gravity CG point is 14.125 inches to the left from the vertical center line
    • Radius of Gear = 144”
    • Radius of pinion = 10.75”
    • Reactions Rb and Rc are supports. Angle Rb-O-G = Angle Rc-O-G= 30 degrees
    • Ra=1,111,887.5#

    Attempt 1:

    • Finding moments about point O.
    • Therefore since direction of reactions Rb and Rc are in the line of action of force, they wouldn't have any moment about the point ‘O’ …. Correct me if I am wrong


    Implies, Pt= (1111887.5*14.125)/144= 109065.3 #

    My Doubts :
    • Is this approach correct ? and is this the value of the force required to hold the cylinder
    • Or do I have to consider the reactions Rb and Rc too ? If that is the case I shall end up having three unknowns, namely Rb,Rc and Pt.
     

    Attached Files:

  2. jcsd
  3. Nov 5, 2012 #2

    haruspex

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    Re: Force to hold a cylinder about its axis with Center of Gravity offset from the ce

    The supports are rollers, turning freely, yes? Then your calculation looks right.
     
  4. Nov 6, 2012 #3
    Re: Force to hold a cylinder about its axis with Center of Gravity offset from the ce

    Yes. Free Rolling !

    Thanks for your help .
     
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