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

subbby

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 Thumbnails

     

haruspex

The supports are rollers, turning freely, yes? Then your calculation looks right.

subbby

Yes. Free Rolling !

Thanks for your help .