Torque to rotate cylinder on two wheels with eccentric center of gravity 2

[kyleShropshire]

I need to calculate the torque on the 20inch diameter wheels to rotate the 30,000lb 120inch diameter cylinder past the 90deg position where the center of gravity is 12.5inches from the rotational center.
Wheels are linked mechanically and driven by single motor.
Coefficient of friction between large cylinder and wheels: 0.4

All angles and dimensions shown in image:

What is the resultant force on the two lower wheels?
I calculated right wheel: 24,826 lb and left wheel: 8,275 lb

On an old design sheet for this system a note says "due to 50deg included contact angle the effective load force is 35,000lb. 35,000 lbs * 0.4 = 14,000 lbs traction. Required force to rotate is 12,500 lbs. This is close but should rotate."

The people who did this are no longer available to ask questions.
How did they come up with those values?

 

[JStephen]

Force down is 30,000 lbs.
Assuming the reaction supporting the weight from those two rollers is radial at the point of contact, the sum of the two contact forces would be 30000/cos(25.377) = 33,200 lbs. If the rollers are linked together, you don't need to know how much on each one.
Moment required to rotate is 30,000 lbs x 12.5" = 375,000 in-lbs.
Tangential force required to rotate is 375,000 in-lbs/60" = 6,250 lbs.
Tangential force available as limited by friction is 0.4 x 33,200 lbs = 13,280 lbs.
Where they got their numbers- I don't know. The 35,000 lbs may just be rounding off, since the 0.4 is pretty approximate anyway.