I’ll try to explain it in a simpler example.
I want to turn my tractor over on its side so I can easily polish the undercarriage. I bring in my good friend Hulk Hogan to turn it over. Using an example of 1000# centered on the rear axle, Hulk can’t lift enough at the existing 2’ axle stub (500#) to turn it over. Hulk adds a 2’ cheater to the axle making it possible for him to exert 250# of upward force to rotate the tractor (250# x 4’=1000# of rotational torque). Shine away.
Next time I need to do this Hulk isn’t available and I can’t lift 500 or even 250#. So I add a 6’ cheater to the axle and I only have to lift 125#, which in my dreams I can do.
The bottom line is that the length of the axle has no effect on the load on the bearing at the end of the transmission case stub. It was 1000# with the factory axle, 1000# with Hulks 2’ cheater and 1000# with my 6’ cheater.
Now this is not a perfect example as because of the rigid transmission case, the “beam” would rotate on the opposite tire, not the center of the transmission as if it was a lever condition. Tractors do not have independent rear suspension.