I drew these up earlier, then deleted, but after yet more "the math doesn't matter" kind of posts I figure I'll post it anyways.
In these illustrations, pretend the implement is a barbell, 100# each end. Really a rotary mower isn't much different from this example but making things as simple as possible is good.
First pic: Ends are on the ground, no tractor involved.
(Not pictured: ground pushing up 100# on the weights, this is called "the normal force"):
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Next pic:
Tractor lifts one end with no top link:
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Funky cylinder thing is the lift arm.
The 100# of the close end is on lift links. The other end's 100# is still on the ground.
Last pic: just like second pic, but top link lifts the other end:
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Other funky cylinder thing at the angle is the top link... part of it. The rest of it extends to the far end weight, whether via piston or chain or what.
There’s no more weight on the lift links - still 100.
Top link has all 100 of the other end.
Final pic: top link pulled strongly enough to raise the end to be level with the top link itself (I got tired of drawing 100's and arrows sorry)
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In this case, the bottom link is now carrying a chunk of the weight of the far end as well as its end.
The tension in the link is shown by the diagonal line going from top right to bottom left; the components are the horizontal compression and the vertical weight pulling down
Top link is carrying some; the higher the angle that the implement has with the bottom link, the more goes onto the bottom link (as soon as the end of the implement goes above the bottom arms, the bottom arms start getting some of the end load).
Consider if the top link lifts the barbell completely vertical (assuming it would fit, ok) - then the top link would end up with zero tension and the bottom arms would be carrying the weight of both ends of the barbell (200#).
That's the situation with a perfectly balanced ballast box.