Thank you guys. Your comments directed me to what to research. Feel free to correct me if I am wrong, no offense taken on construtive critisim.
So, you end up with 7500lb of force by using the formula for the force acting on the opposite side of the rod F2 = (π* d^2 / 4)* P2 ==> (3.14*2^2/4)*2500 ==> 7853lb
Meaning a 2.5"dia cylinder will produce 12271lb!!
And then you used the 7850lbs * sin(7.5ー) = 1024lbs. Got it so far. I didn't get the argument that the cylinder is half way back, but there is so much waste anyway
The actual angle in the 2nd design is 10.79ー, resulting only 1470lbs at the closing end with 2500lb at the opening place (18.67ー). still not good.
I was struglling figuring out why the length of the rod is important as its not included in the calculations but I got it.
So it looks like I have to find a balance. Larger bore, means more force, but slower acting cylinder, while larger rod can achieve more applied force but also takes more time to full extend/retract.
What is an acceptable clamping force? (Don't tell me it depends what I want to use the grapple for!!! I know that!

Considering that a compact tractor cannot lift more than 1800lbs on the front pivot, does the target of 2000lbs sounds good enough?
These formulas are really interesting, I will stay up tonight to finish that!
These are the CAD files that I am making on Freecad 0.17 if you want to see them. You only need the "Master CAD" file if you do not want to edit. This is just temporary, the final will go to the
GrabCAD
Quick question: These are the hydraulic specs of my tractor. Which number do I use for the calculations? The total flow (13.7) of the implement pump (8.2)??
Implement pump (gpm): 8.2
Steering pump (gpm): 5.5
Total flow (gpm): 13.7
Category, 3 pt hitch: CAT 1
Hitch Lift capacity (lb): 1,808
Lift control type: Manual