You also need to give us the distance between the pushrod and the center of the servo, like...how far out on the servo arm the pushrod is.
From all the numbers, we could figure it out...but it'd be twice as much work to get an accurate result than most of us would be willing to spend. Also, as Howard said, the circular effect of the servo is even more work. You could go two routes, exact numbers or fairly close numbers. Exact is WAY harder, and you need calculus to get a good number. Anyone with a semester of engineering physics could do it. I'm sure I could if I stared at a pen and paper for long enough
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Someone please correct me, it's been a while since my last physics class and I'm doing all of this off of the top of my head. A simple formula is T=F*d. The torque is equal to the product of the force and the distance from the center. The servo has 160oz/in of torque...which means 10lb/in. A little trigonometry with your numbers gives me a guess of about 1.375". I got that by taking the total travel of the front of the bed(5.5"), using triangles to figure the vertical travel at the same angle 5" from the front of the bed. That's 2.75", divided by 2=1.375" from the center of the servo. That would mean it can lift 7.273lbs ((10lb/in)/(1.375in)) linearly (the inches cancel out, leaving only pounds...which is why I showed it like that). So for all intents and purposes, using simplified numbers and math, we have an actuator with 7.273lbs of force and a 2.75" travel length. Again, as howard said, to be accurate we need to figure out the angular forces and stuff which will make it a little weaker than that. A good guesstimate for the true force is just above 7lbs, so you need to go lower for a minimum set on the range. For that, I'll do it rounded down to 7lbs, and again at 7.273lbs. Now, this is the part I really need checked:
T(7&7.273lbs)=F*distance(5")
T/D=F
(7&7.273)/5=F
F=1.4lbs & F=1.4546lbs
I think that the servo should be able to lift 1.45 lbs with your setup.