Ahh, so what you are saying is the upward pull of the downward thrust has increased so in order for the aircraft to remain in equilibrium the lift must be decreased. But what about the downward pull of the increased drag?
If I understand correctly, the drag would have decreased. As you climb, like you said lift would decrease and so would the drag associated with the lift. If you climbed at the same airspeed or at a slower airspeed the form/skin friction drag would remain the same or decrease respectively.
I just searched on google and came to this thread. I just wanted to lite a spark to this!
It makes things more clear if we say weight is equal to the F in F=MG. It won't change one bit while in a turn or while being accelerated at 2G. F=MG is your mass, which is constant, and the force of gravity exerted by the earth, which again is constant! The only time it will change is when you change your distance from the center of the body which is exerting the acceleration on you (earth). As you get higher in altitude your "weight" will change, gravity decreases. It is very slight however for civilian aircraft. We can't think of weight and apparent weight while analyzing the same set of forces because you are switching back and forth between frames of reference to try and explain the actual force you are feeling. So for all this I'm going to talk in the frame of reference of weight, which is an inertial frame of reference.
I think the definition of weight should be clear. Now, looking at the forces in the turn we should agree that the weight doesn't change. The aircraft has an acceleration force applied to it. We have to think in 3 dimensions here. Let's take this to a 0G environment so that we can isolate the force in the turn, space. If we fix a vertical pole in space, and space meaning deep space away from ANYTHING that could affect the results of this. Then take a string and attach it to small box with a small rocket engine on the back. As we fire the rocket it will speed up the box on a circular track around the pole, assuming it doesn't wind around the pole. At a constant speed around the pole, the box will have "1" force on it. Since we are in the inertial frame of reference, the box has a centripetal force acting on it toward the center. Centrifugal force does not exist in the inertial frame of reference. (If I were to sit in the box I would be in a non-inertial reference frame and I would feel the centrifugal force (a pseudo-force), which would translate to me feeling an "apparent weight" that is greater than my actual weight"). So that 1 force, since it is obviously unbalanced, will be 'accelerating' the box inwards. The box's weight HAS NOT changed!!
In a turn, the aircraft's weight DOES NOT change!! Once you understand that, you'll be able to clearly understand the forces in a turn.
The concept that in a turn you have a reduced effective wingspan is still correct when talking about lift according to the y axis. If you enter a level gentle turn and do not increase power, you will have to trade off airspeed for some extra lift to counteract the reduced effective wingspan by adding up elevator.
Now adding up elevator can't be looked at as adding "apparent weight" to the aircraft. We still need to remain in an inertial reference frame to analyze this. The horizontal stabilizer and elevator are a separate airfoil, just like the wing, that happens to be attached to an extremity of an aircraft. It will create it's own lift and it's own drag. The most correct way of looking at it is that drag will be the "apparent weight" in the non-inertial reference frame. It won't be the lift of the horizontal stabilizer!! The lift, or lack of lift, will only "point" the aircraft in an attitude. There is a range of lift it will produce from positive to zero to negative lift. It will have induced drag if it's producing lift at all and that's what will be the "apparent weight". If it's at zero lift, it will only be experiencing form/skin friction drag.