Climb is not an equilibrium state, therefore thrust=drag and lift=weight do not apply. One can perfectly well climb without pitching the aircraft up if one has (a) powerful enough engine(s)…
what_next wrote:
Climb is not an equilibrium state, therefore thrust=drag and lift=weight do not apply
Steady climb would be if you discregarded the air resistance on the vertical plane (which will be very small anyway). If you have lift > weight + resistance on vertical plane then you’ll be continuously increasing your vertical speed
It’s a pretty silly question I think, since the “correct” answer isn’t necessarily correct, and will mislead students to think that some part of the prop’s thrust vector must be downwards for a steady climb. Why are these things in the question bank, anyway?
Climb is not an equilibrium state, therefore thrust=drag and lift=weight do not apply
Yes it is an equilibrium state, because no acceleration occurs. But what_next is still right: lift does not equal weight.
It is about the definition of lift. Lift is by definition always at a right angle to the direction of travel. So when climbing, the direction of travel is somewhat upwards and the lift is an equal angle tilted back from the vertical. So it is exactly as italinjon put it in post #02: lift is slightly less than weight in a climb – and also in a descent.
And thrust does not equal drag either, and that is perhaps easier to visualize. Because in a climb the thrust must not only fight the drag, but also part of the weight, depending on how steep the climb is.
Yes it is an equilibrium state
Isn’t the entire universe in equilibrium?
There is nothing fundamentally different between a balance of forces stabilising a mass at a constant velocity in one case, and stabilising a mass at a constant acceleration in another case. The latter is likely to require a vacuum and a perpetual source of constant thrust but that’s an engineering challenge which is irrelevant to the discussion and which can be left to the student to plagiarise from google 
Even an atomic bomb going off is in equilibrium. The radiation pressure is balanced by the acceleration of the various bits of mass departing in all directions. The radiation pressure is “considerable” but that’s not the point, either
Anyway, it’s a typical JAA/EASA PPL/CPL/IR type of stupid question which works only with people who have had no scientific/engineering education and who learn this stuff without questioning the ambiguity. PROB99 that was exactly the person who drafted the question, too.
Just my opinion, you understand 
The lifting of the airplane is originally obtained by the lift vector being greater than the weight thereby causing a vertical acceleration. The increased drag from the wings generating lift translates into reduced speed which balances lift and weight thereby reinstating equilibrium. Or something like that.
Peter wrote:
Isn’t the entire universe in equilibrium?
No, never. Well, eventually at infinite time when the entropy in the universe no longer increases.
I like your universe in equilibrium. Even if it is just your opinion, it is a pleasant one. Even atomic bombs seem slightly less frightening being balanced this way :-)
The question itself is probably as old as hangars. Intuitively an airplane needs more lift than its own weight to go up, but as this is “scientifically” wrong, there is a basis for teasing students and more willing victims with this and then feel clever yourself.
As the person responsible for developing questions and answers for the PPL/LAPL “Principles of Flight” written tests in Denmark, I did indeed consider including the question only weeks ago. But I could not imagine that students answering this question right would turn out to be substantially better pilots for that, so I left it out. After reading this thread, I feel reassured.
huv wrote:
lift is slightly less than weight in a climb – and also in a descent.
No, that is only the special case when the thrust of the engine also is in the same direction as the direction of travel. Lift can perfectly well be much larger than the weight if the thrust is in another direction (down). An airplane with thrust vectoring will not “obey” your oversimplified law
LeSving wrote:
No, that is only the special case when the thrust of the engine also is in the same direction as the direction of travel.
You are right, of course. Even if most things called airplanes will have engines thrusting in approximately the direction of travel, your point is another good reason this is a bad question for a test.
