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What physics is involved with wing flex when in air?

A simplified answer will suffice.

I can understand when on the ground that a heavier wing makes it bend more down.

My feeble understanding: When in air isn't it only positive G that makes it bend up, and weight of wing don't really matter, instead the weight of the body matters, yes? If that is the case the amount of normal force don't really matter, correct?

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  • $\begingroup$ "the amount of normal force don't really matter..." Which normal force might you be referring to in this sentence? $\endgroup$ – BowlOfRed Feb 28 at 22:01
  • $\begingroup$ Normal force. Its equal to lift at AoA of 0. At AoA of 90 its equal to drag. $\endgroup$ – Invariant Feb 28 at 22:04
  • $\begingroup$ Yes, weight distributed well outboard on the wingspan decreases the wing-bending moment in actual flight, unlike the case when the aircraft is parked on the ground. $\endgroup$ – quiet flyer Feb 28 at 22:38
  • $\begingroup$ Some info in this post $\endgroup$ – Koyovis Feb 29 at 0:16
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Here's a simplified, almost trivial answer.

Call "normal force" the component of the net force on the wing that is perpendicular to the wing chord. The net force: gravity (in whatever direction), plus any force due to air moving past it (in whatever direction(s)), plus any force due to mounted ordnance, or mounted thrust-vectored engines, or air-show wing walkers, or giant magnets, or whatever.

The greater that force, the more the wing bends.

The stiffer the wing (spar), the less the wing bends.

The more that force is nearer the tip than the root, the more the wing bends.

The wing just acts like a (complicated) spring, roughly obeying Hooke's law.

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    $\begingroup$ It is important to mention that gravity means just gravity acting on the wing itself. If you add the weight of the fuselage, you'll obviously get zero for level flight, but the point is to get the force between the two. $\endgroup$ – Jan Hudec Feb 29 at 23:28
  • $\begingroup$ Yes. "Net force on the wing," not on anything else. $\endgroup$ – Camille Goudeseune Mar 1 at 3:46
  • $\begingroup$ Although I wrote a simplified answer will suffice, I maybe didn't mean that simple. But its okay, I found a good answer elsewhere. I will mark this as answer at least until somebody answers with a little more details, perhaps some formulas. $\endgroup$ – Invariant Mar 4 at 12:49

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