Difference resultant aerodynamics force on an airfoil and a flat plate From basic airfoil theory the following free body diagram can be determined for a two dimensional asymmetric airfoil:  

Here the direction of the resultant force is governed by the geometry of the airfoil section.
However, I'm unsure on how the direction of the resultant force is affected when instead of an airfoil a two dimensional flat plate is considered.
Do the resultant force $R$ and normal force $N$ just overlay each other as shown below (my suspicion)? 

 A: The direction of the Resultant force $R$ is always dependent on the direction of the $V\infty$
But however, the direction/orientation of the Normal force $N$ is dependent on the orientation of the body itself ($N$ is perpendicular to the body and axial force $A$ is parallel to the body.) 
In the above case, since there is not much of surface interaction, the lift component $L$ is considerably higher than the drag component $D$ at lower angles of attack $\alpha$. For the given $\alpha$, Resultant force $R$ will be in a different orientation to the Normal force $N$ (You cannot ignore the tiny axial force $A$.) As the $\alpha$ increases, $R$ moves more and more closer to $N$ as the axial force $A$ decreases in the magnitude.
$R$ and $N$ will overlay each other when the angle of attack $\alpha$ is $90 ^{\circ}$.
A: That would sure make it easy to figure out the drag on the flat plate if you knew the lift: $$ D = R \sin(\alpha)$$ and $$ L = R\cos(\alpha)$$ so that $$D = L \tan(\alpha)$$  and the lift to drag ratio is simply $$\frac{L}{D}=\cot(\alpha)$$ Common sense should tell you that can't be right.  Back to reality...
