Scale of mechanical systems and force-displacement relation

I have a question which has kept my mind busy recently. Assume that we have a system of two masses (m) as shown in the figure. The force (F) results from the gravity force between two masses (m) apart at distance (d), and is applied on a bar with modulus of elasticity (E), length (L), and cross section (A). Assume two pictures are taken from this system from two different distances. So, although the two pictures show an identical system, one of them looks smaller in size compared of the other one.

One person is now assigned the work of solving these two systems, without being told that these two systems are actually identical. Let’s assume that the size of the system in picture 2 is (n) times of that in picture 1. He compares the two systems and thinks that:

m2=n^3 *m1

d2=n*d1

So he thinks that F2=n^4*F1. He also thinks A2=n^2 * A1, however he doesn’t see any difference between the strains of the two bars. So:

e2=e1 => F1/(A1*E1)=F2/(A2*E2) => … => E2=n^2*E1

So he ends up thinking that the second bar is made of stronger material.

The fact that these are actually two identical systems makes us to think maybe it would be better (or at least worth trying) if we redefine the force as F/A^2 and E as E/A (since these new parameters are equal for both systems). Or, instead of (sigma = E.e) we can say (F/A^2)=(E/A)*e    or    sigma/A=(E/A)*e      which is obviously the same as the traditional equation, but has the advantage of being scale-free.

My question is that, is this right? Does it make any sense at all?