Reference frames and Newton’s second law

Question

I have an understanding Issue.

We were told that apparent weightlessness was due to the absence of Normal Reaction force.The example chosen was a man in a lift...the latter moving with different accelerations.

For each value of acceleration we calculated the value for the normal reaction and saw that it decreased as the lift accelerated downwards and reached zero when the lift accelerated with an acceleration: a=g Consider a downward acceleration less than g...we have :

Mg-R=Ma => R is less than Mg

(With respect to an inertial referential)

I agree with this idea( A balance will have more issues measuring the mass of a body if they are both in a falling elevator)

But we were also told normal reaction was the force responsible for the fact that the mass didn’t run through the floor of the elevator and break it...and since the elevator is still in one piece despite a downward acceleration...with respect to the Non inertial referential centred at the elevator’s floor...we should have

Mg-R =0 => Mg=R

But R cannot be less than and equal to mg at the same time...Help needed

Answer 1

Newton's 2nd Law in the form $$\sum \vec{F} = M\vec{\mathcal{A}}$$ doesn't work in an accelerating reference frame if all you include in the sum of forces are weight and normal forces. You must include an $-M\vec{A}$ term where $\vec{A}$ is the acceleration of the non-inertial frame and the $\vec{\mathcal{A}}$ is acceleration relative to the non-inertial frame.

So, for your situation if down is positive and the elevator is accelerating downward (relative to some inertial frame) with magnitude $A=a$ and the mass is not accelerating relative to the elevator ($\vec{\mathcal{A}}=0$) you have $$\sum\vec{F}=Mg-R-Ma = 0$$ $$R=Mg-Ma$$ as in the first case.