Apparent weight ($N=m(g+a)$)

Apparent weight measures support rather than the gravitational force itself. In an ideal scale model, it is the magnitude $N$ of the normal force exerted on the body. A scale calibrated to display mass divides that support force by its calibration value of $g$.

Take upward as positive in a constant downward gravitational acceleration of magnitude $g$. If a body stays in contact with a horizontal scale and accelerates upward at $a$, Newton's second law gives $N-mg=ma$, hence $N=m(g+a)$. At rest, $N=mg$. In common free fall, $a=-g$, so $N=0$ despite the nonzero gravitational weight $mg$. If the formula would require $N<0$, ordinary supporting contact cannot realize it: the body loses contact instead.