An object's weight is the gravitational force acting on it. Near Earth's surface its magnitude is approximately

$$W=mg.$$

A scale does not directly measure this gravitational force. It measures the contact force between the object and the scale. For a person standing on a scale, the scale reading is the normal force $N$. This reading is called the apparent weight.

When there is no vertical acceleration,

$$N-mg=0,$$

so $N=mg$ and apparent weight equals weight. That familiar equality is a special case, not a general rule.

Accelerating elevator

Choose upward as positive. Newton's second law gives

$$N-mg=ma,$$

so

$$N=m(g+a).$$

If the elevator accelerates upward, $a>0$ and the scale reads more than $mg$. If it accelerates downward, $a<0$ and the scale reads less. The reading depends on acceleration, not on whether the elevator happens to be moving upward or downward.

For example, a $60,\mathrm{kg}$ person in an elevator accelerating upward at $2.0,\mathrm{m/s^2}$ has

$$W=mg\approx589,\mathrm N,$$

but

$$N=60(9.81+2.0)\approx709,\mathrm N.$$

The gravitational force has not increased; the larger scale reading comes from the larger supporting normal force required to accelerate the person upward.

Apparent weightlessness

In free fall the downward acceleration is $g$, so with upward positive $a=-g$. Then

$$N=m(g-g)=0.$$

The scale reads zero: the object has zero apparent weight. Gravity has not disappeared. Its actual weight $mg$ is still nonzero; what has disappeared is the supporting contact force.

This is why an astronaut and a nearby spacecraft can float relative to one another while orbiting Earth. Both are falling together under gravity, so the spacecraft does not need to provide the astronaut with a sustained supporting force. Orbital “weightlessness” is therefore not the same as zero gravity.

A hanging force meter gives an analogous result using tension instead of a normal force: when the suspended object and meter accelerate together, the reading reflects the supporting tension, which need not equal $mg$.

Keeping weight for the gravitational force and apparent weight for the support-force reading avoids confusing a change in what a scale reports with a change in gravity itself.