Spring Scale in a Lift

Problem

You are in a lift standing on a spring scale. The scale reads your weight. The lift starts moving upward. What does the scale read now? And when it slows down at the top? And while it moves at constant speed?

Solution

The scale does not measure the weight $mg$ directly, but the normal force $N$ with which its surface pushes up on your feet (by the third law, equal to the force with which you press on the scale). We write Newton’s second law vertically, taking upward as positive. The weight force $-mg$ and the normal force $+N$ act on the body:

$N - mg = m\,a$

$N = m\,(g + a)$

The reading therefore depends on the sign of the acceleration $a$, not on the velocity.

Starting to move upward (acceleration upward, $a > 0$). The lift, at rest, must set off upward: it accelerates upward. Then $N = m(g + a) > mg$: the scale reads more than your true weight. It is the same “squashed” feeling you get at start-up.

Slowing down at the top (acceleration downward, $a < 0$). While rising, the lift must come to a stop: the (upward) speed decreases, so the acceleration points downward. Then $N = m(g + a) < mg$: the scale reads less than your true weight. You feel a sensation of “lightness”.

Constant speed ($a = 0$). With zero acceleration the resultant is zero: $N = mg$. The scale reads exactly your true weight, exactly as when the lift is at rest.

$\ev{\text{Starting up: } N > mg \qquad \text{Braking at the top: } N < mg \qquad \text{Constant speed: } N = mg}$

The key point is that uniform motion (constant speed, even a high one) is indistinguishable from rest: only the phases of acceleration change the reading.

Links

Topics: Dynamics Concepts: Newton’s second law · Normal force Objects: Lift · Scales