Stand in the lift of a tall shopping mall and press the button for the top floor. As the lift starts, your knees bend a little and you feel heavier. For most of the ride you feel nothing unusual. As the lift slows near the top, you feel lighter for a moment, as if you might float. Going down, the order is reversed. Yet your mass and the pull of the Earth on you do not change during the trip.
What a scale measures
A weighing scale does not measure your mass, and it does not measure the pull of the Earth on you directly. It measures how hard you press on it. By the third law, that is the same as how hard it pushes up on you: the normal force \(N\). The scale is calibrated to show \(N/g\) in kilograms. On a floor at rest, \(N = mg\), and the scale shows your mass.
The reading of the scale is your apparent weight. Your true weight \(mg\) is the pull of the Earth, and it is the same everywhere in the lift. The apparent weight changes whenever the normal force changes.
The scale in an accelerating lift
Stand on a scale in a lift whose acceleration is \(a\), upward positive. You move with the lift, so you have the same acceleration. Two forces act on you: your weight \(mg\) down and the normal force \(N\) up. Take the \(y\) axis upward.
\[ N - mg = ma \quad\Rightarrow\quad N = m(g + a) \]
When the lift accelerates upward, \(a > 0\) and the scale reads more than your mass. When it accelerates downward, \(a < 0\) and the scale reads less. When it moves at a steady speed, up or down, \(a = 0\) and the scale reads your true mass.
The reading depends on the acceleration, not on the velocity. Take a lift moving up at \(4\,\text{m/s}\) and a lift moving down at \(4\,\text{m/s}\), neither speeding up nor slowing down. Both scales show your true mass. A lift that is going up but slowing down has a downward acceleration, so the scale reads less. This is the moment near the top floor when you feel lighter.
This pays off the promise of the mall. You feel your apparent weight, because your legs and feet sense the push of the floor. The push is larger only while the lift speeds up on the way up or slows down on the way down. During the long steady middle of the ride the push is \(mg\), and you feel normal.
A \(50\,\text{kg}\) student stands on a scale in a lift that is going down and speeding up at \(2\,\text{m/s}^2\). Take \(g = 10\,\text{m/s}^2\). What does the scale read, in kilograms?
A person stands on a scale in a lift. Rank these moments by the reading of the scale, largest first.
A lift moves upward at a steady \(5\,\text{m/s}\). A person in it stands on a scale. The scale reads
When the cable snaps
Suppose the cable of the lift snaps and the brakes fail. The lift and everything in it fall freely, with \(a = -g\).
\[ N = m(g - g) = 0 \]
The scale reads zero. The person and the scale fall together with the same acceleration, so the person does not press on the scale at all. This state is called weightlessness. The name is misleading. The Earth still pulls with \(mg\); that pull is exactly what makes everything fall. What has vanished is the normal force, the push that we feel as weight.
Weightless does not mean that gravity has stopped. An astronaut in an orbiting space station is pulled by the Earth with nearly nine tenths of the weight the astronaut has on the ground. The astronaut floats because the astronaut and the station fall freely together, so nothing pushes on the astronaut.
In April 1984 Squadron Leader Rakesh Sharma became the first Indian in space. He flew on a Soviet Soyuz spacecraft to the Salyut 7 space station and spent about eight days in orbit. On board, pencils, water drops and people floated. Nothing had switched gravity off. The station and everything in it were falling around the Earth at the same rate, so nothing pressed on anything else. The crew exercised every day, because muscles that never push against a floor grow weak. Asked by the Prime Minister how India looked from space, Sharma answered, "Saare Jahan Se Achha".
A boy stands on a scale in a lift whose cable has snapped. While the lift falls freely, which statement is correct?
A mall lift speeds up at about \(1\,\text{m/s}^2\). A \(60\,\text{kg}\) person stands on a scale in it as it starts upward. Estimate the reading in kilograms.
A shopkeeper hangs a \(5\,\text{kg}\) bag of rice from a spring balance fixed to the roof of a lift. Take \(g = 10\,\text{m/s}^2\). Find the reading of the balance in three cases. (a) The lift accelerates upward at \(3\,\text{m/s}^2\). (b) It moves down at a steady \(2\,\text{m/s}\). (c) It slows down at \(3\,\text{m/s}^2\) on the way down.
The balance reads the tension \(T\) in its spring. The same argument as for the scale gives \(T = m(g + a)\), with \(a\) upward positive.
(a) \(a = +3\): \(T = 5 \times 13 = 65\,\text{N}\), shown as \(6.5\,\text{kg}\).
(b) Steady speed, \(a = 0\): \(T = 50\,\text{N}\), shown as \(5\,\text{kg}\).
(c) Slowing down on the way down means the acceleration is upward, \(a = +3\): the reading is \(6.5\,\text{kg}\), the same as in (a).
Lift makers keep the acceleration of a passenger lift close to \(1\,\text{m/s}^2\). The push of the floor then changes by no more than about ten percent. Fast lifts in tall buildings reach high speeds by accelerating gently for longer, not by accelerating harder. A larger acceleration would make passengers stagger at the start and feel their stomachs rise at the stop.
- Deciding the reading from the direction of the velocity.Only the acceleration matters. Going up and slowing down has a downward acceleration and a reading below \(mg\).
- Saying that the weight is zero in free fall.The weight \(mg\) is unchanged. The normal force is zero, and so is the reading.
- Forgetting to divide by \(g\) when the scale is marked in kilograms.The scale shows \(N/g\). A normal force of \(720\,\text{N}\) reads \(72\,\text{kg}\).