Pop Video: Not Everyone Stuck in a Lift Appreciates Maths

In this video we learn how your knowledge of Newton’s equations of motion under constant acceleration can help you pass the time when you are stuck in a lift, but also that you need to be quite careful how you use them, partly because other stuck lift passengers may not appreciate your conclusions.


Video Transcript

It turns out that not everyone stuck in a lift appreciates Newton’s equations of motion. A few years ago, I found myself stuck in a lift with five of my colleagues, somewhere between the ground floor and the seventh floor, on our way to a meeting. If we’d been in America, we would have been stuck in an elevator somewhere between the first and eighth floors.

At the ground level, we all stepped in, someone pressed the button labelled seven, the door is closed, and we set off, merrily discussing the pros and cons of using different characters as separators between data elements in our application-to-application protocol for computer messages before the meeting had even officially begun.

One of the colleagues suggested, “I think we should go for a slash.” And several of us again considered the differences between the English language in the UK and the US. Suddenly, there was a bang, and the lift came to a grinding halt. We were stuck. I picked up the “Call for Help” phone in the lift and was told that the engineer would be called but would probably be a couple of hours.

The collective mood was reasonably cheerful, and we set about seeing if we could free ourselves in the meantime. Two of us managed to pull apart the lift door, hoping that we’d be able to clamber out and use the stairs. But unfortunately, there was another pair of doors beyond them, and we couldn’t budge them; they were stuck fast. We weren’t going anywhere, so we held our meeting in the lift and then chatted a bit.

After a while, conversation dried up, and I spotted that there was a tiny gap between the two open doors, below which was the drop to the bottom of the lift shaft. I suggested that if we dropped a penny through the gap and counted how many seconds it took to hit the ground, we could use one of Sir Isaac Newton’s equations of motion under constant acceleration to work out how high up we were.

𝑆 equals 𝑢𝑡 plus a half 𝑎𝑡 squared, where 𝑆 is the distance travelled in metres, 𝑢 is the initial velocity of the object in metres per second, 𝑡 is the time taken in seconds, and 𝑎 is the constant rate of acceleration of the object in metres per second per second.

The initial velocity of the penny would be zero if we dropped it from rest, rather than threw it with a force through the gap. And this would mean that the first term was zero times the time, just zero, so we could ignore it. I remembered that the acceleration due to gravity was about 9.8 metres per second per second. And we would ignore air resistance, which we thought would be fairly negligible for a small penny under these conditions at the sort of height we were. And that would give us the constant acceleration we needed to work with this formula.

So our penny would start its fall at zero metres per second. And every second, it would get 9.8 metres per second faster. After one second, it would be travelling at 9.8 metres per second. After two seconds, it would be travelling at 19.6 metres per second. After three seconds, it would be travelling at 29.4 metres per second, and so on. And for the sake of convenience, we could round the acceleration to 10 metres per second per second to make our calculations easier, because nobody had a calculator with them.

So all we needed to do in order to work out how far off the ground we were was drop the penny, count how many seconds it took to hit the ground, square that number, multiply the answer by 10, and then halve that result. This seemed to amuse the group and raised their spirits a bit. We could do a scientific experiment and would have an interesting twist to add to our anecdote about getting stuck in a lift for two hours in the morning.

So we dropped a penny through the gap and counted how long it took to hit the bottom of the lift shaft. It was about two seconds. And the distance was about half times 10 times two squared; that’s 20 metres. That seemed like quite a fall, and the mood in the lift suddenly changed as we all took this information in.

With one simple calculation, we’ve turned from a group of relaxed people chatting into a group trapped in a confined space under perilous conditions, with five of them wanting to throw me down the lift shaft. Luckily, the small gap wasn’t large enough to enable that to happen.

I suggested that we’d slightly overestimated the height because we’d used a value of 10 metres per second per second for acceleration due to gravity, rather than 9.8. Now, a half times 9.8 times two squared was actually quite easy to work out in our heads. Multiplication is commutative and associative, so we could do a half times two squared first to get two and then multiply that by 9.8.

This meant we were only 19.6 metres high. I’d saved us 40 centimetres. But they still weren’t happy. So I said, “And we counted two seconds, but that could easily have been as low as one and a half seconds, cause it’s so difficult to count such a small amount of time in your head.” That would mean we were only just over 11 metres above the ground. Then someone said that “It’s still like jumping off nearly three double-decker buses stacked on top of each other; they’re about four metres tall each, so it’s not gonna do us much good.” And someone else added, “And we could just as easily have underestimated the time; maybe it was two and a half seconds. How high would we be then?” “About 30 metres,” I calculated. “And that’s nearly eight double-decker buses,” added the first chap.

“Great,” I said. “I think we’ve all learned something here, for example, about upper and lower bounds. When we try to measure anything, there’s a limit to how accurately we can measure it. So it’s wrong to give just one answer. In our case, by considering the margin of error of our timing, plus or minus half a second, we can confidently say that we were between about 11 metres and 30 metres above the ground. That gives us a realistic picture of the accuracy of our measurement.”

Then someone said, “But don’t we need to know how much the penny weighs? Surely heavier things fall faster than lighter things.” This was an interesting point. If you drop a feather and a hammer, it seems obvious that the heavier hammer would hit the ground first, while the lighter feather would gently float to the ground some time later. This is due to air resistance.

There’s some excellent video footage from NASA’s Apollo 15 mission in 1971, in which astronaut David Scott dropped a hammer and a feather on the moon, where there’s no atmosphere to provide air resistance. The two fell at the same rate and hit the ground together. The mass of each object didn’t affect the rate at which it accelerated toward the ground when there was no air resistance.

When an object is falling towards Earth, it has a force due to gravity acting on it, pulling it downwards. But as it gets faster, it has an increasing air resistance force acting against gravity, to reduce the resultant downward force and so reduce the acceleration.

The air resistance is equal to the density of the air, measured in kilograms per cubic metre, times the drag coefficient of the penny, which is a constant specific to that particular penny, times the area of the penny, which is facing downwards as it falls, in square metres, times the square of the velocity, in metres per second, all divided by two.

A reasonable estimate for the air density at around 15 degrees Celsius at ground level is about 1.225 kilograms per cubic metre. A British penny has a diameter of about 20 millimetres and a thickness of about 1.5 millimetres. So if it’s on its side, falling, it exposes an area of about 30 square millimetres. That’s 0.00003 square metres.

Now the drag coefficient’s a bit trickier. Ideally, you’d put a penny in a wind tunnel, take lots of measurements, and work out an estimated empirical value. I don’t have access to a wind tunnel, but a web search has given me some estimates of between about 0.5 and 1.17. So for the sake of argument, I’ll go for 0.8. And when we work that out, we get that the air resistance is equal to 0.0000147 times the square of the velocity.

Since the coin was only falling for a couple of seconds, it didn’t reach a very high velocity, and the air resistance would be pretty small. But as the velocity increases, the air resistance increases. And there will come a point at which it balances the gravitational force, and the penny will just keep falling at that velocity with no more acceleration. That’s called the terminal velocity.

So what’s the terminal velocity of a British penny? Well, there’s a formula to work this out. The terminal velocity is equal to the square root of two times 𝑚 times 𝑔 over 𝜌 times 𝐴 times 𝐶, where 𝑚 is the mass of the coin in kilograms, 𝑔 is the acceleration due to gravity in metres per second per second, 𝜌 is the density of the fluid being fallen through by the object in kilograms per cubic metre, 𝐴 is the area of the object projected onto the fluid, in square metres, and 𝐶 is the drag coefficient, again a constant.

Now, using my assumptions about the coin falling sideways, so projecting a very small area, a drag coefficient again of 0.8, and a mass of 3.5 grams, we get that the terminal velocity is equal to 48.3 metres per second. That’s about 108 miles per hour.

Now I’ve seen several calculations on the Internet that say the terminal velocity for a penny is closer to 11.1 metres per second. But they are for US pennies, which are a bit smaller and lighter than UK pennies. And they assumed that the penny would fall face down all the way, with a much larger surface area projected, and also a drag coefficient of 1.17. They used the same method as me, but based on a different coin and different assumptions for how the coin falls.

I suspect the coin actually flips over a few times when it falls. So my figure is probably an overestimate of the terminal velocity for a UK penny, and theirs is probably an underestimate for the terminal velocity of a US penny. Anyway, the upshot of all of this is that the coin doesn’t accelerate at a constant rate. As each second goes by, we don’t simply add another 9.8 metres per second to the velocity, like we would in a vacuum. We add a bit less each second, until the coin reaches its terminal velocity. The rate of acceleration decreases with time, and the equation of motion we used doesn’t apply.

For the limited height of our experiment, it actually gave us a reasonable estimate. But because the coin was travelling a bit slower than we thought as time went by, we were calculating a bit of an overestimate of our height.

So how high up were we actually? Well, eventually, the lift engineer came and opened the door. And we discovered that we were about halfway between the fourth and fifth floors. Assuming that the height of each storey of the building was about 4.3 metres and there was maybe a one-metre pit beneath the lift, that would have placed the base of the lift just over 20 metres above the ground. The errors in our measurements and the calculations we used had luckily balanced out and given us about the right answer.

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