When I get excited about a movie, my only way of calming down is to do a little physics. That explains why I found myself pondering Hooke's law and Young's modulus while watching a trailer for Spider-Man: Homecoming.
Oh, wait. Before going further, I should provide a spoiler alert, just in case you're the type who doesn't even watch trailers. I consider trailers fair game. You have been warned.
I'm not sure just what's happening in this scene, but it looks exciting. For some reason, the elevator in the Washington Monument is falling. Spider-Man crashes through a window, slides across the floor, and fires a web to stop it. He manages to slow it down a bit, but the elevator soon gives way and pulls him into the shaft. Spidey lands on the floor of the elevator (the ceiling had been blown open, or something) and fires a web at the top of the shaft.
Does it work? Beats me. We'll have to wait to see the movie.
Superheroes provide all kinds of fun reasons to look at physics. The big idea I'll focus on here is, of course, Spider-Man's webs. They act like springs, and physicists love springs. Why? Because they aren't too difficult to model. Basically, the more you stretch a spring, the greater the force it pulls back. Most springs exhibit a linear relationship between stretch distance and spring force. Physicists call this Hooke's law, and express it like this:
In this equation, k is the spring constant and it is a measure of the spring stiffness. The flimsy spring in your retractable pen might have spring constant of just 1 newton per meter, while the much stiffer springs on your car might be 200,000. The variable s represents the amount the spring is compressed or stretched, and F denotes the magnitude of the spring force.
How does this apply to the falling elevator scenario? Well, imagine Spider-Man could fire a super web that doesn't stretch at all. He fires it at the elevator, braces himself, and BOOM. The elevator stops. Awesome! No, not really. The elevator stopped in exactly the distance that Peter Parker's arms moved as the elevator jerked to a halt. The small stopping distance means a large elevator acceleration. (Yes, I know, the elevator slowed to a stop. We still call that acceleration.) Any acceleration over 10 g's or so poses a serious---even deadly---risk. In the end, Spider-Man's super-stiff web might stop the elevator, but kill everyone on it.
So it's not a case that Spidey's webs are probably stretchy. Rather, they have to be stretchy.
Now for some rough estimations and calculations to determine the spring constant of a web. Hold on to your pants because this might get a little wild. Let me divide this elevator's fall into three parts.
First, the elevator falls. Does it reach a state of free fall with an acceleration of -9.8m/s2? I'm not sure. Perhaps the elevator's brakes exert some drag, reducing acceleration to something less than free fall. The trailer does provide a glimpse of the brakes, so I will approximate an initial acceleration of -4.9 m/s2.
Watching the trailer, it looks like 1.5 seconds passes between the time the elevator breaks free and when Spidey's web grabs it. Assuming a constant acceleration, the elevator achieves a downward speed of 7.35 m/s. Since the elevator started from rest, I can use the average speed to find the distance traveled: 5.5 meters. That's also the unstretched length of the web (at least approximately).
Next, the elevator slows down as the web stretches. Judging from the trailer, this appears to take about 2.6 seconds. I don't know from the video if the elevator stops, but for the sake of this calculation, I will assume it does---right before a brake fails and pulls Spider-Man into the shaft. This means the elevator goes from 7.35 m/s to zero m/s in that short time. If I use the average velocity during this interval, the elevator falls 9.55 meters. Ah! But this estimation is wrong. It assumes the elevator exhibits constant acceleration during this time. It would not. As the web stretches, the force from the web increases, thereby increases the acceleration. To compensate (for now), I am going to use a stretch distance of 5 meters.
Now I can use the work energy principle to calculate the spring constant. The basic idea is to assume that both the change in kinetic energy and the change in gravitational potential energy are offset by a change in spring potential energy. The energy stored in a spring is proportional to the square of the stretch distance multiplied by the spring constant. At this point, I only need to estimate the mass of the elevator (and the human passengers) to get a value for this spring constant. Let me guess that the elevator and humans come to 1,500 kg. That gives me a spring constant of 7.76 x 104 N/m.
Does that sound crazy? Let me compare it with something to find out. What if I replace the web with a steel cable? Yes, a steel cable does indeed stretch as you pull on it. Not much, no, but it does stretch. The effective spring constant of a cable depends on three things: its length, its diameter, and its material. In general, the longer and thinner a cable, the stretchier it is. The material speaks to something called Young's modulus, and steel has a value around 200 GPa. Using this value and a cable length of 9.55 meters with a diameter of, say, 5 mm, I get a spring constant of 4 x 105 N/m. Yes, this most definitely is a higher value than the spring constant for the web, meaning it wouldn't stretch as much. But remember, stretching saves lives by decreasing the acceleration.
This is too good to not have homework questions. Ignore them if you want, but I'll leave them here to remind myself of things I can calculate later.
What is the true acceleration of the elevator when it falls? If you determine the dimensions of the elevator, you can use video analysis and angular size to get an approximate distance-time graph and find the acceleration. Good luck with this one.
Estimate the diameter of Spider-Man's web so you can determine its Young's modulus value. How does it compare to carbon nanotubes?
In my calculation, I estimated the stretch distance. Perform the calculation with a method that does not involve estimating the web stretch.
Create a numerical calculation showing both the falling and stopping of the elevator.
Calculate the apparent weight of the passengers as Spidey brings the elevator to a stop.
Estimate the time it takes Spider-Man to fall into the elevator (assume that the web is stretched and pulling down on him during the fall).
