# Tension in Spider-Man's Webs

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## Spider-Man (2002)

The feature film Spider-Man is about a young man who obtains extraordinary powers by being bitten by a genetically altered spider. He begins to experiment and develop his powers and realizes that he can shoot web out of his forearms. Soon after that, he realizes that he can shoot his web onto objects, such as tall buildings. He furthers this power by using his web as a rope, and swinging from place to place.

In Chapter 8, entitled "Scaling the heights", Spider-Man is first learning how to use his web to swing from place to place. He attempts to swing from one roof on to another by shooting his web onto a crane over a far building, and swinging himself onto a lower building's rooftop. Since he's new at this, his plan doesn't work out well. He's able to swing to the far building, but he is unable to stop himself. He reaches his lowest point, and then swings back up a bit, crashing into a billboard.

The height of the building Spider-Man starts off on is 6 stories, or 18 meters high (assuming one story is 3 meters). The height of the building he wants to swing to is 1 story, or 3 meters high. The crane which he shoots his web onto is 7 stories, or 21 meters high. In order to calculate Spider-Man's initial potential energy, we must use the formula

*U* = *mgh*

where *U* represents potential energy, *m* represents mass, *g* represents the acceleration due to gravity (9.8 m/s^{2}), and *h* represents height. In this specific case, *m* is 68 kg, and *h* is 15 meters (not 18 meters, because he's swinging to the roof of a building that's 3 meters off the ground).

*U* = *mghU* = (68 kg)(9.8 m/s

^{2})(15 m)

*U*= 9996 joules

Since Spider-Man starts off not moving, he has only potential energy. At the bottom of his swing, he has only kinetic energy, given by

*K* = ½*mv*^{2}

where *v* is velocity, and once again, *m* is mass. We don't know Spider-Man's velocity at the bottom of his swing. In order to calculate it, we apply the law of conservation of energy, where the sum of energy initially equals the sum of the energy finally. Therefore, kinetic energy at the bottom of the swing is equal to Spider-Man's initial potential energy, or

½*mv*^{2} = *mgh*

Using the magic of algebra, we can solve for *v*, and find that

*v* = sqrt(2*gh*)*v* = sqrt(2(9.8 m/s^{2})(15 m))*v* = 17.15 m/s

The velocity at the bottom of his swing is found to be 17.15 m/s (38.36 mph, or 61.74 km/h).

In order to calculate the tension in Spider-Man's web, we must use centripetal force, the net force acting on Spider-Man. The centripetal force is the force directed toward the center of the circle, that keeps an object moving in a circle, and is given by the formula

*F*_{c} = *mv*^{2}/*r*

Where m is the mass of the object, *v* is the velocity of Spider-Man's swing, and *r* is the radius of the circle formed while Spider-Man swings. In this situation, centripetal force is equal to the tension in the rope subtracted from the weight of Spider-Man. Tension is a force found in objects such as wires, strings, ropes, or webs, in this case. Weight is the force found by multiplying an object's mass by the acceleration due to gravity (mg). Therefore

*mv*^{2}/*r* = *T*–*mg*

and using the magic of algebra once again, we solve for *T*, and find that

*T* = *mv*^{2}/*r* + *mg T* = (68 kg)(17.15 m/s)

^{2}/(18 m) + (68 kg)(9.8 m/s

^{2})

*T*= 1777 newtons

In more conventional units, 1777 newtons is equal to 399 pounds-force, or 181 kilograms-force.

## Spider-Man 2 (2004)

In Chapter 43 of Spider-Man 2, entitled "A Train to Catch", Dr. Octopus seizes the controls of a train full of passengers, and puts the train at maximum speed. He then breaks the controls, leaving the train on course to drive straight off the elevated track. Spiderman is the only person that can save the train and its passengers from the deadly crash. He initially attempts to stop the train with his own strength, but fails miserably and doesn't slow the train down. He then attempts to fire his webs onto the surrounding buildings, creating a makeshift sling of himself and his webs. He is unsuccessful once again, but realizes that he's onto something. Finally, he fires numerous webs onto the surrounding buildings, and again makes a sling of himself. Finally, the train begins to slow down, and Spiderman stops the train right before it would have crashed.

When Dr. Octopus breaks the controls of the train, it's moving at an initial speed of 35.76 m/s (80 mph). Each car of the train is a 2200 series, as according to http://www.chicago-l.org/multimedia/Spiderman2/. The mass of one car of the train is 21,500 kg (47,399 lbs), as according to http://www.chicago-l.org/trains/roster/2200.html, and since the train has 6 cars total, its total mass is 129,000 kg (284,396 lbs). Assuming there are 20 passengers in each car of the train, and assuming they all have a mass of 67.5 kg (149 lbs), the total mass of the train is 137,100 kg (302,254 lbs). Finally, the time the train was stopped was found to be 46 seconds. Since we know that the train ends up completely stopped, we can calculate the acceleration of the train over the time that it is stopped, using the formula *a* = (*v*−*v*_{o})/*t*: *a* represents the acceleration, v is the final velocity, *v*_{o} is the initial velocity, and *t* is the time over which the object is accelerated.

*a* = (*v*−*v*_{o})/*t a* = (0–35.76 m/s)/46 s

*a*= −0.78 m/s

^{2}

In this case, the acceleration is negative, since the train is being slowed down rather than sped up.

Using the acceleration and the total mass of the train and its passengers, we can calculate the force Spiderman exerts to stop the train (tension in his web). We can calculate this by applying Newton's Second Law of Motion, where Σ*F* = *ma*. Σ*F* is the net force (in this case, tension), *m* is the mass, and *a* is the acceleration.

Σ**F** = *m***a**

Σ**F** = (137,100kg)(−0.78 m/s^{2})

Σ**F** = −106,938 newtons

In more conventional units, 106,938 newtons is equal to 24,041 pounds-force, or 10,905 kilograms-force.

As a bonus, we decided to calculate the total distance the train traveled from the time Dr. Octopus destroyed the controls to the time Spider-Man stopped it. We can do this using the formula

*s* = *v*_{o}*t* + ½*at*^{2}:

*s* is distance, *a* represents the acceleration, *v*_{o} is the initial velocity, and *t* is the time over which the distance is traveled.

*s* = *v _{o}t* + ½

*at*

^{2}

*s*= (35.76 m/s)(46 s) + ½(−0.78 m/s

^{2})(46 s)

*s*

^{2}= 819.72 meters

## Bonus Analysis: Spider-Man's Free Fall

In the movie Spider-Man 2, a young student (Peter Parker) is struggling with his discovered abilities and trying to keep it a secret. He also has to juggle a separate life outside his spider-like abilities and keep people from finding out who he really is. He encounters numerous challenges throughout the movie both as being a real man and Spider-Man.

In chapter 13, titled "Web Failure" we see Spider-Man swinging above many buildings. He unexpectedly loses his power to shoot his web and falls. Spider-Man experiences free-fall for 4 seconds before he hits the roof of a building. During the free-fall, Spider-Man experiences an acceleration due to gravity of 9.8 m/s^{2}. By having the time and the acceleration due to gravity, we can use a simple kinematic equation to find the distance Spider-Man falls before hitting the roof of the building.

*y* = ½*at*^{2}

*y* = ½(9.8 m/s^{2})(4 s)*y* = 78.4 m

If we ignore the force of drag acting upward on Spider-Man as he falls, we can also determine Spider-Man's vertical speed at the instant before he hits the roof of the building. We use the formula:

*v*^{2} = *v _{o}*

^{2}+ 2

*as*

*v*

^{2}= (2)(9.8 m/s

^{2})(78.4 m)

*v*= 39.2 m/s

Daniel Saronson, Michael Robbins, Gafei Szeto, David Rozenberg -- 2005

Follow up studies:

- Doing whatever a spider can. M. Bryan, J. Forster and A. Stone.
*Journal of Physics - Special Topics*. Department of Physics and Astronomy. University of Leicester. 31 October 2012.

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