# Coefficient of Friction for Skin: The Incredible Hulk

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## Purpose

To determine the coefficient of friction between The Incredible Hulk's skin and sand.

## Introduction

### Movie Synopsis

Director Ang Lee's Hulk is a 2003 movie adaptation of the thirty-year old Marvel comic book series about a young six-foot, two-hundred-pound scientist, Bruce Banner, who accidentally exposes himself to gamma radiation, and consequentially, he acquires an extra 8 feet and a few extra tons whenever he loses his temper. In addition, his skin gains a ridiculous greenish tone, and becomes bulletproof. As the Hulk, he can jump across a distance of 3-4 miles and a height of one kilometer. He also gains super strength and can lift more than 100 tons in the comics. The angrier he is, the stronger and larger he becomes. Therefore, the movie and the comic book series almost break every law of physics and human biology. Almost.

The storyline starts off with Dr. David Banner (Paul Kersey as "Young David Banner", Nick Nolte as "Father"), a genetics researcher obsessed with "improving"” human DNA to the point at which he modifies his own genes. His son, Bruce Banner (Eric Bana), unfortunately inherits this genetic discrepancy, and contracts blue patches on his skin whenever he experiences intense emotions. David, under extreme guilt, desperately tries to find a cure for the child's conditions until the government orders him to shut down his project. In an act of rage, Banner sets the facilities gamma reactor to self-destruct, and accidentally killed his wife in an attempt to murder his son. The elder Banner was taken away to a mental hospital while the younger was sent to foster care.

Twenty years later, Dr. Bruce Banner is working in the same scientific field as his father, and experiments with nanobots, energized by gamma radiation from a device called the Gammasphere, to regenerate living tissue. In a laboratory accident, the Gammasphere malfunctions, and emits radiation into Banner's body, thus activating his altered genes and manifesting him into the monster known as the Hulk. After discovering Banner's condition, his love interest, Betty Ross (Jennifer Connelly), has the army general Thaddeus Ross confine him to an industrial-military complex where he can be researched and treated. Afterwards, Banner breaks out of the complex as the Hulk, and confronts the US military with his bare hands. This is when we made our analysis. (View a video clip of the scenes to be analyzed here.)

### Given Data

The Hulk immobilizes four fully armed Abrams M1 tanks, and two RAH-66 Comanche helicopters in the remote desert before arriving at San Francisco, and freefalling from an F-22 fighter plane at flight level higher than 950 (95,000 feet or 28.96 kilometers). In one instance, he lifts up one of the Abrams and swings it in 3 full circular revolutions before finally releasing it into the sky. Interestingly and miraculously, the troop inside the tank survives the entire ordeal.

According to Wikipedia, the Abrams tank weighs about about 61.4 tonnes (or 61,400 kilograms), and is about 9.76 meters long, with its 5.89 meter gun included. According to IMDb The Hulk himself weighs about 3,400 pounds or 1657 kg when he's at his maximum mass. His onscreen height is about 15 feet or 4.6 meters. Using this data, and the tangential velocity we are about measure, we can determine the centripetal acceleration of, and the centripetal force applied by the tank.

### Physics Formulas

Centripetal Acceleration equals to the square of the revolving body's tangential velocity (at the moment of release), divided by the distance between its center of mass, and the center of the circular path (the Hulk). We can measure velocity knowing that velocity = distance/time, or in this case, arc length/time.

Centripetal Acceleration = (Velocity)^{2}/Radius

According to Newton's Second Law, the product of an object's mass and acceleration is equivalent to the force applied to it.

Force = (Mass)(Acceleration)

Thus, the product of the tank's mass and centripetal acceleration is equivalent to its centripetal force.

Centripetal Force = (Mass)(Velocity)^{2}/Radius

A centripetal force has t come from something. In our case, it is the friction force between the Hulk's bare feet and the ground beneath him. The force of friction is equal to the normal force applied by the ground multiplied by a friction coefficient. The friction coefficient is a constant value and is usually unknown until experimentally determined.

Frictional Force = (Coefficient)(Normal Force)

Next, we can determine the height (h), and horizontal distance (s) the tank traveled by using the final tangential velocity and the formulas for projectile motion. Projectile motion refers to the motion of an object projected into the air at an angle, thus moving in two different dimensions: x (horizontal) and y (vertical).

The initial velocity (our tangential velocity) can be divided into two velocity vectors: vertical and horizontal velocities. The vertical velocity v_{y} can be defined as the product of the initial velocity and the sine of the angle of trajectory. The horizontal velocity v_{x} is defined as the product of the initial velocity and the cosine of the angle of trajectory.

v_{y} = (initial velocity) sin θ

v_{x} = (initial velocity) cosθ

Another formula used for projectile motion is:

s = s_{0} + v_{0}t + ½ at^{2}

y = y_{0}+ v_{0}t + ½ gt^{2}

We can use this formula to find the time (t) it took for the tank to reach maximum height.

The v_{0} is the initial velocity and equals to zero at maximum height. a is the acceleration (g or 9.8 m/s^{2} in this case) and s_{0} equals to the height at which the object was thrown y_{0}. The final s is the maximum height y.

The next commonly used equation for projectiles is distance equals to the product of horizontal velocity and time. Thus, we can multiply the "t" value we'll determine by v_{x} to compute the horizontal distance, s.

s = v_{x}t

Finally we can determine the Hulk's time of free fall from the fighter plane simply by using the same formula as before:

y = y_{0} + v_{0}t + ½ at^{2}

However, this time y equals to zero because the hulk hits sea level and the hulk starts falling from a height of y_{0}. v_{0} also equals to zero because the Hulk does not begin falling with an initial velocity.

## Procedure:

- First, we timed the first rotation of the Abrams tank as the Hulk swung it, as well as counting the number of frames for that rotation. This is necessary to find the time associated with each frame (any portion of the swing could be timed and be used to find time per frame; we just used 1 rotation as a standard or to be consistent).We used this method (as explained in analysis) to find the time per frame 15 times. Then, we averaged the all the times to get a more moderate time per frame measurement. Record all measurements.
- Second, since counting frames is rather easier and more efficient than using a stopwatch, we counted the number of frames for each quarter of rotation (π/2 or 90°).

## Analysis

Centripetal force or any type of moving force must be opposed by a greater frictional force in order to prevent slipping. However, at the instant that Hulk immediately releases the Abrams tank, his body has stopped rotating altogether. This indicates to us that both the acceleration and net force at that instant is zero. Therefore, we can set the force of friction (this equal but opposite force cancels out the centripetal force) and centripetal force equal to solve for the coefficient μ.

First, we found the time each frame is worth by using the following formula: time taken/number of frames. Since the number of frames is constant (50 frames for the first rotation), we averaged all the times taken by a stop watch and then divided the average time by 50 frames. The average time we found for the first rotation was 1.82 s. 1.82 s divided by 50 frames then gave us 0.036 s per frame.

Frame | Time | Rotation | Distance (m) | Velocity (m/s) |
---|---|---|---|---|

0 | 0.000 | 0.00 | 0.000 | 10.167 |

25 | 0.910 | 0.25 | 9.252 | 14.524 |

35 | 1.274 | 0.50 | 18.504 | 26.755 |

44 | 1.602 | 0.75 | 27.756 | 31.772 |

51 | 1.857 | 1.00 | 37.008 | 33.890 |

59 | 2.148 | 1.25 | 46.260 | 33.890 |

66 | 2.403 | 1.50 | 55.512 | 31.772 |

75 | 2.73 | 1.75 | 64.764 | 31.772 |

82 | 2.985 | 2.00 | 74.016 | 42.363 |

87 | 3.167 | 2.25 | 83.268 | 58.483 |

91 | 3.313 | 2.50 | 92.520 | 58.483 |

96 | 3.495 | 2.75 | 101.772 | 63.544 |

99 | 3.604 | 3.00 | 111.024 | 84.725 |

Then, we used Graphical Analysis to help us convert each frame (each transition in row/shown frame represent a 90° turn by hulk) to its corresponding time by inserting into the calculated column of Time

time = (frame count)/(27.74 frames/sec)

Second, we find the corresponding accumulative distances traveled by the tank for each 90° change in rotation. This is accomplished by multiplying the percent of a rotation by the circumference of the circle formed by the tank in circular motion (the tank length is the radius): In other words, (2π)(9.76 m)(rotation) is used for the calculated column of distance (meters).

In order to find the corresponding velocities at each start or end of a 90° turn by Hulk, we used another calculated column for velocity that uses the function ds/dt.

Using the time and distance acquired, we created the distance vs. time graph below:

To find centripetal force use Newton's Second Law:

F = ma

a_{c} = v^{2}/r

F_{c} = mv^{2}/r

The mass (m) in the equation represents the mass of the Abrams tank that is being swung around ~ 61.4 tonnes or 61,400 kg. The velocity (v) represents the release velocity ~ 84.725 m/s (found from Graphical Analysis); and the radius (r) is the tank gun itself (5.89meters) because the gun muzzle starts from the center of rotation and ends approximately at the center of the tank.

F_{c} = mv^{2}/rF _{c} = (61,400 kg)(84.725 m/s)^{2}/(5.89 m)F _{c} = 74,342,591 N |

To find the friction force use the equation described in the introduction.

ƒ = μN

μ = ƒ/N

On level ground, the normal force equals the weight. In this case, the weight of the Hulk (most angry version) ~ 1657 kg; plus the weight of the Abrams tank ~ 61,400 kg.

N = W = (m_{hulk} + m_{tank}) g

N = (1657 kg + 61,400 kg) (9.81 m/s^{2})

N = 618,589 N

Set ƒ = F_{c}, and solve for μ.

μ = ƒ/N

μ = (74,342,591 N)/(618,589 N)

μ = 120

## Conclusion

Coefficient of friction of Hulk's skin versus the sand is 120. The coefficient of friction for ordinary human skin is approximately 0.8. The Hulk has some pretty crazy, sticky skin.

# Bonus: How Far Did the Tank Fly?

## Abstract

Our purpose is to find the distance the tank traveled after Hulk released the tank from his furious swings.

## Procedure

- Take a screen shot
- Use a virtual ruler to measure the picture in pixels

## Analysis

Before any calculations, we must first find the ratio between the expected dimensions (reality) and those measured by the virtual ruler in the above screen shot. We measured Hulk’s height in the screen shot to be 246 pixels tall. Since we are given Hulk’s height to be 15 ft tall or 4.572 meters in the movie, we can set 4.572 m equal to 246 pixels to determine the ratio.

4.572 meters = 246 pixels

1 meter = 53.806 pixels

1 pixel = 0.0186 meters

To find the distance the tank traveled, we first needed to determine the angle at which it was thrown.

First we draw the right triangle that we will use to determine the angle in which the Hulk threw the Abrams tank. We chose point A to be our origin (where the end of the muzzle is). Then we drew a line along the muzzle to the end of the tank. This line was 111 pixels wide by 35 pixels high. Since these are the sides of a right triangle we used the tangent function to determine the angle of release.

tan θ = opposite/adjacent = y/x

sinθ = 35/111

θ = 17.5°

Using the release velocity we found in the previous section (84.725 m/s) and the angle we just found (17.5°), we can determine the initial vertical velocity

v_{y} = 84.725 sin(17.5°) = 25.48 m/s

Use this value to determine the maximum height, which occurs when v_{y} = 0.

v_{y}^{2} = v_{0y}^{2}–2g(y–y_{0})

0 = (25.48 m/s)^{2}–2(9.81 m/s^{2}) y

1856.83 = 2(9.81 m/s^{2}) y

y = 33.09 m

We can use this to figure out the time it takes for the tank to fall from max height to the ground and then multiply the time by 2 to find the total time the tank was in the air.

y = ½ at^{2}

33.09 m = (9.81 m/s^{2})(t^{2})/2

t = 2.60 s

t_{total} = 2t = 5.20 s

Now use the horizontal velocity

v_{x} = 84.725 cos(17.5°) = 80.80 m/s

to find the horizontal distance traveled by the tank by:

s = v_{x}t

s = (80.80 m/s)(5.20 s)

s = 420 m

## Conclusion

The Hulk throws the Abrams tank, weighing 61,400 kg, 420 m away. Or in other words, Hulk's got pretty big muscles. On the side note, the guy in the tank somehow managed to survive this throw.

# Double Bonus: Free Fall Hulk

## Background

The Hulk was fighting with dozens of fighter jets at San Francisco Bay. When an F-22 raptor jet was about to hit a bridge unintentionally, Hulk jumped on to the aircraft to prevent it from crashing into the bridge. Then, the pilot tried to give Hulk a taste of "thin air" by flying as high as it could.

## Abstract

Our purpose is to prove whether the distance that the pilot claimed to have traveled (with Hulk hanging on) is truly compatible with the time we found for the free fall

## Procedure

- Using a stop watch, time the Hulk's free fall immediately after he lets go of the aircraft. Stop the watch when Hulk strikes the water. Repeat this 15 times and average the times.
- Note that the pilot mentions his flight level to be 950, which is 95,000 ft or 29,000 m.

## Analysis

We found the time that Hulk took to fall from the 950 flight level to be approximately 48.2 seconds. We used the following formula to find the distance Hulk traveled during the free fall:

y = ½ gt^{2}

y = (9.81m/s^{2})(48.2 s)/2

y = 11,400 m

## Conclusion

Since the distance that we calculated from our timing of the free fall is only half as great as the distance claimed by the pilot, we can safely say that either the pilot was scared out of his mind or the movie director didn't want the free fall to take up too much time of his movie.

## Sources of Error

- When we tried to count the number of frames per 90 degree turn, we are watching hulk in front of him. To know when exactly a 90 degree turn is truly complete is impossible.
- The desert in the movie is not a uniform plane; there are hills and dunes here and there. Thus our projectile analysis was inaccurate since the tank landed at a slightly higher elevation.
- Our analyses owed much of its errors to air resistance, air resistance and more air resistance. In the free fall analysis, for instance, the Hulk is supposedly falling through 30 kilometers of atmosphere, with the jet stream included. In addition, the air resistance in the projectile analysis could have reduced the time, and distances in both x and y directions.
- Furthermore, the Hulk did not release the tank after one full rotation; rather it was slightly more than one rotation. We assumed that the tank was released exactly 90 degrees into the page, but in actuality it was released a few degrees to the left. Therefore the velocity we found was slightly inaccurate.

Simon Chen, Anthony Ho and Jason Lu -- 2005

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