By Rhett Allain
I probably won’t see the Avengers movie for quite some time. (I never seem to get out to the theaters.) So, what do you do when you can’t see it? Blog about it. I assume you have seen the trailer, right? Here it is.
There are lots of things to look at in this short clip, but I like The Hulk. What kind of forces would The Hulk exert on the ground during a jump? Before guessing at some of the parameters, let me start with a general case. Here is the jumping Hulk in three positions (not drawn to scale).
I didn’t draw the forces, but during the jumping part there would be the gravitational force and the force of the ground pushing up. The best approach in a problem like this is the work-energy principle. Why? Because we are dealing with changes in position instead of changes in time. If The Hulk is treated as a point particle, then during any phase the work-energy principle can be written as:
A quick hat tip to the Matter and Interactions curriculum for getting me to think about the work-energy principle in terms of point particles versus real systems. However, going from position 2 to 3, it doesn’t matter how I treat The Hulk. For that motion, the only force on him is the gravitational force (mg). Also, at the end of this motion (position 3), The Hulk has no velocity. The work-energy principle for this motion says:
I could solve for the speed The Hulk has when he leaves the ground, but for now I think I only need the kinetic energy. Notice that the work done by the gravitational force is negative. This is because gravity is pulling in the opposite way The Hulk is moving. Also, the work depends on the change in vertical position of The Hulk’s center of mass (which I have labeled as h – s2).
Now what about the “jumping” part of The Hulk’s jump? Now there are two forces doing work on The Hulk when treated as a point particle (which he is not).
Since I already have an expression for the kinetic energy at point 2, I can solve for the force the ground pushes on The Hulk (using mathemagic).
A quick note about the force the ground pushes on The Hulk: Really, this force does zero work. Why? Because work is the dot product of Force and displacement. Since the point this force pushes doesn’t move (the ground), the work is zero. Here is where we cheat. If we pretend The Hulk is just the Point, then I can calculate the work done by this force as though it were applied to the center of mass. But now I have the force the ground exerts on The Hulk and thus the force The Hulk uses to jump. I need to estimate some of these values.
This is a tough one. The Hulk is shown in so many different ways, he can have a huge range of values for his mass. Let me assume that he is the same density as a normal human and a size as shown in theAvengers movie. Here is a nice shot to start with.
If I assume Hawkeye is a normal-sized human (around 1.8 meters tall), then The Hulk would be about 2.5 meters tall. This is pretty much a guess since The Hulk is sort of bent over, but I am going with it. But what about the mass? Suppose a human and The Hulk were cylinders. In this case, could draw the following.
So, here I am assuming that there is some relationship between the height and radius of a person modeled as a cylinder. For a normally proportioned human male, I will use the ratio a1 to relate the height and radius. If I assume a normal human has a mass of 70 kg, then:
If The Hulk were just like a normal human, but bigger, he would have the same value for a1 as a normal human. However, I think for The Hulk, he is a bit bulkier than a human. Let me just say that for The Hulk, his ratio is 1.25a1. This would put his mass at:
So, now I just need to put in my values for the heights and the mass of the human. This puts the mass of The Hulk at 293 kg (that is 645 pounds). BAM! That is huge. The important thing to remember, just because The Hulk is 40 percent taller than a normal human doesn’t mean his mass if 40 percent larger.
While I am talking about mass, there is something that always bothered me. Bruce Banner is a pretty normal-looking human, right? But then he turns into The Hulk (I guess The is his first name since it is always capitalized). So, if he goes from 70 kilograms as a human to almost 300 kg as The Hulk, where does the extra mass come from? What if this is conversion of energy to mass from Einstein’s E = mc2? This would take 2.7 x 1019 Joules of energy. Where does that come from? The total power output from the Sun is about 4 x 1026 Watts. However, only about 1.7 x 1017 Watts hits the Earth. If The Hulk used ALL of this solar energy, it would take over two and a half minutes in order to capture enough energy to “transform.” I guess this could be the “getting angry time.”
But what if The Hulk doesn’t change mass? In this case, he would still be 70 kg, but have a different density. Solving for the density, I get:
Using the same values for heights, this would put The Hulk’s density at 0.24 times the density of a human. If I assume about 1,000 kg/m3 for a human, the Hulk would have a density of 240 kg/m3. Just to compare, this is similar to the density of cork (as listed by Wikipedia). Crazy. But what am I doing here? This isn’t a post about The Hulk’s mass, is it? No. Back to jumping stuff.
First, let me estimate the values for s1 and s2 (from the diagram). I will just go ahead and put the height of the center of mass for The Hulk while standing at 1.4 meters. With a simple test, it seems that my center of mass decreases to about 40 percent for an extreme jump. If the jumping The Hulk takes a similar position, his value for s1 would be about 0.56 meters.
Here is a shot of The Hulk jumping.
Maybe I just got lucky – but it seems I found this location in New York on Google Earth.
From this, I am going to estimate that The Hulk jumped to a height of at least 400 feet (120 meters). I still can’t believe I found that building.
Now, it is just plugging in my values. I get an average force of 4.08 x 105 Newtons. That is how hard The Hulk pushes on the ground and how hard the ground pushes on The Hulk. Yes, this is an average force. But this is also the smallest overall force. If I consider a non-constant force, that means that there will be some part of the jump with a lower force but also some part with a greater force. I want to use the case with the best possible jump with the lowest overall value of the force.
Would He Break the Concrete?
This is really the question I was aiming for. My suspicion is that during a jump, The Hulk would push so hard on the concrete that it would crack. How can I find out if the concrete (or whatever the surface may be) survives? I need to look at the compressive strength. This is the maximum pressure a material can withstand before failing.
In order to calculate the pressure on the ground during this jump, I need an estimate for the size of The Hulk’s feet. They look pretty big to me. Let me just ballpark estimate these based on my own feet. For me, one foot is about 0.25 meters long and about 0.09 meters wide. If I assume that The Hulk is 1.4 times tall (and bigger feet) then he would have a foot length of 0.35 m. But let me add in a biggie factor of 1.25 times (just like I did for his “radius”). This would give him feet that are 0.44 meters by 0.16 meters. The total foot-covered area would then be 0.14 m2.
This would result in an average pressure on the ground of (4.08 x 105 Newtons)/(0.14 m2) = 2.9 x 106N/m2 = 2.9 Mega Pascals (MPa). According to The Engineering ToolBox, concrete has a compressive strength maybe around 10 MPa. Well, I guess he would still crack the road during this type of jump. I suspect that the force versus displacement curve is not constant. If I had to sketch it, it might look like this:
If the peak goes over a force that would produce a larger pressure, the road could crack. Also, I assumed The Hulk jumped flat-footed. I don’t jump that way (but I am not The Hulk).