I love Fox Trot, as I suspect many physicists do. Bill Amend was a physics major in college and many of the Fox Trot strips have science themes and humor built in them.

When I saw the above strip recently, some questions came to mind:

What volume of tube is needed to provide the depicted motion in the strip?

If the estimated volume is not realistic, then what would the motion look like for a realistic volume for the tube?

Here's how I'm going to approach answering these questions: I'm going to model Jason's jump as a body in free-fall with an acceleration less than \(9.8 m/s^2\), and I'll determine the acceleration by estimating the buoyant force provided by the tube.

The first thing I want to do is draw a free body diagram. Consider a spherical Jason:

*Up*. Note he points out that the buoyant force doesn't depend on whether the balloons are in the house or tied to the house by strings. Similarly, in this comic the buoyant force applies with the tube wrapped around Jason: it still points up.

So let's look at what we have (yes, the forces are vectors, but it's in one dimension and the vector notation isn't looking so great here right now):

$$ W = M g$$

$$ B = \rho_{air} V g$$

where \(V\) is the volume of the tube. Now let's apply Newton's second law:

$$ F_{net} = M a = B - W = \rho_{air} V g - M g$$

Okay, now that I have a free body diagram and the application of Newton's second law let me think for a second about what I wanted to know.

I wanted to know the volume of the tube required to produce the depicted motion. So, I need to estimate something reasonable for what motion is being depicted. I'm going to make some assumptions:

- The height of the springboard is 3 meters.
- Based on that height, I'm estimating that Jason falls 2.0 meters from the board. (Two sig figs? No, not really, but I want two digits in my solution.)
- I'll neglect the small upward jump at the beginning. It looks small and maybe he is simply pulling his legs up which makes the upward part of the trajectory more obvious. With this assumption the initial velocity will be zero.
- I'll estimate Jason's mass to be 30 kg. (I don't need it now, but I'll put it here with all my other assumptions.) I'm also assuming the tube's mass is negligible.
- The air density is \(1.2 kg/m^3\) based on the temperature of a hot summer day.
- The important question is what is the time interval being depicted. I'm going to take a cue from the artist and guess that since there are seven panels depicted after the jump, the time interval is roughly seven times longer than it would have been without the jump. Is that reasonable? I don't know, but it's my interpretation of the comic.

$$\Delta y = {1 \over 2} g t^2$$

Since I assumed that the time interval is going to be 7 times longer than without the tube, I should find what the time to fall 2 meters is without the tube. In this case \(g= 9.8\: m/s^2\):

$$t = \sqrt{ 2 \Delta y \over g} = \sqrt{4 m \over 9.8 m/s^2}= 0.64\: s$$

So the length of the fall with the tube would be \(7 \times 0.64 s = 4.5\: s\). With that time, I want to solve for the acceleration with the tube:

$$a = 2\Delta y \over t^2 = 4.0 m \over (4.5 s)^2 = 0.2\: m/s^2$$

That's a pretty low acceleration. Cool, now I can solve for the volume of the tube:

$${a \over g}M + M = \rho_{air} V$$

Solving for \(V\):

$$V = 26\: m^3$$

Hmm...is that big? Yes it is. I have one of the jumbo sized hoberman spheres that I bring in to class to show what one cubic meter is. There is no what the tube has that large of a volume. Okay, then what would the acceleration (and the time of the fall) be if the tube has a realistic volume. Let's estimate the volume to be around \(1.0\: m^3\). So that means the acceleration Jason will experience is:

$$a = ({\rho_{air} V - M \over M}) g$$

which for our estimated tube volume gives an acceleration of \( 0.96\: g\ = 9.4\: m/s^2\). The time to fall 2 meters at this acceleration is 0.65 seconds. (Now you see why I wanted 2 sig figs?)

Note that the change in acceleration is linear with changing tube volume, but that the time for falling a given distance is NOT linear. I'd go into that more, but this has already taken me longer to write than I was intending.

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