## Answer: Boa Constrictor New Problem: “But Who’s Counting?”

Answer: Boa Constrictor
Here’s the question.
Consider a differential element of length along the boa constrictor. It’s being pulled by a tension $T$ on the left and a tensions $T + dT$ on the right. These pull in different directions by an angle $d \theta$. Then the maximum force friction can exert on this differential element is $dF_f = \mu T d \theta$. This is also the amount that $T$ can increase over the differential element, so $dT = \mu T d \theta$. This gives $T(\theta) = T_{joe} e^{\mu \theta}$. So if Debbie is strong enough to pull the snake, all the snake has to do is wrap around a little bit further, because how strong Debbie has to be to pull the snake grows exponentially with the angle it’s wrapped.

New Problem: But Who’s Counting?
What’s the best strategy for the following game?

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### 4 Responses to “Answer: Boa Constrictor New Problem: “But Who’s Counting?””

1. Nik Says:

Get anything above 6 at any point, slot it into the ten thousand’s place.

2. Answer: But Who’s Counting? « Arcsecond Says:

[…] Arcsecond The Humblest Ever Blog About Everything « Answer: Boa Constrictor New Problem: “But Who’s Counting?” […]

3. nc Says:

for the boa constrictor problem, does theta = 0 work as a solution?

4. meichenl Says:

nc,

Yes, theta = 0 does work. In that case, you get T_debbie = T_joe, which is exactly what you expect if the angle is zero – meaning the snake’s not wrapped around you at all and they’re just playing tug of war with it.

In fact, theta<0 also works. This describes the situation where the snake is wrapped around you by the same angle theta, but Joe is pulling harder than Debbie, so the force of friction points to opposite way.