The Buddy Bear Crawl; or, Finally a Practical Use for That PHYS-151 Problem!

Posted on by donnorum

What is the buddy bear crawl? This:

Buddy Bear Crawl IRL
Or rather, it’s like this, only the person underneath has their rucksack switched to the front, and you have a ruck on, too, so they’re probably grabbing you around the neck.

A wild “classic problem from introductory physics” appears! (Seriously, this was one hell of an AHA! moment for me.)

(Specifically, Young & Friedman 5-38. [Question #2 in the linked pset.])

So, for dragging something or someone along a flat surface at an angle Θ above the horizontal, you have:

F = \frac{\mu _{k}mg }{\cos \theta +\mu _{k} \sin \theta }

Note that the above is for a constant velocity – to minimize the force required, set the derivative to 0:

\frac{\mathrm{d} }{\mathrm{d} \theta }\left ( \frac{\mu _{k}mg}{\cos \theta +\mu _{k}\sin \theta } \right ) = 0\rightarrow \frac{\mathrm{d} }{\mathrm{d} \theta }\left ( \cos \theta +\mu _{k}\sin \theta \right )^{-1} = 0

\frac{\sin \theta -\mu _{k}\cos \theta }{\left ( \cos \theta + \mu _{k}\sin \theta\right )^{2}}=0\rightarrow \sin \theta = \mu _{k} \cos \theta

\tan \theta = \mu _{k}\rightarrow \theta = \arctan \mu _{k}

What’s the coefficient of kinetic friction for dragging a person + rucksack along the ground? Based on the Journal of Strength and Conditioning Research, Volume 27 / Issue 5, May 2013, p. 1175-8, I’ll go with 0.33, giving:

\theta = \arctan 0.33 \approx 18^{\circ }

So, the next time you’ve got to pull someone wearing a rucksack (filled with bricks) while you’re wearing a rucksack (filled with bricks), while crawling like a bear the entire length of the soccer pitch in a public park (at three in the morning, after hydro burpees), remember – eighteen (18) degrees above the horizontal.

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Published by donnorum

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