# Fingering the center of gravity.

You can even tape a weight on one end, or at any other point on the stick, and it still works. A two-dimensional version: Use a large dinner plate, with objects on it, perhaps even a glass of water. Use three fingers spread wide to support it. Bring the fingers together! (Your fingertips should be dry.) Don't use expensive china. Practce over a bed.

So why does this work?

A mathematical analysis of this can be found in Arnold Sommerfeld's Lectures on Theoretical Physics, Volume 1, Mechanics. Academic Press, 1964, pp. 83-85. It's a good exercise in forces, torques, and friction. But you would rather puzzle this out yourself before you consult that reference.

The secret is our old friend friction. Specifically it's the fact that the coefficient of kinetic (sliding) friction is usually smaller than the coefficient of static friction. The coefficient of kinetic friction is usually a constant, independent of velocity. The coefficient of static friction may also be treated as constant. Of course, both coefficients are dependent on the nature of the contact surfaces.

fs ≤ μsFn.
fk = μkFn.

fk is the force due to kinetic friction.
fs is the force due to static friction.
Fn is the normal force, perpendicular to the sliding surfaces.

Your fingers apply upward (normal) forces on the stick. Initially these forces are very unlikely to be exactly equal, for they aren't equidistant from the center of gravity of the stick. The finger with the greatest load will also have the greater force due to sliding friction. So when your fingers move toward each other, the one with the least load slides against the stick and the other doesn't slide. But as the sliding finger moves closer to the center of gravity it supports a greater fraction of the load so the size of the sliding friction increaes until it is equal to the force of static friction on the other finger. Then that other finger begins to slide, and immediately the friction acting on it decreases to a lower value. But it, too, increases as it moves closer to the center of graviy of the stick, and the process repeats, alternating between stationary for each finger. When one slides, the other one doesn't.

Here's the details.

Taking torques about finger A we find A = (b/(a+b))W.

Taking torques about finger B we find B = (a/(a+b))W.

Let A > B initially. The finger at B slides along the stick.

B moves until a = b, but continues moving until b1 ≤ a.

In general, the friction equations are fB,k = μka(W/(a+b)) and fA,s = μsb(W/(a+b)).

For b = b1 these are equal.

So, then, μka = μsb1

and a/b1 = μsk ≥ 1

and the stick moves at finger A.

The force due to friction fA,s drops to FA,k ≤ FA,s.

So when finger B is at b1, the force due to sliding friction at B exceeds the force due to static friction at A.

B comes to rest and FB,k changes to FB,s.

This process repeats at each turning point. A and B approach in geometric progression the center of mass of the stick, for which a = b = 0, since the quotient μks occurs each time. In the final state the stick balance is an equilibrium over the juxtaposed fingers.

Follow-up question: Suppose you put a glove on one hand, so that the coefficients of friction are significantly different for the two fingers used. How will the outcome be different?

• Donald Simanek, 2017.