Recently, I've encountered a really curious thing, namely:

Support in the air a straight stick of length, say, 1 meter, using your index fingers in such a way that the left one is e.g. 30cm from the center of mass and the right one is e.g. 20cm. Now start slowly moving your fingers to each other.

I noticed that the fingers always meet at the point where the center of mass is. What is the physics explanation for this?

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    $\begingroup$ That's actually a quite famous simple experiment. $\endgroup$ – valerio Sep 24 '16 at 22:08
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    $\begingroup$ First of 5 fun physics phenomena in Youtube video by Veritasium. $\endgroup$ – Qmechanic Sep 24 '16 at 22:12
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    $\begingroup$ Oddly it's one that I don't recall ever seeing or hearing about before. The world of physics is so broad and detailed that you can become reasonably 'expert' and still be missing some simple corners. $\endgroup$ – dmckee --- ex-moderator kitten Sep 24 '16 at 23:12
  • $\begingroup$ Now try starting both fingers at the center of mass and moving them outwards simultaneously $\endgroup$ – BlueRaja - Danny Pflughoeft Sep 25 '16 at 5:04
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    $\begingroup$ @immibis: This is a physical phenomenon, not a psychological one. You could build a robot to do it, and find the same result. $\endgroup$ – BlueRaja - Danny Pflughoeft Sep 25 '16 at 10:02

The explanation is the interplay between the friction forces and the distribution of the weight of the ruler as the fingers move along it.

The two fingers do not share the weight $W$ of the ruler equally. The one which is closer to the centre of mass (CM) bears the greater share. If they are distances $x$ and $y$ from the CM, then by balancing moments we find that they bear weights $L$ and $R$ such that $L+R=W$ and $xL=yR$.

The finger which bears the greater weight $N$ (ie that nearer the CM) has the greater (static) friction force $F=\mu_s N$ with the ruler, where $\mu_s$ is the coefficient of static friction. So if you push your fingers together just enough to make one slide, the ruler will slide over the one bearing less weight - ie the finger further from the CM. This brings the CM closer to that finger, which increases the weight that it bears and the friction force on it, and reduces the forces on the other finger.

When your fingers reach the same distance from the CM they bear the same weight. In ideal conditions they will then both slide towards the CM at the same time. In practice the fingers slide alternately, first one then the other, until they meet at the CM.

This alternation could happen because the coefficient of kinetic friction $\mu_k$ (on the sliding finger) is slightly less than the coefficient of static friction $\mu_s$ (on the stationary finger). The sliding finger then "overshoots" the static friction balance-point ($\mu_s L=\mu_s R$) and continues until the kinetic friction force on this finger becomes slightly greater than the static friction force on the stationary finger ($\mu_sL=\mu_kR$).

Mathematically, suppose we start with $x<y$. Then $L>R$ so the right finger slides toward the centre, increasing $R$ and decreasing $L$. It stops where $\mu_s L=\mu_k R$. The new position $y'$ also satisfies $xL=y'R$ so $y'=\frac{\mu_k}{\mu_s}x$. The ruler then slides over the left finger, and the process repeats to new positions $x'=\frac{\mu_k}{\mu_s}y'=(\frac{\mu_k}{\mu_s})^2 x$ etc. So the successive "stopping" distances of each finger from the CM get smaller in the ratio $(\frac{\mu_k}{\mu_s})^2$.

Another explanation of the alternating motion of the fingers is the "see-sawing" action of the rod as it rotates about one "carrying" finger and then the other. (Thanks to alephzero for pointing this out in the comment below.) Such motion is more likely to happen if the fingers move relatively quickly and you do not keep them perfectly horizontal, so that the stick starts rocking. This effect could also be described mathematically, but might not be worthwhile because it relies too much on unknown human factors.

The link Friction Demo with a Meterstick provided by valiero92 claims that the 'trick' depends on a uniform coefficient of friction.

I don't think this claim is true. Provided that the coefficient of friction at the CM is non-zero, when one finger reaches the CM it will bear all of the weight of the stick. This finger will stop sliding and will become the "carrying" finger. The friction force on the other finger will become zero regardless of how much larger the coefficient of friction is at that finger.

Possible complications :

(i) The coefficient of friction is slightly different for each finger, eg because one is more greasy or more sweaty than the other. This includes the case mentioned by alephzero in his comment below, that the sliding finger hits a sticky part of the rod.

If this is simply a section where the coefficient of friction is suddenly much higher, and the Law of Friction $F=\mu N$ still applies, then my explanation above regarding the Friction Demo holds. The 'trick' still works, but the distance which each finger slides will be different.

However, if the Law of Friction does not apply to this 'glue,' and it is able to provide a maximum horizontal force $G$, then the sliding finger will pass the CM if $G>\mu_k W$. The 'trick' will then fail.

(ii) The speed of the moving finger is high compared to the rotational speed of the rod about the "carrying" finger. If the sliding finger does not keep perfectly level with the carrying finger, it takes a finite time for the rod to tilt and redistribute the weight. This finger might overshoot the CM before the friction force on that finger has reached its maximum value. The rod then overbalances and falls.

This complication is more likely to occur when both fingers are close to the CM. Torque $\tau$ is proportional to distance of the pivoting finger from the CM, whereas moment of inertia $J$ varies from $\frac13ML^2$ at one end to $\frac{1}{12}ML^2$ at the centre, so the angular acceleration $\alpha=\tau/J$ decreases as the pivot reaches the centre. If the vertical distance between the fingers is constant, then the angle which the rod must turn through increases as they get closer together.

(iii) Excessive force might be applied so that both fingers slide against the rod at the same time - as suggested by mike30 in his comment below. This will happen if the applied forces are $>\mu_s L$ and $>\mu_s R$ respectively. In order for the rod to stop on one finger before it reaches the CM, the applied forces must each be less than $\mu_k W$.

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    $\begingroup$ If there is a patch of sticky glue on the rod, corresponding to a high coefficient of friction over a small region, a finger of one hand will get "stuck" in the glue and the other hand can move past the center of mass. Also, the rod also has inertia, so in real life this a dynamics problem, not a pseudo-statics problem - nothing physically impossible happens when one finger move past the center of mass and all the weight of the rod is taken on one finger. The rod then has a finite rotary acceleration around one finger, as it falls off your hands. $\endgroup$ – alephzero Sep 25 '16 at 0:08
  • $\begingroup$ I think the 'glue' complication cannot be related to the usual law of friction $F=\mu N$, because the normal reaction $N$ at this finger will become zero when the other finger reaches the CM. I agree about the dynamical problem, and have noted it in my answer. $\endgroup$ – sammy gerbil Sep 25 '16 at 0:23
  • $\begingroup$ And what if we slide each finger at different speed? $\endgroup$ – mike30 Sep 26 '16 at 16:26
  • $\begingroup$ @mike30 : Do you mean the fingers move at different times? Or at the same time? The former does not make any difference. The latter possibility raises an interesting point. I will update my answer to address that. $\endgroup$ – sammy gerbil Sep 26 '16 at 17:48
  • $\begingroup$ At the same time, but at different speed. $\endgroup$ – mike30 Sep 26 '16 at 17:51

That's cute.

Here's the thing: there is friction between each finger and the ruler, and those are the only horizontal forces acting on the ruler. So by adding them up we can find the horizontal acceleration of the ruler.


The friction at each finger is proportional to the normal force between that finger and the ruler, and when the fingers are asymmetrically positioned, the the normal force of each is not equal (we're talking static equilibrium here). Which means that (when the fingers are asymmetrically positioned and moving) the horizontal force is not zero, so the ruler wants to accelerate.

In the usual physical case (where $\mu_k < \mu_s$) the motion of the ruler will be dominated by static friction, and will alternate back and forth in a series of motions governed by a geometric series.

The analysis will be smoother in the $\mu_k = \mu_s$ approximation where the forces vary continuously, and I suspect we will get some kind of highly damped oscillator.

This is a very interesting problem in the sense that the physics is all stuff that appears in the first semester of instructions (friction (both static and kinetic), relative motion, and static equilibrium) so it is very accessible, but at the same time it contains subtleties enough to requires some integrated thinking to put it all together.

I may work this out in some detail and report back.


The stick is supported on your fingers and they share the weight by a ratio proportional to revers of their distance from CG of stick.

When you start to move your fingers together the stick will feel the friction forces proportional to the load on your fingers so the finger that is near to CG will carry the stick by first letting its skin roll back till the torque stress in your figure is greater than the force need to slide the stick over the other finger, and the finger farther from CG will skid under the stick close to CG.

Meanwhile the skin on your active finger having overcome the initial jerk off force of friction now unwinds and overshoots the pushing of stick over your passive finger causing the stick to lean down over passive finger and exert extra force of friction by dynamic rotation and the fact that now the passive finger is nearer to CG.

Therefore the Passive finger becomes the active finger and its skin rotates around the bone and is loaded with torque energy. The direction of action reverses and passive figure now turned to active will carry the stick on its top letting it skid on the other finger.

This seesaw action will repeat till the two fingers meet under the CG.
In real life because we balance our fingers by sensory motor control in our brain our fingers and brain will introduce small delays and miscalculations, so this test can be a measure of our auto reflex acuity as well.


All the previous answers attempt a physics answer, but I think mostly they are greatly underestimating the power of the human brain. The fingers meet at the center of mass because of the iterative feedback loop and automatic adjustment your brain makes to keep the stick balanced. Try the same experiment with a robot instead, the physics alone cannot account for the auto-correcting PID-type process that happens in your nervous system as your fingers move together.

  • $\begingroup$ Can you point to some evidence that the same experiment with a robot does not give the same result? $\endgroup$ – user300 Sep 25 '16 at 20:27
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    $\begingroup$ The human brain and it's sense organs are even complex enough to fool themselves into believing they're controlling something unconsciously when they're not. Have a read of the up-voted answers for some context ;) $\endgroup$ – Lamar Latrell Sep 25 '16 at 20:57
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    $\begingroup$ You don't need a robot. Do it blind-folded. $\endgroup$ – Bill N Sep 26 '16 at 20:06
  • $\begingroup$ Another simple demonstration that this answer is false: let one of the supports be a fixed point. Move your finger (the second support) slowly to avoid inertial effects. Without "controlling" the fixed support you will get the same result. Or make the second support a little car (so all you control is the horizontal velocity, not the vertical force). Still the same thing happens. $\endgroup$ – Floris Jan 24 '17 at 15:23

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