Question

There is a demonstration where the person holds a yardstick on 2 fingers and brings the fingers closer together until they reach the center of mass. I was asked to explain the phenomenon. I think I have the explanation, but I was wondering if the answer is somewhere so I can see whether my hypothesis is correct. Any ideas?

Answer 1

Why don't you post your theory? :)

I'm thinking we can explain it by looking at all the physical properties that matter in that situation. Sound good?

Like, gravity is always present. That's why the yardstick is pushed onto the fingers.

There are positive coefficients of friction (static and dynamic/kinetic) between the yardstick and fingers, which - combined with the gravitational force putting a normal force on your fingers that are not accelerating with respect to the yardstick - is why it won't slide at first and will resist movement.

When your fingers do slide, it's because your fingers have a force exerted on them that is greater than the coefficient of static friction multiplied by the normal forces on your fingers.

Now, the trick to performing this demonstration is that the static and dynamic/kinetic friction forces remain the same for both fingers. This way, both fingers can exceed static friction to move at once. That is the most general description of what must happen. The more complex part is how to make sure that that happens.

The following paragraphs assumes that the yardstick is flat and that the demonstration uses the yardstick horizontally.

Basically, if your fingers have the same normal force and coefficients of friction, then they have the same friction force, and so they will cancel so that the ruler has no net force and does not move.

So, when you want to do this easily, you just try to get your fingers to have the same coefficients of friction with the ruler at all parts symmetrically. Best way to due this is to make sure that the yardstick is clean, and your fingers are equally dry/wet. Putting a substance on your fingers might help, such as chalk, so that the coefficients of frictions are more constant (not getting more wet or dry, like fingers do). Also, make sure that the normal force is the same. Use an evenly weighted ruler and slide fingers symmetrically about the center of mass. This will make sure that the fingers are supporting an equal portion of the yardstick's weight, so normal forces are equal.

Any questions? Or your theory? And I think hypotheses are more for assuming a statements to be true (or false maybe?) to test them. :)

Answer 2

Simply put, my theory was that one side will move when the force exerted overcame the resistive force of friction. That finger will then slide until there is enough ruler on the other side to exert a certain leverage, causing the other end to lift up, thereby increasing friction on the first finger and decreasing it on the other finger, allowing the other finger to slide. This pattern continues, so that the ruler is kind of rocking back and forth until there is equal leverage on both sides, which means the fingers are at the center of gravity. In the demonstration, one finger moves, then the other. They do not slide at the same time. @theEric Thanks for your ideas!

Answer 3

That sounds good! :) I'd have to see the demonstration to know for sure, but I don't think the ruler has to move at all!

I'm really hoping someone else can shed some insight and either disprove or validate what I say. I'm not completely sure myself!

The stationary finger will hold the yardstick in place with its static friction negating the moving finger's kinetic friction. As the finger positions move, they hold different amounts of the yardstick's weight, so the normal force on each finger changes and so the friction force on each finger changes.

As the moving finger approaches the center of mass, it supports more of the weight. (I'll show you why soon!) That means its kinetic friction increases.

The stationary finger will support less weight as the other moves, so it's maximum static friction is lessened.

When the kinetic friction of the moving finger exceed the static friction of the stationary finger, the stationary finger will begin to move, and will then have the lesser kinetic friction.

If the other finger does not stop, then the yardstick will have a net force and will move the yardstick. But the demonstrator can stop their finger. Otherwise, the demonstrator would have to wait until the yardstick is accelerated to the velocity of the demonstrator's higher-friction finger until it is at rest. \(a=\dfrac{F_\text{net}}{m}\). The less-friction moving finger will be supporting more weight and so its friction force on the stick will increase while the greater-friction finger's friction will decrease. All the while, the less-friction finger's friction will oppose the motion of the yardstick by reducing \(F_\text{net}\). I think that is what happens, at least.

\(\huge\sf Why\ the\ weight\ on \ each\ changes.\)


We can look at the torques involved. If there is no rotation, then the torques add up to be \(0\). The friction forces are nearly parallel to the fulcrum, so they are causing only negligible torque (you need force with a component that is perpendicular to the fulcrum to cause torque, and there is very little of that).
In the next post, I'll alter the drawing I posted a little bit ago.