Center of Mass Experiment

I'm curious to know thw mathematical explanation to the following experiment:

Support the stick by resting each of its ends on a finger. Slowly slide your fingers together until they meet. The fingers will meet under the stick's center of gravity.WHY?

It is very important for me to know a mathematical answer to this experiment, beacuse my final grade depends on this report.

I'll be very grateful for your help. Thank You

The center of gravity is that point on the stick where the downward force of gravity is felt. Suppose in moving your fingers one of them gets a bit closer to the center of gravity. In that case that finger has a shorter lever with which to lift the stick than does the other finger. Therefore that finger has to exert more force on the stick to keep its end from dropping. With a greater force between the finger and the stick there will be more friction at that point and the more distant finger will slip on the stick, closing the difference in distance. In other words, whichever finger is more distant from the center of gravity looses its traction and slips closer. Of course this results in the fingers meeting at the center of gravity.

This information is brought to you by M. Casco Associates, a company dedicated to helping humankind reach the stars through understanding how the universe works. My name is James D. Jones. If I can be of more help, please let me know.

JDJ

Now I understand what is going on in this experiment. But the problem is I have to explain this mathematically, and I have no clue where to start, what to take into consideration and what equationsor formulas to use.

Please help me to do this, I'll be very grateful.

Once again thanks for your reply, and I'll be waiting.

Suppose that the stick has a mass of 1kg. That means its weight in Newtons is 1 times the acceleration due to gravity. That gives us a weight of 9.8 Newtons which must be divided up between the fingers in contact with the stick. The portion of the weight felt on each finger will be 9.8 times the distance from the opposite finger to the center of gravity of the stick, divided by the total distance between the fingers.

Call the weight on the left finger Wl and that on the right finger Wr. Let the distance from the left finger to the center of gravity by Dl and that from the right finger be Dr. The Wl=9.8*Dr/(Dl+Dr) and Wr=9.8*Dl(Dl+Dr).

Let's assume a coefficient of friction between finger and stick of 0.1, meaning that to slide a finger along the stick requires 0.1 times the weight supported by that finger. Then the force necessary for the left finger to slip is Wl*0.1 and for the right finger it is Wr*.01.

Now when we bring our fingers together, the force exerted along the direction of the stick is equal for both fingers, otherwise the stick as a whole would move one way or the other. With equal forces along the stick, that finger with the reduced weight will first reach sufficient force to cause slipping, as soon as the applied force exceeds 0.1 times the weight on the finger. This causes the relative position of the fingers relative to the center of mass to change, transferring the load to the slipping finger. As soon as the weight on the slipping finger brings the friction force there up to greater than the friction for slipping on the other finger, the first finger stops sliding and the opposite finger begins to slip.

The requirement that the frictional force be equal means that

.98*Dr/(Dl+Dr)=.98*Dl/(Dl+Dr).

This reduces to Dl=Dr meaning that the fingers remain equidistant from the center of gravity.

That is pretty much the math story.

Could you please answer the following questions:

(These questions are related to the weight of the stick on the right finger Wr)

Wr = (Wt * dl)/(dl+dr)

1. Why do we ignore the already covered distance by the right finger?

2 When we multiply Wt (total weight) by dl(distance between the other finger and C.G.)why do we use other finger's distance?

3 Then, at the end, why do we divide the numerator(Wt * dl )by (the sum of the distance between fingers)

Thank you

The assumption is that the experiment is performed such that the stick remains level, that is it does not rotate. In that case the sum of all torque on the stick must be zero. Torque is the product of a force times a moment arm times the sine of the angle between the force and the arm. In our experiment the stick is held level so that the angle between the upward forces and the stick itself is 90 degrees so the sine is 1.0.

In our situation gravity acting directly downward at the center of gravity causes a torque about the left finger tending to rotate the right end of the stick downward. The magnitude of that torque is dl*Wt. In order to prevent that rotation the right finger must provide a counter torque about the left finger. The moment arm for that torque is dl+dr, so (dl+dr)*Wr must equal dl*Wt. Therefore Wr=dl*Wt/(dr+dl). Of course we could have chosen to measure torques about the right finger and found the corresponding force required of the left finger.