|
Photo Courtesy of Wikimedia Commons |
Support a yardstick (or meter stick) horizontally atop your open palms, with one hand near each end. (Or you can rest it on two fingers, or the backs of your hands).
|
If you slowly slide your hands towards each other, the yardstick will have to respond in some fashion. Generally it will try to stay in the same place on your hand, restrained by frictional forces, until your hands have moved too far and it is forced to slip. Then it will slip on one side or the other, or sometimes both. Usually it will slip on one side briefly, then on the other, then back to the first, so that when your hands finally meet, they are very nearly in the center of the yardstick.
|
What's interesting is that your hands will end up at the center of the yardstick even if they don't start out at approximately equal distances from the center. If one hand is slightly closer to the center, the stick will slip more against the other hand. You can even start with one hand very nearly at the center, and at the end both hands will end up at the center of the stick.
Why is this? What determines where the stick ends up relative to your hands? And how can you change the outcome so that your hands end up, say 1/3 of the way from one end?
Part A
What are the forces acting on the Yardstick before the hands start moving?
Solution
System:
Interactions:
Model:
Approach:
Diagrammatic Representation
A force diagram of the yardstick looks like this. Gravity pulls downward at the center of mass with force mg (where m is the yardstick's mass) and is resisted by the normal forces F1 and F2 of the hands
|
Mathematical Representation
We now write the equations of Newton's 2nd Law for the center of mass and of torque balance for the system. Since the system is at rest, we can put the axis wherever we choose. The simplest point is at the center of mass , since that removes one term from the expression for torque (single-axis). The torque must be zero, since the yardstick is not rotating.
\begin
[ \vec
= \sum{\vec
\times \vec{F}} = 0]\end
\begin
[ | \tau | = - a F_
+ b F_
= 0 ]\end
Force balance in the vertical direction gives:
\begin
[ mg = F_
+ F_
]\end
Using the first equation to eliminate one of the normal forces and substituting into the second gives, after a little algebra:
\begin
[ F_
= \frac
mg ]\end
\begin
[ F_
= \frac
mg ]\end
The closer one hand is to the center of mass, the greater the normal force on it.
Part B
Now imagine that the hands start moving slowly toward the center. There is frictional force preventing easy sliding of the yardstick on the hands, with coefficient of friction μ , but eventually the stick must slide against the hands. What happens?
The yardstick will try to remain stationary, with the skin on the hands moving against the underlying tissue until it can go no farther. We know that the force of friction will prevent motion until the force exceeds Fmax given by
\begin
[ F_
= \mu N ]\end
where N is the normal force.
Solution
System:
Interactions:
Model:
Approach:
Diagrammatic Representation
We again sketch the situation, this time adding frictional forces.
|
Mathematical Representation
The magnitude of the sum of the horizontal forces is
\begin
[ F_
= F_
- F_
= \mu F_
- \mu F_
]\end
Substituting from Part A for the normal forces F1 and F2 gives
\begin
[ F_
= \mu mg \frac
]\end
If a > b , then the normal force F2 is greater than the normal force F1, and hence the frictional force due to the hand a distance b from the center of mass is greater. Therefore that hand will remain fixed relative to the yardstick and the other hand will slide, bringing it closer to the center of mass of the yardstick.
As it does so, the distance a will change, and the distribution of normal forces will change – that on the left hand will begin to increase, and that on the right hand will decrease by the same amount. The two normal forces will become equal when the new distance between the left hand and the center of mass equals b. At this point, the normal forces will be equal, and so will the frictional forces.
As you continue to press your hands towards each other, the forces will eventually cause the yardstick to slip against your hand. In an ideal world, with the same normal force on each hand and the same coefficient of friction, the forces on both hands will be equal and you would expect both hands to move toward the center. Inevitably, however, all things will not be equal – the coefficient of friction will vary from place to place on the yardstick, for instance, and one side will move first. But, by the same logic as above, the hand closer to the center of mass will have more normal force on it, and hence the frictional force will increase and will eventually stop the hand from moving.
We then have the re-appearance of the initial state, with one hand closer to the center of mass than the other, and consequently having more normal force, so it is the hand farther from the center of mass that now moves until the hands are again equidistant from the center of mass.
The hands and the yardstick proceed in this way, with one hand moving a little bit closer, then the stick stopping and the alternate hand catching up, until the hands meet, at which point the center of mass will be very nearly overhead, to within the range of one of these motions.
dx = 0 ]\end