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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).
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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.
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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
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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.
\begin
[ \vec
= \sum{\vec
\times \vec{F}}]\end
\begin
[ | \tau | = - a F_
+ b F_
]\end
Force balance in the vertical direction gives:
\begin
[ mg = F_{1}} + F_
]\end
Part B
Suppose the cyclist described in Part A is coasting along at a constant speed when suddenly a car pulls out in front of them. The cyclist hits the brakes and locks both wheels (they stop rotating and begin to skid). The bike skids straight forward. If the cyclist and bike are decelerating at 0.55 g, what is the size of the normal force exerted by the ground on each wheel?
Solution
System:
Interactions:
Model:
Approach:
Diagrammatic Representation
We again sketch the situation and construct a coordinate system.
Unable to render embedded object: File (bikepartb.jpg) not found.
Mathematical Representation
We now write the equations of Newton's 2nd Law for the center of mass and of torque balance about the center of mass for the bicycle.
The bicycle's center of mass is accelerating linearly in the negative x direction, but the bicycle is not rotating about its center of mass. Thus the torques about the center of mass must balance.
Because the bicycle is accelerating, the only point moving with the bike that can be chosen as an axis of rotation is the center of mass. Try repeating the problem with another axis (such as the point of contact of the front wheel with the ground) and you should find that the torques do not balance).
\begin
[ \sum F_
]
[\sum F_
= - mg + N_
+ N_
= 0 ]
[\sum \tau = N_
L_
- N_
- F_{f,{\rm front}} h - F_{f,{\rm rear}}h = 0]\end
We have only three equations and four unknowns, but because the friction forces have the same moment arm about the center of mass, we can substitute for their sum. Thus, using torque balance, we can find:
\begin
[ N_
= \frac{N_
- ma_
h}{L_{\rm front}}]\end
We can then substitute for Nrear using Newton's 2nd Law for the y direction:
\begin
[ N_
= \frac{mg L_
- ma_
Unknown macro: {x}h}{L_
+L_{\rm front}} = \mbox
]
\end
Note that ax is negative in our coordinate system.
Which means that:
\begin
[ N_
= mg-N_
+ ma_
h}{L_
+L_{\rm front}} =\mbox
]\end