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Excerpt
hiddentrue

What happens to a yardstick (or meter stick) supported by two fingers as those fingers are slowly moved toward each other?

Composition Setup
Image Added
Photo Courtesy of Wikimedia Commons
 Original Image by RJFJR
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). 
Image Added
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 friction, 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. 
Image Added
 Image Added
 Image Added
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?
 Deck of Cards
idbigdeck
 Wiki Markup
{excerpt:hidden=true}What happens to a yardstick (or meter stick) supported by two fingers as those fingers are slowly moved toward each other? {excerpt}

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|!800px-Yardstick.jpg!|
|Photo Courtesy of Wikimedia Commons
Original Image by *RJFJR*|


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). 

|!Yardstick 01.PNG!|

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. 

|!Yardstick 02.PNG!
!Yardstick 03.PNG!
!Yardstick 04.PNG!|

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?





{deck:id=bigdeck}
{card:label=Part A}

h2. Part A

What are the forces acting on the Yardstick before the hands start moving?
h4. Solution

{toggle-cloak:id=sysa} *System:* {cloak:id=sysa} The Yardstick is a [rigid body] subject to [single-axis torque].{cloak}

{toggle-cloak:id=inta} *Interactions:* {cloak:id=inta}External forces due to [gravity] and the two hands.{cloak}

{toggle-cloak:id=moda} *Model:*  {cloak:id=moda}[Single-Axis Rotation of a Rigid Body] and [Point Particle Dynamics].{cloak}

{toggle-cloak:id=appa} *Approach:* 

{cloak:id=appa} 

{toggle-cloak:id=diaga} {color:red} *Diagrammatic Representation* {color}

{cloak:id=diaga}
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 *F{~}1{~}* and *F{~}2{~}* of the hands

|!Yardstick Force Diagram 01.PNG!|



{cloak:diaga}

{toggle-cloak:id=matha} {color:red} *Mathematical Representation* {color}

{cloak:id=matha}

We now write the equations of [Newton's 2nd Law|Newton's Second 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. The torque must be zero, since the yardstick is not rotating.

 
{latex}
 Card
labelPart A
Part A
What are the forces acting on the Yardstick before the hands start moving?
Solution
 Toggle Cloak
idsysa
 System: 
 Cloak
idsysa
 The Yardstick is a  subject to .
 Toggle Cloak
idinta
 Interactions: 
 Cloak
idinta
External forces due to  and the two hands.
 Toggle Cloak
idmoda
 Model: 
 Cloak
idmoda
 and .
 Toggle Cloak
idappa
 Approach: 
 Cloak
idappa
 Toggle Cloak
iddiaga
  Diagrammatic Representation 
 Cloak
iddiaga
A Force Diagram of the yardstick looks like this. gravity (interaction) 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
Image Added
 Cloak
diaga
diaga
 Toggle Cloak
idmatha
  Mathematical Representation 
 Cloak
idmatha
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.
 Latex
\begin{large} \[ \vec{\tau} = \sum{\vec{r} \times \vec{F}} = 0\]\end{large}
{

 Latex
}

\\

{latex}\
\begin{large}\[ | \tau | = - a F_{1} + b F_{2} = 0 \]\end{large}
{latex}


Force balance in the vertical direction gives:

{latex}
Force balance in the vertical direction gives:
 Latex
\begin{large}\[ mg = F_{1} + F_{2} \]\end{large}
{latex}


Using 
 
 the 
 
 first 
 
 equation 
 
 to 
 
 eliminate 
 
 one 
 
 of 
 
 the 
 
 normal 
 
 forces 
 
 and 
 
 substituting 
 
 into 
 
 the 
 
 second 
 
 gives, 
 
 after 
 
 a 
 
 little 
 
 algebra:


{
 Latex
}
\begin{large}\[ F_{1} = \frac{b}{a + b} mg \]\end{large}
{latex}

\\

{latex}

 Latex
\begin{large}\[ F_{2} = \frac{a}{a + b} mg \]\end{large}
{latex}


The 
 
 closer 
 
 one 
 
 hand 
 
 is 
 
 to 
 
 the 
 
 center 
 
 of 
 
 mass, 
 
 the 
 
 greater 
 
 the 
 
 normal 
 
 force 
 
 on 
 
 it.



{cloak:matha}
{cloak:appa}
{card}
{card:label=Part B}

h2. 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 *F{~}max{~}* given by

{latex}
 Cloak
matha
matha
 Cloak
appa
appa
 Card
labelPart B
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
 Latex
\begin{large}\[ F_{\rm max} = \mu N \]\end{large}
{latex}

where *N* is the normal force.
 

h4. Solution

{toggle-cloak:id=sysb} *System:* {cloak:id=sysb} The Yardstick is a [rigid body] subject to [single-axis torque] and frictional forces.{cloak}

{toggle-cloak:id=intb} *Interactions:* {cloak:id=intb}External forces due to [gravity] and both frictional and normal forces from the two hands.{cloak}

{toggle-cloak:id=modb} *Model:*  {cloak:id=modb}[Single-Axis Rotation of a Rigid Body] and [Point Particle Dynamics].{cloak}

{toggle-cloak:id=appb} *Approach:*  

{cloak:id=appb}
{toggle-cloak:id=diagb} {color:red} *Diagrammatic Representation* {color}

{cloak:id=diagb}
We again sketch the situation, this time adding frictional forces.

|!Yardstick Force Diagram 02.PNG!|

{cloak:diagb}

{toggle-cloak:id=mathb} {color:red} *Mathematical Representation* {color}

{cloak:id=mathb}

The magnitude of the sum of the horizontal forces is


{latex}
where N is the normal force.
Solution
 Toggle Cloak
idsysb
 System: 
 Cloak
idsysb
 The Yardstick is a  subject to  and .
 Toggle Cloak
idintb
 Interactions: 
 Cloak
idintb
External forces due to  and both frictional and normal forces from the two hands.
 Toggle Cloak
idmodb
 Model: 
 Cloak
idmodb
 and .
 Toggle Cloak
idappb
 Approach: 
 Cloak
idappb
 Toggle Cloak
iddiagb
  Diagrammatic Representation 
 Cloak
iddiagb
We again sketch the situation, this time adding friction.
Image Added
 Cloak
diagb
diagb
 Toggle Cloak
idmathb
  Mathematical Representation 
 Cloak
idmathb
The magnitude of the sum of the horizontal forces is
 Latex
\begin{large} \[ F_{\rm total} = F_{f1} - F_{f2} = \mu F_{1} - \mu F_{2}\]\end{large}
{latex}


Substituting 
 
 from 
 
 Part 
 
 A 
 
 for 
 
 the 
 
 normal 
 
 forces 
 *
F
{~}
1
{~}* 
 and 
 *
F
{~}
2
{~}* gives


{latex}
 gives
 Latex
\begin{large}\[ F_{\rm total} = \mu mg \frac{
a
b - 
b
a}{a + b} \]\end{large}
{latex}




We can then substitute for _N_~rear~ using Newton's 2nd Law for the _y_ direction:

{latex}\begin{large}\[ N_{\rm front} = \frac{mg L_{\rm rear} - ma_{x}h}{L_{\rm rear}+L_{\rm front}} = \mbox{718 N}\]
\end{large}{latex}

{note}Note that _a_~x~ is negative in our coordinate system.{note}

Which means that:

{latex}\begin{large}\[ N_{\rm rear} = mg-N_{\rm front} = \frac{mg L_{\rm front} + ma_{x}h}{L_{\rm rear}+L_{\rm front}} =\mbox{213 N} \]\end{large}{latex}

{cloak:mathb}
{cloak:appb}
{card}
{card:label=Part C}

h2. Part C

What is the largest coefficient of kinetic friction allowed between the front tire and the ground if the rear tire is to remain on the ground during a skid?

h4. Solution

{toggle-cloak:id=sysc} *System:* {cloak:id=sysc} Cyclist plus bicycle are treated as a single [rigid body].{cloak}

{toggle-cloak:id=intc} *Interactions:* {cloak:id=intc}External forces from the earth (gravity) and the ground (normal force and friction).  {cloak}

{toggle-cloak:id=modc} *Model:*  {cloak:id=modc}[Single-Axis Rotation of a Rigid Body] and [Point Particle Dynamics].{cloak}

{toggle-cloak:id=appc} *Approach:* 

{cloak:id=appc}

 The problem is the same as Part B, except that we take the limit as _N_~rear~ approaches zero.  This gives the equations:

{latex}\begin{large} \[ \sum F_{x} = - \mu_{k}N_{\rm front} = ma_{x} \]
\[\sum F_{y} = - mg + N_{\rm front} = 0 \]
\[\sum \tau = N_{\rm front} L_{\rm front} - \mu_{k}N_{\rm front} h  = 0\]\end{large}{latex}

The equation of torque balance is the key here.  If the torques are to sum to zero, we see that we must have:

{latex}\begin{large} \[ \mu_{k} = \frac{L_{\rm front}}{h} = 0.88 \]\end{large}{latex}

This is an upper limit, since a larger force of friction will result in a negative torque, tending to raise the rear wheel.  A lower force of friction will create a positive torque, tending to press the rear wheel against the ground (we should not have neglected the rear normal force if &mu;~k~ < 0.88). 

{tip}A coefficient of friction of 0.88 is high for kinetic friction between tire and road, but reasonable for static friction (realized if the front tire is rolling without slipping).  For this reason, braking hard with the front wheel can be dangerous.{tip}

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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.
 Cloak
mathb
mathb
 Cloak
appb
appb
 Card
labelPart C
Part C
How can we make the hands meet at some other point than the center of the yardstick?
Solution
 Toggle Cloak
idsysc
 System: 
 Cloak
idsysc
 The Yardstick is a  subject to  and .
 Toggle Cloak
idintc
 Interactions: 
 Cloak
idintc
External forces due to  and both frictional and normal forces from the two hands. 
 Toggle Cloak
idmodc
 Model: 
 Cloak
idmodc
 and .
 Toggle Cloak
idappc
 Approach: 
 Cloak
idappc