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

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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).

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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 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.

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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?

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id	bigdeck

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{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} {table:cellspacing=0|cellpadding=8|border=1|frame=void|rules=cols} {tr:valign=top} {td:width=310|bgcolor=#F2F2F2} {live-template:Left Column} {td} {td} {composition-setup}{composition-setup} |!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 [torque (single-axis)].{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 (single-axis)]. The torque must be zero, since the yardstick is not rotating. {latex}

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label	Part A

Part A

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

Solution

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id	sysa

System:

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The Yardstick is a subject to .

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Interactions:

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External forces due to and the two hands.

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Model:

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and .

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Approach:

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Diagrammatic Representation

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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

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	diaga

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Mathematical Representation

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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}

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}\


\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}

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	matha

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	appa

Card

label	Part 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 F_max 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 [torque (single-axis)] 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

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id	sysb

System:

Cloak

id	sysb

The Yardstick is a subject to and .

Toggle Cloak

id	intb

Interactions:

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id	intb

External forces due to and both frictional and normal forces from the two hands.

Toggle Cloak

id	modb

Model:

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and .

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id	appb

Approach:

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Diagrammatic Representation

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We again sketch the situation, this time adding friction.

Image Added

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	diagb
	diagb

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Mathematical Representation

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

{latex} \\ If *a > b* , then the normal force *F{~}2{~}* is greater than the normal force *F{~}1{~}*, 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} {cloak:appb} {card} {card:label=Part C} h2. Part C How can we make the hands meet at some other point than the center of the yardstick? h4. Solution {toggle-cloak:id=sysc} *System:* {cloak:id=sysc} The Yardstick is a [rigid body] subject to [torque (single-axis)] and frictional forces..{cloak} {toggle-cloak:id=intc} *Interactions:* {cloak:id=intc}External forces due to [gravity] and both frictional and normal forces from the two hands. {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} Since the hands slip or stick based upon their distance from the [center of mass], and they therefore end up under that point, the obvious way to make the hands finish at a different location than the center of the stick is to move the center of mass. One can use a stick with a mass that varies with position to achieve this. The easiest way, however, is simply to put a large mass at one end of the yardstick. If you're doing this as a demonstration, you can tape a mass at one end, or stick a ball of clay there. If the mass added is *M* at the extreme end of the stick, then the new cenmter of mass will be at a distance *D* from the end without the mass, where {latex}\begin{large}\[D = \frac{m {\frac{L}{2}}+ML}{m + M} = \left( \frac{L}{2}\right) \left( \frac{m + 2M}{m + M} \right) \]\end{large}{latex} XXXXXX {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} {cloak} {card} {deck} {td} {tr} {table} {live-template:RELATE license}

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.

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	mathb
	mathb

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	appb
	appb

Card

label	Part C

Part C

How can we make the hands meet at some other point than the center of the yardstick?

Solution

Toggle Cloak

id	sysc

System:

Cloak

id	sysc

The Yardstick is a subject to and .

Toggle Cloak

id	intc

Interactions:

Cloak

id	intc

External forces due to and both frictional and normal forces from the two hands.

Toggle Cloak

id	modc

Model:

Cloak

id	modc

and .

Toggle Cloak

id	appc

Approach:

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id	appc

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