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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}

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|!bike1800px-Yardstick.pngjpg!|
|Photo Courtesy of Wikimedia Commons
Original Image by *RJFJR*|
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{card:label=Part A}

h2. Part A

A cyclist sits on their bike at rest, supported only by the two wheels of the bike.  The combined mass of the bike plus cyclist is 95 kg.  The center of mass of the bike plus cyclist is 0.75 m above the ground, 0.42 m forward of the center of the rear wheel and 0.66 m behind the center of the front wheel.  What is the size of the normal force exerted by the ground on each wheel?

h4. Solution

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

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

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

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We first sketch the situation and construct a coordinate system.

!bikeparta.jpg!

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

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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.  We select the point of contact of the rear wheel with the ground to remove an unknown from the torque equation.  

{latex}\begin{large} \[ \sum F_{y} = - mg + N_{\rm front} + N_{\rm rear} = 0 \]
\[\sum \tau = N_{\rm front} (L_{\rm front}+L_{\rm rear}) - mg L_{\rm rear}  = 0\]\end{large}{latex}

{note}Newton's Law for the _x_ direction is trivial (0 = 0), so we have ignored it.{note}

The torque equation immediately gives:

{latex}\begin{large}\[ N_{\rm front} = \frac{mgL_{\rm rear}}{L_{\rm front}+L_{\rm rear}} = \mbox{362 N}\]\end{large}{latex}


Force balance in the _y_ direction then gives:

{latex}\begin{large}\[ N_{\rm rear} = mg - N_{\rm front} = \frac{mgL_{\rm front}}{L_{\rm front}+L_{\rm rear}} = \mbox{569 N}\]\end{large}{latex}

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{card:label=Part B}

h2. 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? 

h4. Solution

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

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

!bikepartb.jpg!

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

{cloak:id=mathb}

We now write the equations of [Newton's 2nd Law|Newton's Second Law] for the center of mass and of torque balance about the center of mass for the bicycle.  

{note}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.{note}

{warning}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).{warning}

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

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:

{latex}\begin{large}\[ N_{\rm front} = \frac{N_{\rm rear}L_{\rm rear} - ma_{x}h}{L_{\rm front}}\]\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}

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{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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