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?

Solution

System:

Interactions:

Model:

Approach:

Diagrammatic Representation

We first sketch the situation and construct a coordinate system.

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

\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}

The torque equation immediately gives:

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

Force balance in the y direction then gives:

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

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.

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.

\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}

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

We can then substitute for N_rear using Newton's 2nd Law for the y direction:

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

Which means that:

\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}

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?

Solution

System:

Interactions:

Model:

Approach:

 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}