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Excerpt

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Determine how the weight of a bicycle plus rider is divided between the wheels in various circumstances.

Composition Setup

Deck of Cards

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Card

label	Part A

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

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id	sysa

System:

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Cyclist plus bicycle are treated as a single .

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id	inta

Interactions:

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External forces from the earth (gravity) and the ground (normal force and friction).

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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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We first sketch the situation and construct a coordinate system.

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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. 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}

Note
Newton's Law for the x direction is trivial (0 = 0), so we have ignored it.

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}

Force balance in the y direction then gives:

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

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Card

label	Part B

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

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id	sysb

System:

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id	sysb

Cyclist plus bicycle are treated as a single .

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id	intb

Interactions:

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id	intb

External forces from the earth (gravity) and the ground (normal force and friction).

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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 and construct a coordinate system.

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	diagb

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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 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.

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).

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}

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}

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}

Note
Note that a_x is negative in our coordinate system.

Which means that:

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\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}

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Card

label	Part C

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

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id	sysc

System:

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Cyclist plus bicycle are treated as a single .

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id	intc

Interactions:

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id	intc

External forces from the earth (gravity) and the ground (normal force and friction).

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Model:

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and .

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Approach:

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Versions Compared

Old Version 36

New Version 37

Key

Part A

Solution

Part B

Solution

Part C

Solution