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Suppose a cyclist 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.  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.  If the cyclist is decelerating at 0.55_g_, what is the size of the normal force exerted by the ground on each wheel? 

System:  Cyclist plus bicycle are treated as a single [rigid body] subject to external forces from the earth (gravity) and the ground (normal force and friction).  

Model:  [Rotation and Translation of a Rigid Body].

Approach:  We first sketch the situation and construct a coordinate system.

PICTURE

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 [lever 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}

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}