Part A
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 for the center of mass and of torque balance about the center of mass for the bicycle.
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.
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).
\begin
[ \sum F_
= - F_{f,{\rm front}} - F_{f,{\rm rear}} = ma_
]
[\sum F_
= - mg + N_
+ N_
= 0 ]
[\sum \tau = N_
L_
- N_
L_
- F_{f,{\rm front}} h - F_{f,{\rm rear}}h = 0]\end
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:
\begin
[ N_
= \frac{N_
L_
- ma_
h}{L_{\rm front}}]\end
We can then substitute for Nrear using Newton's 2nd Law for the y direction:
\begin
[ N_
= \frac{mg L_
- ma_
Unknown macro: {x}h}{L_
+L_{\rm front}} = \mbox
]
\end
Note that ax is negative in our coordinate system.
Which means that:
\begin
[ N_
= mg-N_
= \frac{mg L_
+ ma_
h}{L_
+L_{\rm front}} =\mbox
]\end
Part B
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?
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: The problem is the same as part A, except that we take the limit as Nrear approaches zero. This gives the equations:
\begin
[ \sum F_
= - \mu_
N_
= ma_
]
[\sum F_
= - mg + N_
= 0 ]
[\sum \tau = N_
L_
- \mu_
Unknown macro: {k}N_
h = 0]\end
The equation of torque balance is the key here. If the torques are to sum to zero, we see that we must have:
\begin
[ \mu_
= \frac{L_{\rm front}}
= 0.88 ]\end
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 μk < 0.88).
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.