\begin{problem*}{5.45}
A car rounds a banked curve as in Fig. 5.13.  The radius of curvature
of the road is $R$, the banking angle is $/theta$, and the coefficient
of static friction is $\mu_s$.
\Part{a} Determine the range of speeds the car can have without
slipping up or down the road.
\Part{b} Find the minimum value of $\mu_s$ such that the minimum
speed is zero.
\Part{c} What is the range of speeds possible if $R = 100\U{m}$,
$\theta = 10.0\dg$, and $\mu_s = 0.100$ (slippery conditions).
\end{problem*} % problem 5.45

\begin{solution}
Looking at Fig. 5.13 (text page 137) and adding friction, we see that
the forces on the car are friction $\vect{F}_f$, gravity $\vect{F}_g$,
and a normal force $\vect{F}_N$.  Let the vertical direction be
\jhat\ and the centerward direction to be \ihat, and the direction
centerward-down parallel to the surface of the road by \khat.  Let us
assume at first that $\vect{F}_N$ is in the $-\khat$ direction and at
its maximum possible value of $F_f = \mu_s F_N$.
\begin{align}
 \sum F_\jhat &= F_N\cos\theta - mg + F_f\sin\theta= 0 \\
 F_N (\cos\theta + \mu_s\sin\theta) &= mg \\
 F_N &= \frac{mg}{\cos\theta + \mu_s\sin\theta} \\
 \sum F_\ihat &= F_N\sin\theta - F_f\cos\theta
           = F_N (\sin\theta - \mu_s\cos\theta) \\
	   &= \frac{mg}{\cos\theta + \mu_s\sin\theta} (\sin\theta - \mu_s\cos\theta) \\
	   &= mg \frac{\tan\theta - \mu_s}{1 + \mu_s\tan\theta}
           = m\frac{v^2}{R} \label{eqn.45.Fi}\\
 v &= \sqrt{ Rg\frac{\tan\theta - \mu_s}{1 + \mu_s\tan\theta} } \label{eqn.45.v}
\end{align}

\Part{a}
The work above shows that the minimum speed a car can have while going
around the turn is given by eqn \ref{eqn.45.v}, because that is the
case when friction is maximized in the $-\khat$ direction.  The
maximum speed that the car can have can be found by simply reversing
the sign of the frictional force above (so that $\vect{F}_f$ points in
the $+\khat$ direction), which we achieve by replacing any $\mu_s$s in
eqn \ref{eqn.45.v} with $(-\mu_s)$.  For any speeds between these
$F_f$ will be less than its maximum value of $\mu_s F_N$, and the car
will still not slip.  So
\begin{equation}
 \ans{ \sqrt{ Rg\frac{\tan\theta - \mu_s}{1 + \mu_s\tan\theta} }
       \le v \le
       \sqrt{ Rg\frac{\tan\theta + \mu_s}{1 - \mu_s\tan\theta} } }
\end{equation}

\Part{b}
If the speed is 0, then $\vect{F}_f$ will be in the $-\khat$ direction
(opposing the $+\khat$ portion of $\vect{F}_g$).  Summing the forces
in the \khat direction we have
\begin{align}
 \sum F_\khat &= F_g\sin\theta - F_N
               = mg(\sin\theta - \mu_s\cos\theta)
               = 0 \\
 \mu_s &= \ans{\tan\theta}
\end{align}
Or we could go use eqn \ref{eqn.45.Fi}, our sum of forces in the
\ihat\ direction.
\begin{equation}
 \sum F_\ihat = mg \frac{\tan\theta - \mu_s}{1 + \mu_s\tan\theta}
              = m\frac{v^2}{R} = 0
\end{equation}
And set the numerator to $0$, which gives the same formula for $\mu_s$.

\Part{c}
Plugging into our ans for \Part{a} we have
\begin{align}
 \sqrt{ 100\U{m} \cdot 9.8\U{m/s}^2 \frac{\tan 10.0\dg - 0.100}
                                    {1 + 0.100 \cdot \tan 10.0\dg} }
       &\le v \le
 \sqrt{ 100\U{m} \cdot 9.8\U{m/s}^2 \frac{\tan 10.0\dg + 0.100}
                                    {1 - 0.100 \cdot \tan 10.0\dg} } \\
 \ans{ 8.57\U{m/s\ } } & \ans{ \le v \le 16.6\U{m/s} }
\end{align}
\end{solution}
