Chapter 12.1, Problem 12.52P

A curve in a speed track has a radius of 1000 ft and a rated speed of 120 mi/h. (See Sample Prob. 12.7 for the definition of rated speed.) Knowing that a racing car starts skidding on the curve when traveling at a speed of 180 mi/h, determine (a) the banking angle θ, (b) the coefficient of static friction between the tires and the track under the prevailing conditions, (c) the minimum speed at which the same car could negotiate the curve.

Fig. P12.52

To determine

Find the banking angle $\theta$.

Answer to Problem 12.52P

The banking angle $\theta$ is $\underset{\_}{43.9°}$.

Explanation of Solution

Given information:

The radius $\left(\rho \right)$ of the track is 1,000 ft.

The rated speed $\left(v_R\right)$ is 120 mi/h.

The frictional force $\left(F_0\right)$ at rated speed is 0.

The speed (v) at prevailing condition is 180 mi/h.

Calculation:

Write the general equation weight (W).

$W=mg$

Here, m is the mass and g is the acceleration due to gravity.

Write the general equation of acceleration (a) in curved path.

$a=\frac{v^2}{\rho }$

Here, v is the speed and $\rho$ is the radius of curved path.

Sketch the free body diagram and kinetic diagram of the racing car as shown in Figure (1).

Refer Figure (1):

Consider the racing car moves at rated speed.

Find the banking angle $\theta$.

Apply Newton’s law of equation along x-axis.

$\begin{array}{l}\Sigma F_x=m{a}_x\\ F_0+W\sin \theta =ma\cos \theta \\ F_0=ma\cos \theta -W\sin \theta \end{array}$

Substitute 0 for $F_0$, $\frac{v_R^2}{\rho }$ for a, and mg for W.

$\begin{array}{l}0=m\left(\frac{v_R^2}{\rho }\right)\cos \theta -\left(mg\right)\sin \theta \\ mg\sin \theta =\frac{m{v}_R^2}{\rho }\cos \theta \\ \frac{\sin \theta }{\cos \theta }=\frac{v_R^2}{\rho g}\\ \tan \theta =\frac{v_R^2}{\rho g}\end{array}$

Substitute 120 mi/h for $v_R$, $32.2{\text{ft/s}}^2$ for g, and 1,000 ft for $\rho$.

$\begin{array}{l}\tan \theta =\frac{{\left(120\text{mi/h}\times \frac{5,280\text{ft}}{1\text{mi}}\times \frac{1\text{h}}{3,600\text{s}}\right)}^2}{\left(1,000\right)\left(32.2\right)}\\ \tan \theta =0.96199\\ \theta =43.9°\end{array}$

Thus, the banking angle $\theta$.is $\underset{\_}{43.9°}$.

To determine

Find the coefficient of static friction between the tires and the track under the prevailing conditions.

Answer to Problem 12.52P

The coefficient of static friction between the tires and the track under the prevailing conditions is $\underset{\_}{0.390}$.

Explanation of Solution

Calculation:

Refer Figure (1):

Consider the racing car moves in prevailing condition.

Apply Newton’s law of equation along y-axis.

$\begin{array}{l}\Sigma F_x=m{a}_x\\ F+W\sin \theta =ma\cos \theta \\ F=ma\cos \theta -W\sin \theta \end{array}$

Substitute $\frac{v^2}{\rho }$ for a and mg for W.

$F=m\left(\frac{v^2}{\rho }\right)\cos \theta -\left(mg\right)\sin \theta$ (1)

Apply Newton’s law of equation along y-axis.

$\begin{array}{l}\Sigma F_y=m{a}_y\\ N-W\cos \theta =ma\sin \theta \\ N=ma\sin \theta +W\cos \theta \end{array}$

Substitute $\frac{v^2}{\rho }$ for a and mg for W.

$N=m\left(\frac{v^2}{\rho }\right)\sin \theta +\left(mg\right)\cos \theta$ (2)

Find the coefficient of static friction $\left({\mu }_s\right)$ between the tires and the track under the prevailing conditions.

Write the general equation of normal force(N).

$\begin{array}{l}F={\mu }_{s}N\\ {\mu }_s=\frac{F}{N}\end{array}$ (3)

Substitute Equation (1) and (2) in Equation (3).

$\begin{array}{c}{\mu }_s=\frac{m\left(\frac{v^2}{\rho }\right)\cos \theta -\left(mg\right)\sin \theta }{m\left(\frac{v^2}{\rho }\right)\sin \theta +\left(mg\right)\cos \theta }\\ =\frac{m\left(v^2\cos \theta -\rho g\sin \theta \right)}{m\left(v^2\sin \theta +\rho g\cos \theta \right)}\\ =\frac{v^2\cos \theta -\rho g\sin \theta }{v^2\sin \theta +\rho g\cos \theta }\end{array}$

Substitute 180 mi/h for v, $43.9°$ for $\theta$, $32.2{\text{ft/s}}^2$ for g, and 1,000 ft for $\rho$.

$\begin{array}{c}{\mu }_s=\frac{{\left(180\text{mi/h}\times \frac{5,280\text{ft}}{1\text{mi}}\times \frac{1\text{h}}{3,600\text{s}}\right)}^2\cos 43.9°-\left(1,000\right)\left(32.2\right)\sin 43.9°}{{\left(180\text{mi/h}\times \frac{5,280\text{ft}}{1\text{mi}}\times \frac{1\text{h}}{3,600\text{s}}\right)}^2\sin 43.9°+\left(1000\right)\left(32.2\right)\cos 43.9°}\\ =\frac{27.89199\times {10}^3}{71.52908\times {10}^3}\\ =0.390\end{array}$

Thus, the coefficient of static friction between the tires and the track under the prevailing conditions is $\underset{\_}{0.390}$.

To determine

Find the minimum speed at which the same car could negotiate that curve.

Answer to Problem 12.52P

The minimum speed at which the same car could negotiate that curve is $\underset{\_}{78.8\text{mi/h}}$.

Explanation of Solution

Calculation:

Write the general equation of normal force (N) in minimum speed.

$\begin{array}{l}F=-{\mu }_{s}N\\ {\mu }_s=-\frac{F}{N}\end{array}$ (3)

Substitute Equation (1) and (2) in Equation (3).

$\begin{array}{l}{\mu }_s=-\frac{m\left(\frac{v^2}{\rho }\right)\cos \theta -\left(mg\right)\sin \theta }{m\left(\frac{v^2}{\rho }\right)\sin \theta +\left(mg\right)\cos \theta }\\ {\mu }_s=-\frac{m\left(v^2\cos \theta -\rho g\sin \theta \right)}{m\left(v^2\sin \theta +\rho g\cos \theta \right)}\\ {\mu }_s\left(v^2\sin \theta +\rho g\cos \theta \right)=-\left(v^2\cos \theta -\rho g\sin \theta \right)\\ v^2{\mu }_s\sin \theta +\rho g{\mu }_s\cos \theta =\rho g\sin \theta -v^2\cos \theta \end{array}$

$\begin{array}{l}{v}^2{\mu }_s\sin \theta +v^2\cos \theta =\rho g\sin \theta -\rho g{\mu }_s\cos \theta \\ v^2\left({\mu }_s\sin \theta +\cos \theta \right)=\rho g\left(\sin \theta -{\mu }_s\cos \theta \right)\\ v^2=\frac{\rho g\left(\sin \theta -{\mu }_s\cos \theta \right)}{{\mu }_s\sin \theta +\cos \theta }\end{array}$

Substitute 0.390 for ${\mu }_s$, $43.9°$ for $\theta$, $32.2{\text{ft/s}}^2$ for g, and 1,000 ft for $\rho$.

$\begin{array}{l}{v}^2=\frac{\left(1,000\right)\left(32.2\right)\left(\sin 43.9°-0.390\cos 43.9°\right)}{0.390\sin 43.9°+\cos 43.9°}\\ v=\frac{13,278.85801}{0.990977}\\ v=115.757\text{ft/s}\times \frac{1\text{mi}}{5,280\text{ft}}\times \frac{3,600\text{s}}{1\text{h}}\\ v=78.8\text{mi/h}\end{array}$

Thus, the minimum speed at which the same car could negotiate that curve is $\underset{\_}{78.8\text{mi/h}}$.