Chapter 5, Problem 32P

To determine

The minimum coefficient of static friction needed if the car is to navigate the curve without slipping.

Answer to Problem 32P

The minimum coefficient of static friction needed if the car is to navigate the curve without slipping is $0.204$.

Explanation of Solution

Figure 1 shows the free body diagram of the car.

Write an expression to calculate the net horizontal force.

$\begin{array}{c}N\sin \theta +f_s\cos \theta =m{a}_x\\ N\sin \theta +{\mu }_{s}N\cos \theta =m\left(\frac{v^2}{r}\right)\end{array}$ (I)

Here, the normal force is $N$, the angle is $\theta$, the static frictional force is $f_s$, the coefficient of static friction is ${\mu }_s$, the mass of the car is $m$, the horizontal acceleration is $a_x$, the speed is $v$ and the radius of path is $r$.

Write an expression for the net vertical force.

$\begin{array}{c}N\cos \theta -mg-f_s\sin \theta =0\\ N\cos \theta -mg-{\mu }_{s}N\sin \theta =0\end{array}$ (II)

Here, the acceleration due to gravity is $g$.

Neglect the frictional force and rewrite equation (I).

$\begin{array}{c}N\sin \theta +\left(0\right)\cos \theta =m\left(\frac{v^2}{r}\right)\\ N\sin \theta =m\left(\frac{v^2}{r}\right)\end{array}$ (III)

Neglect the frictional force and rewrite equation (II).

$\begin{array}{c}N\cos \theta -mg-\left(0\right)\sin \theta =0\\ N\cos \theta -mg=0\end{array}$ (IV)

Rearrange equation (IV) to find $N$.

$N=\frac{mg}{\cos \theta }$ (V)

Substitute equation (V) in equation (III). 

$\begin{array}{c}\left(\frac{mg}{\cos \theta }\right)\sin \theta =m\left(\frac{v^2}{r}\right)\\ g\tan \theta =\left(\frac{v^2}{r}\right)\end{array}$ (VI)

Rewrite equation (VI) to find $\theta$.

$\theta ={\tan }^{-1}\left(\frac{v^2}{gr}\right)$ (VII)

Now consider the presence of frictional force.

Rearrange equation (I) to find $N$.

$N=\frac{m{v}^2}{R\left(\sin \theta +{\mu }_s\cos \theta \right)}$ (VIII)

Rearrange equation (II) to find $N$.

$N=\frac{mg}{\cos \theta -{\mu }_s\sin \theta }$ (IX)

Compare equation (VIII) and (IX).

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

Rearrange equation (X) to find ${\mu }_s$.

${\mu }_s=\frac{v^2\cos \theta -gr\sin \theta }{v^2\sin \theta +gr\cos \theta }$ . (XI)

Conclusion:

Substitute $15.0\text{m/s}$ for $v$, $9.80{\text{m/s}}^2$ for $g$, and $75.0\text{m}$ for $r$ in equation (VII) to find $\theta$.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{{\left(15.0\text{m/s}\right)}^2}{\left(9.80{\text{m/s}}^2\right)\left(75.0\text{m}\right)}\right)\\ ={\tan }^{-1}\left(0.306\right)\\ =17.0°\end{array}$

Substitute $20.0\text{m/s}$ for $v$, $9.80{\text{m/s}}^2$ for $g$, $75.0\text{m}$ for $r$, and $17.0°$ for $\theta$ in equation (XI) to find ${\mu }_s$.

$\begin{array}{c}{\mu }_s=\frac{{\left(20.0\text{m/s}\right)}^2\cos \left(17.0°\right)-\left(9.80{\text{m/s}}^2\right)\left(75.0\text{m}\right)\sin \left(17.0°\right)}{{\left(20.0\text{m/s}\right)}^2\sin \left(17.0°\right)+\left(9.80{\text{m/s}}^2\right)\left(75.0\text{m}\right)\cos \left(17.0°\right)}\\ =\frac{382.5{\text{m}}^2{\text{/s}}^2-214.9{\text{m}}^2{\text{/s}}^2}{116.9{\text{m}}^2{\text{/s}}^2+702.9{\text{m}}^2{\text{/s}}^2}\\ =0.204\end{array}$

Thus, the minimum coefficient of static friction needed if the car is to navigate the curve without slipping is $0.204$.