Chapter 6, Problem 40PQ

(a)

To determine

Give estimation on weight and area of a typical car.

Answer to Problem 40PQ

A typical car can have weight around $906\text{kg}$ and area $2{\text{m}}^{\text{2}}$.

Explanation of Solution

The average of weight of car can be taken as two thousand pounds. Express two thousand pounds in kilogram.

    $2000\text{pounds}\left(\frac{0.453\text{kg}}{1\text{pound}}\right)=906\text{kg}$

Normally a car can have $6\text{feet}$ width and $3\text{feet}$ height. Express area in square meter.

    $\left(6\text{feet}\times 3\text{feet}\right)\left(\frac{0.0929{\text{m}}^{\text{2}}}{1\text{foot}}\right)\approx 2{\text{m}}^{\text{2}}$

Conclusion:

Therefore, a typical car can have weight around $906\text{kg}$ and area $2{\text{m}}^{\text{2}}$.

(b)

To determine

Draw the graph of $v^2$ against $F_D$ and check whether the shape of graph conveys sensible data.

Answer to Problem 40PQ

Graph of $v^2$ against $F_D$ is drawn and the conveying result is consistent with equation for drag force.

Explanation of Solution

The graph of $v^2$ with $F_D$ is drawn below.

    

Write the equation for drag force.

    $F_D=\frac{1}{2}C\rho A{v}^2$

Here, $F_D$ is the drag force, $C$ is the drag coefficient, $\rho$ is the density of air, and $v$ is the speed of air.

Substitute $0.4$ for $C$, $1.3\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, and $2{\text{m}}^{\text{2}}$ for $A$ in the above equation.

    $\begin{array}{c}{F}_D=\frac{1}{2}\left(0.4\right)\left(1.3\text{kg}/{\text{m}}^{\text{3}}\right)\left(2{\text{m}}^{\text{2}}\right)v^2\\ \approx 0.5v^2\end{array}$                                                             (I)

The equation tells that the drag force is varying with the square of speed of car. It can be seen that the graph is also conveying the same result. Graph shows that $F_D$ is directly proportional to $v^2$.

Conclusion:

Therefore, the graph of $v^2$ against $F_D$ is drawn and the conveying result is consistent with equation for drag force.

(c)

To determine

The value of Rolling frictional force of car and add the estimate to graph.

Answer to Problem 40PQ

Rolling frictional force is $180\text{N}$ and it appears as a straight line in the graph.

Explanation of Solution

Write the equation to find the rolling friction.

    $F_r={\mu }_r\left(mg\right)$

Here, $F_r$ is the rolling frictional force, ${\mu }_r$ is the coefficient of friction, $m$ is the mass of car, and  $g$ is the gravitational acceleration.

Conclusion:

Substitute $0.02$ for ${\mu }_r$, $900\text{kg}$ for $m$, and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the above equation to find $F_r$.

    $\begin{array}{c}{F}_r=\left(0.02\right)\left(900\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\\ =180\text{N}\end{array}$                                                                 (II)

It can be seen that $F_r$ is independent of $v$. So, it appears as a straight line in the graph.

Therefore, the rolling frictional force is $180\text{N}$ and it appears as a straight line in the graph.

(d)

To determine

The speed at which drag force equals rolling frictional force and the range of speed which the drag force can be neglected.

Answer to Problem 40PQ

The speed at which drag force equals rolling frictional force is $19\text{m}/\text{s}$ and the drag force can be neglected for speed less than around $10\text{m}/\text{s}$.

Explanation of Solution

Equate the right hand sides of equations (1) and (II).

    $\begin{array}{c}0.5v^2=180\text{N}\\ v^2=360\text{N}\end{array}$

Rewrite the above expression in terms of $v$.

    $v=19\text{m}/\text{s}$

It can be seen that lower speeds, drag force will be lesser than rolling frictional force. So drag force can be neglected at lower speed. It is better to limit the speed within $10\text{m}/\text{s}$. Note that this value is not a result of exact calculation.

Conclusion:

Therefore, the speed at which drag force equals rolling frictional force is $19\text{m}/\text{s}$ and the drag force can be neglected for speed less than around $10\text{m}/\text{s}$.

(e)

To determine

The change in answer on the assumption that $F_d$ is proportional to $v$ instead of $v^2$.

Answer to Problem 40PQ

$F_d$ would decrease more if is proportional to $v$ instead of $v^2$.

Explanation of Solution

$F_d$ is directly proportional to $v$. On comparing the value of $F_d$ for same value of speed in $v$ and $v^2$. Lower value of $F_d$ is obtained while the variation is proportional to $v$. In other words, it tells that it would be more safe to ignore the frictional force too at the lower speeds.

Therefore, the $F_d$ would decrease more if is proportional to $v$ instead of $v^2$.