Chapter 1, Problem 44SP

A truck is moving north at a speed of 70 km/h. The exhaust pipe above the truck cab

sends out a trail of smoke that makes an angle of $20°$ east of south behind the truck. If

the wind is blowing directly toward the east, what is the wind speed at that location?

[Hint: The smoke reveals the direction of the truck with-respect-to the air.]

To determine

The speed of wind at the location considering the wind is blowing toward east and if the truck moves north at a speed of $70\text{ km/h}$ and the exhaust pipe sends smoke at an angle $20°$ east of south.

Answer to Problem 44SP

Solution:

$25\text{ km/h}$

Explanation of Solution

Given data:

The speed of truck is $70\text{ km/h}$ toward north.

The exhaust pipe sends smoke at an angle $20°$ east of south.

The wind is blowing toward the east.

Formula used:

Consider two vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$ such that $\overrightarrow{P}=a_1\widehat{i}+a_2\widehat{j}$ and $\overrightarrow{Q}=b_1\widehat{i}+b_2\widehat{j}$. The sum of vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$ is calculated as

$\begin{array}{c}\overrightarrow{P}+\overrightarrow{Q}=a_1\widehat{i}+a_2\widehat{j}+b_1\widehat{i}+b_2\widehat{j}\\ =\left(a_1+b_1\right)\widehat{i}+\left(a_2+b_2\right)\widehat{j}\end{array}$

The magnitude of vector $\overrightarrow{P}+\overrightarrow{Q}$ is calculated as

$\left|\overrightarrow{P}+\overrightarrow{Q}\right|=\sqrt{{\left(a_1+b_1\right)}^2+{\left(a_2+b_2\right)}^2}$

Here, $\left|\overrightarrow{P}+\overrightarrow{Q}\right|$ is magnitude of the vector $\overrightarrow{P}+\overrightarrow{Q}$.

The angle of vector $\overrightarrow{P}+\overrightarrow{Q}$ with the horizontal is calculated as

$\tan \theta =\frac{a_2+b_2}{a_1+b_1}$

Here, $\theta$ is the angle of the vector $\overrightarrow{P}+\overrightarrow{Q}$ with the horizontal.

The scalar components of vector $\overrightarrow{P}+\overrightarrow{Q}$ are calculated as

$\begin{array}{l}{S}_x=\left|\overrightarrow{P}+\overrightarrow{Q}\right|\cos \theta \\ S_y=\left|\overrightarrow{P}+\overrightarrow{Q}\right|\sin \theta \end{array}$

Here, $S_x$ and $S_y$ represent the horizontal and vertical scalar components of vector $\overrightarrow{P}+\overrightarrow{Q}$, respectively.

Vector $\overrightarrow{P}+\overrightarrow{Q}$ is expressed as

$\overrightarrow{P}+\overrightarrow{Q}=S_x\widehat{i}+S_y\widehat{j}$

The resultant vector of two vectors represented by adjacent sides of a parallelogram is expressed as

$\overrightarrow{T}=\overrightarrow{P}+\overrightarrow{Q}$

Here, $\overrightarrow{P}$ and $\overrightarrow{Q}$ indicate the two adjacent sides of a parallelogram and $\overrightarrow{T}$ is resultant vector of vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$.

Explanation:

Understand that the smoke is emitted from the truck in a direction exactly opposite to the direction of motion of truck. According to the problem, the truck moves north, therefore the smoke is initially emitted in the south direction.

Draw a diagram for the problem considering north as +y direction, south as –y direction, east as +x direction, and west as –x direction:

Here, $\overrightarrow{A}$ is the velocity vector for the original smoke emitted by the truck, $\overrightarrow{R}$ is the resultant velocity vector for smoke through the exhaust pipe, which is at an angle $20°$ east of south, due to the wind, and $\overrightarrow{B}$ is the velocity vector that indicates the direction of wind, which is toward east.

Consider the expression for angle with the horizontal:

$\varphi =90°-\beta$

Here, $\beta$ is the angle made by the exhaust velocity with the vertical and $\varphi$ is the angle made by the exhaust velocity with the horizontal.

Substitute $20°$ for $\beta$

$\begin{array}{c}\varphi =90°-20°\\ =70°\end{array}$

Consider the expression for horizontal scalar component of the velocity vector of air through the exhaust pipe:

$R_x=\left|\overrightarrow{R}\right|\cos \varphi$

Here, $\left|\overrightarrow{R}\right|$ is the magnitude of resultant velocity vector and $R_x$ is the horizontal component of the resultant velocity.

The angle is measured anticlockwise from the horizontal, therefore, consider the angle as negative.

Substitute $-70°$ for $\varphi$

$\begin{array}{c}{R}_x=\left|\overrightarrow{R}\right|\cos \left(-70°\right)\\ =\left|\overrightarrow{R}\right|\cos 70°\end{array}$

Consider the expression for vertical scalar component of the velocity vector of air through the exhaust pipe.

$R_y=\left|\overrightarrow{R}\right|\sin \varphi$

Here, $R_y$ is the vertical scalar component of the resultant velocity of air.

Substitute $-70°$ for $\varphi$

$\begin{array}{c}{R}_y=\left|\overrightarrow{R}\right|\sin \left(-70°\right)\\ =-\left|\overrightarrow{R}\right|\sin 70°\end{array}$

Consider the Cartesian form of resultant velocity of air:

$\overrightarrow{R}=R_x\widehat{i}+R_y\widehat{j}$

Here, $\overrightarrow{R}$ is the resultant displacement vector representing the velocity of air.

Substitute $-\left|\overrightarrow{R}\right|\sin 70°$ for $R_y$ and $\left|\overrightarrow{R}\right|\cos 70°$ for $R_x$

$\begin{array}{c}\overrightarrow{R}=\left(\left|\overrightarrow{R}\right|\cos 70°\right)\widehat{i}+\left(-\left|\overrightarrow{R}\right|\sin 70°\right)\widehat{j}\\ =\left(\left|\overrightarrow{R}\right|\cos 70°\right)\widehat{i}-\left(\left|\overrightarrow{R}\right|\sin 70°\right)\widehat{j}\end{array}$

Consider the velocity of smoke of truck, which is equal to magnitude of the velocity of truck, that is, $70\text{ km/h}$, and the smoke is moving in the south direction. The diagram indicates that the vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ are perpendicular to each other and therefore, they indicate the adjacent sides of a parallelogram. Write the formula for resultant velocity of smoke using parallelogram law of vector addition:

$\begin{array}{l}\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}\\ \overrightarrow{B}=\overrightarrow{R}-\overrightarrow{A}\end{array}$

The velocity vector $\overrightarrow{A}$ is $70\text{ km/h}$ in the south direction. Hence, substitute $\left(-70\widehat{j}\right)\text{ km/h}$ for $\overrightarrow{A}$ and $\left(\left|\overrightarrow{R}\right|\cos 70°\right)\widehat{i}-\left(\left|\overrightarrow{R}\right|\sin 70°\right)\widehat{j}$ for $\overrightarrow{R}$.

$\begin{array}{c}\overrightarrow{B}=\left[\left(\left|\overrightarrow{R}\right|\cos 70°\right)\widehat{i}-\left(\left|\overrightarrow{R}\right|\sin 70°\right)\widehat{j}\right]-\left(-70\widehat{j}\right)\text{ km/h}\\ =\left[\left(\left|\overrightarrow{R}\right|\cos 70°\right)\widehat{i}-\left(\left|\overrightarrow{R}\right|\sin 70°\right)\widehat{j}\right]+\left(70\widehat{j}\right)\text{ km/h}\\ =\left(\left|\overrightarrow{R}\right|\cos 70°\right)\widehat{i}+\left(70-\left|\overrightarrow{R}\right|\sin 70°\right)\widehat{j}\end{array}$

The resultant velocity of the wind is in the east direction. Therefore, the vertical component of $\overrightarrow{B}$ is zero and the coefficient of $\widehat{j}$ is equated as zero.

$\begin{array}{c}\left(70\text{ km/h}-\left|\overrightarrow{R}\right|\sin 70°\right)=0\\ \left|\overrightarrow{R}\right|\sin 70°=70\text{ km/h}\\ \left|\overrightarrow{R}\right|=\frac{70\text{ km/h}}{\sin 70°}\\ \left|\overrightarrow{R}\right|=74.4\text{ km/h}\end{array}$

According to the problem, the truck moves in north direction, therefore, it doesn’t have any horizontal velocity. Therefore, the velocity of wind is equal to the horizontal component of resultant smoke velocity. The horizontal component of vector $\overrightarrow{R}$ is equal to the value of velocity of the wind. Hence,

$\overrightarrow{B}=R_x$

Substitute $\left|\overrightarrow{R}\right|\cos 70°$ for $R_x$

$B_x=\left|\overrightarrow{R}\right|\cos 70°$

Substitute $74.4\text{ km/h}$ for $\left|\overrightarrow{R}\right|$

$\begin{array}{c}{B}_x=\left(74.4\text{ km/h}\right)\cos 70°\\ \simeq 25\text{ km/h}\end{array}$

Conclusion:

The speed of wind at the location is $25\text{ km/h}$ toward east.