Chapter 11.4, Problem 11.131P

To determine

The speed $\left(v_w\right)$ and direction $\left(\alpha \right)$ of the wind.

Answer to Problem 11.131P

The speed $\left(v_w\right)$ and direction $\left(\alpha \right)$ of the wind are $\underset{\_}{22.33\text{km/h}}$ and $\underset{\_}{9.04°}$.

Explanation of Solution

Given Information:

The small boat travel north at the speed ${\left(v_B\right)}_y$ of $15\text{km/h}$ at an angle $\left(\theta \right)$ of $50°$ with centerline of the boat.

A short time later the boat travels east at the speed ${\left(v_B\right)}_x$ of $25\text{km/h}$ at an angle $\left(\theta \right)$ of $50°$.

Calculation:

Consider case (1) while the boat travel north,

Write the speed of ship $\left(v_B\right)$ in vector notation:

${\left(v_B\right)}_y=15j$

Write the relative velocity $\left(v_{A/w}\right)$ of the wind with respective to boat in terms of vector notation.

${\left(v_{w/B}\right)}_1={\left(v_{w/B}\right)}_1\left(-\sin \theta i+\cos \theta j\right)$

Substitute $50°$ for $\theta$.

$\begin{array}{c}{\left(v_{w/B}\right)}_1={\left(v_{w/B}\right)}_1\left(-\sin \left(50°\right)i+\cos \left(50°\right)j\right)\\ =-{\left(v_{w/B}\right)}_1\sin \left(50°\right)i+{\left(v_{w/B}\right)}_1\cos \left(50°\right)j\end{array}$

Write the velocity vector $\left(v_W\right)$ of the wind:

$v_W=v_{wx}i+v_{wy}j$

Calculate velocity vector $\left(v_W\right)$ of the wind using the relation below:

$\begin{array}{c}{v}_W=v_B+{\left(v_{B/w}\right)}_1\\ =v_B-{\left(v_{w/B}\right)}_1\end{array}$

Substitute $v_{wx}i+v_{wy}j$ for $v_W$, $15j$ for $v_B$ and $-{\left(v_{w/B}\right)}_1\sin \left(50°\right)i+{\left(v_{w/B}\right)}_1\cos \left(50°\right)j$ for ${\left(v_{w/B}\right)}_1$.

$\begin{array}{c}{v}_{wx}i+v_{wy}j=\left(15j\right)-\left(-{\left(v_{w/B}\right)}_1\sin \left(50°\right)i+{\left(v_{w/B}\right)}_1\cos \left(50°\right)j\right)\\ v_{wx}i+v_{wy}j=\left(15j\right)+{\left(v_{w/B}\right)}_1\sin \left(50°\right)i-{\left(v_{w/B}\right)}_1\cos \left(50°\right)j\end{array}$ (1)

Separate the i and j component in the equation (1).

$i:v_{wx}i=0+{\left(v_{w/B}\right)}_1\sin \left(50°\right)i$ (2)

$j:v_{wy}j=\left(15j\right)-{\left(v_{w/B}\right)}_1\cos \left(50°\right)j$ (3)

Consider case (2) while the boat travel east,

Write the speed of ship $\left(v_B\right)$ in vector notation:

${\left(v_B\right)}_x=25i$

Write the relative velocity ${\left(v_{w/B}\right)}_2$ of the wind with respective to boat in terms of vector notation.

${\left(v_{w/B}\right)}_2={\left(v_{w/B}\right)}_2\left(\cos \theta i+\text{sin}\theta j\right)$

Substitute $50°$ for $\theta$.

$\begin{array}{c}{\left(v_{w/B}\right)}_2={\left(v_{w/B}\right)}_2\left(\cos \left(50°\right)i+\sin \left(50°\right)j\right)\\ ={\left(v_{w/B}\right)}_2\cos \left(50°\right)i+{\left(v_{w/B}\right)}_2\sin \left(50°\right)j\end{array}$

Write the velocity vector $\left(v_W\right)$ of the wind as below:

$v_W=v_{wx}i+v_{wy}j$

Calculate velocity vector $\left(v_W\right)$ of the wind using the relation below:

$\begin{array}{c}{v}_W=v_B+{\left(v_{B/w}\right)}_2\\ =v_B-{\left(v_{w/B}\right)}_2\end{array}$

Substitute $v_{wx}i+v_{wy}j$ for $v_W$, $25j$ for $v_B$ and ${\left(v_{w/B}\right)}_2\sin \left(50°\right)i+{\left(v_{w/B}\right)}_2\cos \left(50°\right)j$ for ${\left(v_{w/B}\right)}_2$.

$\begin{array}{c}{v}_{wx}i+v_{wy}j=\left(25i\right)-\left({\left(v_{w/B}\right)}_2\cos \left(50°\right)i+{\left(v_{w/B}\right)}_2\sin \left(50°\right)j\right)\\ v_{wx}i+v_{wy}j=\left(25i\right)-{\left(v_{w/B}\right)}_2\cos \left(50°\right)i-{\left(v_{w/B}\right)}_2\sin \left(50°\right)j\end{array}$ (4)

Separate the i and j component in the equation (1).

$i:v_{wx}i=\left(25i\right)-{\left(v_{w/B}\right)}_2\cos \left(50°\right)i$ (5)

$j:v_{wy}j=0-{\left(v_{w/B}\right)}_2\sin \left(50°\right)j$ (6)

Equate equation (2) and (5).

$\begin{array}{l}0+{\left(v_{w/B}\right)}_1\sin \left(50°\right)i=\left(25i\right)-{\left(v_{w/B}\right)}_2\cos \left(50°\right)i\\ {\left(v_{w/B}\right)}_1\sin \left(50°\right)+{\left(v_{w/B}\right)}_2\cos \left(50°\right)=25\end{array}$ (7)

Equate equation (3) and (6).

$\begin{array}{l}\left(15j\right)-{\left(v_{w/B}\right)}_1\cos \left(50°\right)j=0-{\left(v_{w/B}\right)}_2\sin \left(50°\right)j\\ {\left(v_{w/B}\right)}_1\cos \left(50°\right)-{\left(v_{w/B}\right)}_2\sin \left(50°\right)=15\end{array}$ (8)

Solve the equation (7) and (8) for ${\left(v_{w/B}\right)}_1$ and ${\left(v_{w/B}\right)}_2$.

The value ${\left(v_{w/B}\right)}_1$ and ${\left(v_{w/B}\right)}_2$ are $28.79\text{km/h}$ and $4.58\text{km/h}$ respectively.

Calculate the x component velocity $\left(v_{wx}\right)$ of the wind:

Substitute $28.79\text{km/h}$ for ${\left(v_{w/B}\right)}_1$ in equation (2).

$\begin{array}{c}{v}_{wx}=28.79\sin \left(50°\right)\\ =22.05\text{km/h}\end{array}$

Calculate the y component velocity $\left(v_{wy}\right)$ of the wind:

Substitute $28.79\text{km/h}$ for ${\left(v_{w/B}\right)}_1$ in Equation (3).

$\begin{array}{c}{v}_{wy}=\left(15\right)-28.79\cos \left(50°\right)\\ =15-18.51\\ =-3.51\text{km/h}\end{array}$

Write velocity $\left(v_w\right)$ of the wind using the relation:

$v_w=\sqrt{v_{wx}^2+v_{wy}^2}$

Substitute $22.05\text{km/h}$ for $v_{wx}$ and $-3.51\text{km/h}$ for $v_{wy}$.

$\begin{array}{c}{v}_w=\sqrt{{\left(22.05\right)}^2+{\left(-3.51\right)}^2}\\ =\sqrt{486.203+12.320}\\ =22.33\text{km}/\text{h}\end{array}$

Calculate the angle made by the resultant of $v_R$ using the relation:

$\varphi ={\tan }^{-1}\left(\frac{v_{wy}}{v_{wx}}\right)$

Substitute $22.05\text{km/h}$ for $v_{wx}$ and $-3.51\text{km/h}$ for $v_{wy}$.

$\begin{array}{c}\varphi ={\tan }^{-1}\left(\frac{-3.51}{22.05}\right)\\ ={\tan }^{-1}\left(0.159\right)\\ =9.04°\end{array}$

Therefore, the speed $\left(v_w\right)$ and direction $\left(\alpha \right)$ of the wind are $\underset{\_}{22.33\text{km/h}}$ and $\underset{\_}{9.04°}$.