Chapter 3, Problem 10P

To determine

The location of the sailboat using graphical method and component method of vector addition.

Answer to Problem 10P

The magnitude of the location of the sailboat is $\text{29}\text{.3 NM}$ and the angle is $\text{342}\text{.8}°$.

Explanation of Solution

Starting from the origin, each vector is drawn beginning at the end of its preceding vector.

The graphical representation of the path of the sailboat is given below:

Write the expression for y-component of the position of the sailboat

    $r_{\text{y}}=r_1\sin {\theta }_1+r_2\sin {\theta }_2+r_3\sin {\theta }_3+r_4\sin {\theta }_4+r_5\sin {\theta }_5$        (I)

Here, $r_{\text{y}}$ is the y-component of the resultant vector, $r_1,r_2,r_3,r_4,r_5$ are the magnitudes of the vectors and ${\theta }_1,{\theta }_2,{\theta }_3,{\theta }_4,{\theta }_5$ are the angles of the vectors.

Write the expression for x-component of the position of the sailboat

    $r_{\text{x}}=r_1\cos {\theta }_1+r_2\cos {\theta }_2+r_3\cos {\theta }_3+r_4\cos {\theta }_4+r_5\cos {\theta }_5$        (II)

Here, $r_{\text{x}}$ is the x-component of the resultant vector.

Write the expression for magnitude of the resultant vector

    $r=\sqrt{{\left(r_{\text{x}}\right)}^2+{\left(r_{\text{y}}\right)}^2}$        (III)

Here, $r$ is the magnitude of the resultant vector.

Write the expression for angle of the resultant vector

    $r={\tan }^{-1}\left(\frac{r_{\text{y}}}{r_{\text{x}}}\right)$        (IV)

Here, $r$ is the magnitude of the resultant vector.

Conclusion:

Substitute $45\text{ NM}$ for $r_1$, $20\text{ NM}$ for $r_2$, $30\text{ NM}$ for $r_3$, $10\text{ NM}$ for $r_4$ and $\text{62 NM}$ for $r_5$, $0°$ for ${\theta }_1$, $-60°$ for ${\theta }_2$, $0°$ for ${\theta }_3$, $60°$ for ${\theta }_4$ and $180°$ for ${\theta }_5$  in (I) to find $r_{\text{y}}$

    $\begin{array}{c}{r}_{\text{y}}=\left\{\begin{array}{c}\left(45\text{ NM}\right)\sin \left(0°\right)+\left(20\text{ NM}\right)\sin \left(-60°\right)+\left(30\text{ NM}\right)\sin \left(0°\right)+\\ \left(10\text{ NM}\right)\sin \left(60°\right)+\left(62\text{ NM}\right)\sin \left(180°\right)\end{array}\right\}\\ =-8.66\text{ NM}\end{array}$

Substitute $45\text{ NM}$ for $r_1$, $20\text{ NM}$ for $r_2$, $30\text{ NM}$ for $r_3$, $10\text{ NM}$ for $r_4$ and $\text{62 NM}$ for $r_5$, $0°$ for ${\theta }_1$, $-60°$ for ${\theta }_2$, $0°$ for ${\theta }_3$, $60°$ for ${\theta }_4$ and $180°$ for ${\theta }_5$  in (II) to find $r_{\text{x}}$

    $\begin{array}{c}{r}_{\text{x}}=\left\{\begin{array}{c}\left(45\text{ NM}\right)\cos \left(0°\right)+\left(20\text{ NM}\right)\cos \left(-60°\right)+\left(30\text{ NM}\right)\cos \left(0°\right)+\\ \left(10\text{ NM}\right)\cos \left(60°\right)+\left(62\text{ NM}\right)\cos \left(180°\right)\end{array}\right\}\\ =28\text{ NM}\end{array}$

Substitute $28\text{ NM}$ for $r_{\text{x}}$ and $-8.66\text{ NM}$ for $r_{\text{y}}$ in equation (III) to find $r$

    $\begin{array}{c}r=\sqrt{{\left(28\text{ NM}\right)}^2+{\left(-8.66\text{ NM}\right)}^2}\\ =29.309\text{ NM}\\ \approx \text{29}\text{.3 NM}\end{array}$

Substitute $28\text{ NM}$ for $r_{\text{x}}$ and $-8.66\text{ NM}$ for $r_{\text{y}}$ in equation (IV) to find $\theta$

    $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-8.66\text{ NM}}{28\text{ NM}}\right)\\ =-17.19°\\ \approx \text{342}\text{.8}°\end{array}$

Thus, the magnitude of the location of the sailboat is $\text{29}\text{.3 NM}$ and the angle is $\text{342}\text{.8}°$.