The two vectors $\overrightarrow{\mathbf{a}\mathbf{}}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\overrightarrow{\mathbf{b}}$in Fig.3.28 have equal magnitudes of 10.0mand the angles are $\mathit{\theta }\mathbf{=}\mathbf{30}\mathbf{°}$and $\mathit{\theta }\mathbf{=}\mathbf{105}\mathbf{°}$. Find the (a) x and (b) y components of their vector sum$\overrightarrow{\mathbf{r}}$, (c) the magnitude of $\overrightarrow{\mathbf{r}}$, and (d) the angle$\overrightarrow{\mathbf{r}}$makes with the positive direction of the x axis

This problem involves the addition of two vectors $\overrightarrow{\mathbf{a}}$and $\overrightarrow{\mathbf{b}}$ . Using the following relations x and y components of the vector sum$\overrightarrow{\mathbf{r}}$can be found and similarly the magnitude and the angle respectively.

localid="1656155400549" $r=\overrightarrow{\left|r\right|}=\sqrt{{\left(x\right)}^2+{\left(y\right)}^2}$ (i)

$\theta ={\tan }^{-1}\left(\frac{y}{x}\right)$ (ii)

Given

Two vectors and $\overrightarrow{\mathrm{a}}=10\mathrm{m}\mathrm{and}\overrightarrow{\mathrm{b}}=10\mathrm{m}$

two angles and ${\theta }_1=30°\mathrm{and}{\mathrm{\theta }}_2=105°$

x component of the vector sum r is given by

$$\begin{gathered} r_x=a\cos {\theta }_1+b\cos {\theta }_2 \\ \overrightarrow{r}=10\mathrm{m}\times \cos \left(30\right)+10\mathrm{m}\times \left(135\right) \\ {\mathrm{r}}_{\mathrm{x}}=1.59\mathrm{m} \end{gathered}$$

y component of vector sum $\overrightarrow{r}$is

$$\begin{gathered} r_y=\mathrm{a}{\mathrm{sin\theta }}_1+{\mathrm{bsin\theta }}_2 \\ \overrightarrow{\mathrm{r}}=10\mathrm{m}\times \sin \left(30\right)+10\mathrm{m}\times \sin \left(135\right) \\ {\mathrm{r}}_{\mathrm{y}}=12.1\mathrm{m} \end{gathered}$$

Using equation (i) the magnitude of vector$\overrightarrow{r}$is written as

$$\begin{gathered} r=\overrightarrow{\left|r\right|}=\sqrt{\left(1.59{\mathrm{m}}^2\right)+{\left(12.1\mathrm{m}\right)}^2} \\ r=12.2\mathrm{m} \end{gathered}$$

The angle$\overrightarrow{r}$ makes with the positive direction of the x axis is

${\tan }^1\left(\frac{12.1}{1.59}\right)=82.5°$