91Ó°ÊÓ

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

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

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

Given

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

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

x component of the vector sum r is given by

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

y component of vector sum $\stackrel{→}{r}$is

$$\begin{gathered} r_y=\mathrm{a}{\mathrm{²õ¾\pm ²Ôθ}}_1+{\mathrm{b²õ¾\pm ²Ôθ}}_2 \\ \stackrel{→}{\mathrm{r}}=10\mathrm{m}×\sin \left(30\right)+10\mathrm{m}×\sin \left(135\right) \\ {\mathrm{r}}_{\mathrm{y}}=12.1\mathrm{m} \end{gathered}$$

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

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

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

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