Question

(a) What is the sum of the following four vectors in unitvector notation? For that sum, what are (b) the magnitude, (c) the angle in degrees, and (d) the angle in radians?

$$\begin{gathered} \overrightarrow{\mathbf{E}}\mathbf{=}\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{m}\mathbf{)}\mathbf{}\mathbf{at}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{900}\mathbf{rad}\mathbf{)} \\ \mathbf{}\overrightarrow{\mathbf{F}}\mathbf{=}\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{m}\mathbf{)}\mathbf{}\mathbf{at}\mathbf{-}\mathbf{75}\mathbf{°} \\ \mathbf{}\overrightarrow{\mathbf{G}}\mathbf{=}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{m}\mathbf{}\mathbf{)}\mathbf{at}\mathbf{+}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{20}\mathbf{}\mathbf{rad}\mathbf{)} \\ \mathbf{}\overrightarrow{\mathbf{H}}\mathbf{=}\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{m}\mathbf{}\mathbf{)}\mathbf{at}\mathbf{-}\mathbf{210}\mathbf{°} \end{gathered}$$

Short answer

(a) Sum of the four vectors is $1.28\hat{\mathrm{i}}+6.60\hat{\mathrm{j}}$

(b) The magnitude of sum of the four vectors is$6.72\hspace{0.33em}\mathrm{m}.$

(c) The angle in degrees is $79.0°.$

(d) The angle in radians is $1.38\mathrm{rad}.$

Step-by-step solution

Step 1: Given

Following vectors are given

$$\begin{gathered} \overrightarrow{\mathrm{E}}=(6.00\mathrm{m})\mathrm{at}+(0.900\mathrm{rad}) \\ \overrightarrow{\mathrm{F}}=(5.00\mathrm{m})\mathrm{at}-75° \\ \overrightarrow{\mathrm{G}}=(4.00\mathrm{m})\mathrm{at}+(1.20\mathrm{rad}) \\ \overrightarrow{\mathrm{H}}=(6.00\mathrm{m})\mathrm{at}-210° \end{gathered}$$

Determining the concept

With the help of given magnitude and the angle the conversion of vectors into unit vector is possible.After addition of four vectors, answer will come out in the form of unit vector and magnitude and angle of that unit vector can be determined.

Required formulae are as follow:

$\overrightarrow{\mathrm{A}}=\mathrm{x}\hat{\mathrm{i}}+\mathrm{y}\hat{\mathrm{j}}\left(\mathrm{i}\right)$

Formula for magnitude:

$\left|\overrightarrow{\mathrm{A}}\right|=\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}\left(\mathrm{ii}\right)$

Formula for angle:

role="math" localid="1657022472163" $\mathrm{\theta }={\tan }^{-1}\frac{\mathrm{y}}{\mathrm{x}}\left(\mathrm{iii}\right)$

Where,x ,y, $\overrightarrow{A}$ are vectors and $\mathrm{\theta }$is the angle between x and y .

Step 3:(a) Determination of sum of the four vectors

Now, convert vectors in the unit vector form,

$$\begin{gathered} \overrightarrow{\mathrm{E}}=\mathrm{x}\hat{\mathrm{i}}+\mathrm{y}\hat{\mathrm{j}} \\ \mathrm{x}=6.00\times \cos \left(0.9^{°}\right) \\ =3.73 \\ \mathrm{y}=5.00\times \sin \left(-{75}^{°}\right) \\ =4.70 \end{gathered}$$

Therefore, $\overrightarrow{E}$ in vector form is,

$\overrightarrow{\mathrm{E}}=3.73\hat{\mathrm{i}}+4.70\hat{\mathrm{j}}$

Similarly,$\overrightarrow{F}$ can be written as

$\overrightarrow{\mathrm{F}}=1.29\hat{\mathrm{i}}-4.83\hat{\mathrm{j}}$

The vector$\overrightarrow{G}$ can be written as,

$\overrightarrow{\mathrm{G}}=1.45\hat{\mathrm{i}}+3.73\hat{\mathrm{j}}$

The vector$\overrightarrow{\mathrm{H}}$ can be written as,

$\overrightarrow{\mathrm{H}}=-5.20\hat{\mathrm{i}}+3.00\hat{\mathrm{j}}$

To find the sum, add all the above vectors using equation (i). Therefore,

$$\begin{gathered} \overrightarrow{\mathrm{A}}=\overrightarrow{\mathrm{E}}+\overrightarrow{\mathrm{F}}+\overrightarrow{\mathrm{G}}+\overrightarrow{\mathrm{H}} \\ =\left(3.73+1.29+1.45+5.20\right)\hat{\mathrm{i}}+\left(4.7-4.83+3.73+3.00\right)\hat{\mathrm{j}} \\ =1.28\overrightarrow{\mathrm{i}}+6.60\hat{\mathrm{j}} \end{gathered}$$

Therefore, sum of unit vectors is $1.28\mathrm{i}⃗+6.60\hat{\mathrm{j}}$.

(b) Determination of magnitude

Use the equation (ii) to find the magnitude.

$$\begin{gathered} \left|\overrightarrow{\mathrm{A}}\right|=\sqrt{1.{28}^2+6.{60}^2} \\ =6.72\mathrm{m} \end{gathered}$$

Therefore, magnitude of $\overrightarrow{A}$is$6.72\mathrm{m}$ .

(c) Determination of angle in degrees

Use the equation (iii) to find the angle in degrees.

$$\begin{gathered} \mathrm{\theta }={\tan }^{-1}\left(\frac{\mathrm{y}}{\mathrm{x}}\right) \\ ={\tan }^{-1}\left(\frac{6.60}{1.28}\right) \\ =79.0^{°} \end{gathered}$$

Therefore, angle in degrees is ${\text{θ=79.0}}^{°}$.

Step 6:(d) Determination of angle in radians

Now, convert the angle into radians.

$$\begin{gathered} \mathrm{\theta }=\frac{79.0\times \mathrm{\pi }}{180}\mathrm{red} \\ =1.38\mathrm{rad} \end{gathered}$$

Therefore, angle in radian is $1.38\mathrm{rad}$.