Chapter 1, Problem 50P

Consider the three displacement vectors $\stackrel{\to }{A}=(3\hat{i}-3\hat{j})\text{ m}$, $\stackrel{\to }{B}=(\hat{i}-4\hat{j})\text{ m}$, and $\stackrel{\to }{C}=(2\hat{i}+5\hat{j})\text{ m}$. Use the component method to determine (a) the magnitude and direction of the vector $\stackrel{\to }{D}=\stackrel{\to }{A}+\stackrel{\to }{B}+\stackrel{\to }{C}$ and (b) the magnitude and direction of $\stackrel{\to }{E}=-\stackrel{\to }{A}-\stackrel{\to }{B}+\stackrel{\to }{C}$.

To determine

The magnitude and direction of the vector $\overrightarrow{D}$.

Answer to Problem 50P

The magnitude of the vector $\overrightarrow{D}$ is $6.32\text{m}$ and direction of the vector $\overrightarrow{D}$ is $342°$.

Explanation of Solution

Write the expression for the given vector $\overrightarrow{D}$.

  $\overrightarrow{D}=\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}$        (I)

Here, $\overrightarrow{A},\overrightarrow{B}$, $\overrightarrow{C}$ is the vectors.

Write the expression for magnitude of the given vector $\overrightarrow{D}$.

  $\left|\overrightarrow{D}\right|=\sqrt{D_x^2+D_y^2}$        (II)

Here, $\left|\overrightarrow{D}\right|$ is the magnitude of the vector, $D_x$ is the x component of the vector, and $D_y$ is the y component of the vector

Write the expression for the angle of the vector.

  $\theta ={\tan }^{-1}\left(\frac{D_y}{D_x}\right)$        (III)

Conclusion:

Substitute $\left(3\widehat{i}-3\widehat{j}\right)\text{m}$ for $\overrightarrow{A}$, $\left(\widehat{i}-4\widehat{j}\right)\text{m}$ for $\overrightarrow{B}$, and $\left(2\widehat{i}+5\widehat{j}\right)\text{m}$ for $\overrightarrow{C}$ in the equation (I) to find $\overrightarrow{D}$.

  $\begin{array}{c}\overrightarrow{D}=\left(3\widehat{i}-3\widehat{j}\right)\text{m}+\left(\widehat{i}-4\widehat{j}\right)\text{m}+\left(2\widehat{i}+5\widehat{j}\right)\text{m}\\ =\left(6\widehat{i}-2\widehat{j}\right)\text{m}\end{array}$

Substitute $6\text{m}\widehat{i}$ for $D_x$ and $-2\text{m}\widehat{j}$ for $D_y$ in the equation (II) to find $\left|\overrightarrow{D}\right|$.

  $\begin{array}{c}\left|\overrightarrow{D}\right|=\sqrt{{\left(6\text{m}\widehat{i}\right)}^2+{\left(-2\text{m}\widehat{j}\right)}^2}\\ =6.32\text{m}\end{array}$

Substitute $6\text{m}\widehat{i}$ for $D_x$ and $-2\text{m}\widehat{j}$ for $D_y$ in the equation (III) to find $\theta$.

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-2}{6}\right)\\ =-18.4°\\ \theta =+360°-18.4°\\ =342°\end{array}$

Therefore, the magnitude of the vector $\overrightarrow{D}$ is $6.32\text{m}$ and direction of the vector $\overrightarrow{D}$ is $342°$.

To determine

The magnitude and direction of the vector $\overrightarrow{E}$.

Answer to Problem 50P

The magnitude of the vector $\overrightarrow{E}$ is $12.2\text{m}$ and direction of the vector $\overrightarrow{E}$ is $99.5°$.

Explanation of Solution

Write the expression for the given vector $\overrightarrow{E}$.

  $\overrightarrow{E}=-\overrightarrow{A}-\overrightarrow{B}+\overrightarrow{C}$        (IV)

Write the expression for magnitude of the given vector $\overrightarrow{E}$.

  $\left|\overrightarrow{E}\right|=\sqrt{E_x^2+E_y^2}$        (V)

Here, $\left|\overrightarrow{E}\right|$ is the magnitude of the vector, $E_x$ is the x component of the vector, and $E_y$ is the y component of the vector

Write the expression for the angle of the vector.

  $\theta ={\tan }^{-1}\left(\frac{E_y}{E_x}\right)$        (VI)

Conclusion:

Substitute $\left(3\widehat{i}-3\widehat{j}\right)\text{m}$ for $\overrightarrow{A}$, $\left(\widehat{i}-4\widehat{j}\right)\text{m}$ for $\overrightarrow{B}$, and $\left(2\widehat{i}+5\widehat{j}\right)\text{m}$ for $\overrightarrow{C}$ in the equation (IV) to find $\overrightarrow{E}$.

  $\begin{array}{c}\overrightarrow{E}=-\left(3\widehat{i}-3\widehat{j}\right)\text{m}-\left(\widehat{i}-4\widehat{j}\right)\text{m}+\left(2\widehat{i}+5\widehat{j}\right)\text{m}\\ =\left(-2\widehat{i}+12\widehat{j}\right)\text{m}\end{array}$

Substitute $-2\text{m}\widehat{i}$ for $E_x$ and $12\text{m}\widehat{j}$ for $E_y$ in the equation (V) to find $\left|\overrightarrow{E}\right|$.

  $\begin{array}{c}\left|\overrightarrow{E}\right|=\sqrt{{\left(-2\text{m}\widehat{i}\right)}^2+{\left(12\text{m}\widehat{j}\right)}^2}\\ =12.2\text{m}\end{array}$

Substitute $-2\text{m}\widehat{i}$ for $E_x$ and $12\text{m}\widehat{j}$ for $E_y$ in the equation (VI) to find $\theta$.

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{12}{-2}\right)\\ =-80.5°\\ \theta =180°-80.5°\\ =99.5°\end{array}$

Therefore, the magnitude of the vector $\overrightarrow{E}$ is $12.2\text{m}$ and direction of the vector $\overrightarrow{E}$ is $99.5°$.