Chapter 3, Problem 31P

(a)

To determine

The magnitude and the direction of $\overrightarrow{\text{D}}=\overrightarrow{\text{A}}+\overrightarrow{\text{B}}+\overrightarrow{\text{C}}$.

Answer to Problem 31P

The magnitude of $\overrightarrow{\text{D}}=\overrightarrow{\text{A}}+\overrightarrow{\text{B}}+\overrightarrow{\text{C}}$ is $2.83\text{m}$ in the direction of $315°$ counterclockwise from the positive $x$-axis.

Explanation of Solution

Section 1:

To determine: The magnitude of $\overrightarrow{\text{D}}=\overrightarrow{\text{A}}+\overrightarrow{\text{B}}+\overrightarrow{\text{C}}$.

Answer: The magnitude of $\overrightarrow{\text{D}}=\overrightarrow{\text{A}}+\overrightarrow{\text{B}}+\overrightarrow{\text{C}}$ is $2.83\text{m}$.

Given information:

The three displacement vectors are $\overrightarrow{\text{A}}=\left(3\widehat{i}-3\widehat{j}\right)\text{m}$, $\overrightarrow{\text{B}}=\left(\widehat{i}-4\widehat{j}\right)\text{m}$ and $\overrightarrow{C}=\left(-2\widehat{i}+5\widehat{j}\right)\text{m}$.

Formula to calculate the magnitude of $\overrightarrow{\text{D}}=\overrightarrow{\text{A}}+\overrightarrow{\text{B}}+\overrightarrow{\text{C}}$ is,

$D=\sqrt{{\left(A_{\text{x}}+B_{\text{x}}+C_{\text{x}}\right)}^2+{\left(A_{\text{y}}+B_{\text{y}}+C_{\text{y}}\right)}^2}$

Substitute $3\text{m}$ for $A_{\text{x}}$, $-3\text{m}$ for $A_{\text{y}}$, $1\text{m}$ for $B_{\text{x}}$, $-4\text{m}$ for $B_{\text{y}}$, $-2\text{m}$ for $C_{\text{x}}$ and $5\text{m}$ for $C_{\text{y}}$ in above expression.

$\begin{array}{c}D=\sqrt{{\left(3\text{m}+1\text{m}+\left(-2\text{m}\right)\right)}^2+{\left(-3\text{m}+\left(-4\text{m}\right)+5\text{m}\right)}^2}\\ =2.83\text{m}\end{array}$

Conclusion:

Therefore, the magnitude of $\overrightarrow{\text{D}}=\overrightarrow{\text{A}}+\overrightarrow{\text{B}}+\overrightarrow{\text{C}}$ is $2.83\text{m}$.

Section 2:

To determine: The direction of $\overrightarrow{\text{D}}=\overrightarrow{\text{A}}+\overrightarrow{\text{B}}+\overrightarrow{\text{C}}$.

Answer: The direction of $\overrightarrow{\text{D}}=\overrightarrow{\text{A}}+\overrightarrow{\text{B}}+\overrightarrow{\text{C}}$ is $315°$ counterclockwise from positive $x$-axis.

Given information:

The three displacement vectors are $\overrightarrow{\text{A}}=\left(3\widehat{i}-3\widehat{j}\right)\text{m}$, $\overrightarrow{\text{B}}=\left(\widehat{i}-4\widehat{j}\right)\text{m}$ and $\overrightarrow{C}=\left(-2\widehat{i}+5\widehat{j}\right)\text{m}$.

Formula to calculate the direction of $\overrightarrow{\text{D}}=\overrightarrow{\text{A}}+\overrightarrow{\text{B}}+\overrightarrow{\text{C}}$ is,

$\theta ={\tan }^{-1}\left(\frac{A_{\text{y}}+B_{\text{y}}+C_{\text{y}}}{A_{\text{x}}+B_{\text{x}}+C_{\text{x}}}\right)$

Substitute $3\text{m}$ for $A_{\text{x}}$, $-3\text{m}$ for $A_{\text{y}}$, $1\text{m}$ for $B_{\text{x}}$, $-4\text{m}$ for $B_{\text{y}}$, $-2\text{m}$ for $C_{\text{x}}$ and $5\text{m}$ for $C_{\text{y}}$ in above expression.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{\left(-3\text{m}+\left(-4\text{m}\right)+5\text{m}\right)}{\left(3\text{m}+1\text{m}+\left(-2\text{m}\right)\right)}\right)\\ ={\tan }^{-1}\left(\frac{-2\text{m}}{2\text{m}}\right)\\ =-45°\end{array}$

Thus, the angle measured from counterclockwise is $360°-45°=315°$

Conclusion:

Therefore, the direction of $\overrightarrow{\text{D}}=\overrightarrow{\text{A}}+\overrightarrow{\text{B}}+\overrightarrow{\text{C}}$ is $315°$ counterclockwise from positive $x$-axis.

(b)

To determine

The magnitude and the direction of $\overrightarrow{\text{E}}=-\overrightarrow{\text{A}}-\overrightarrow{\text{B}}+\overrightarrow{\text{C}}$.

Answer to Problem 31P

The magnitude of $\overrightarrow{\text{E}}=-\overrightarrow{\text{A}}-\overrightarrow{\text{B}}+\overrightarrow{\text{C}}$ is $13.42\text{m}$ in the direction of $296.57°$ counterclockwise from the positive $x$-axis.

Explanation of Solution

Section 1:

To determine: The magnitude of $\overrightarrow{\text{E}}=-\overrightarrow{\text{A}}-\overrightarrow{\text{B}}+\overrightarrow{\text{C}}$.

Answer: The magnitude of $\overrightarrow{\text{E}}=-\overrightarrow{\text{A}}-\overrightarrow{\text{B}}+\overrightarrow{\text{C}}$ is $13.42\text{m}$.

Given information:

The three displacement vectors are $\overrightarrow{\text{A}}=\left(3\widehat{i}-3\widehat{j}\right)\text{m}$, $\overrightarrow{\text{B}}=\left(\widehat{i}-4\widehat{j}\right)\text{m}$ and $\overrightarrow{C}=\left(-2\widehat{i}+5\widehat{j}\right)\text{m}$.

Formula to calculate the magnitude of $\overrightarrow{\text{E}}=-\overrightarrow{\text{A}}-\overrightarrow{\text{B}}+\overrightarrow{\text{C}}$ is,

$E=\sqrt{{\left(\left(-A_{\text{x}}\right)+\left(-B_{\text{x}}\right)+C_{\text{x}}\right)}^2+{\left(\left(-A_{\text{y}}\right)+\left(-B_{\text{y}}\right)+C_{\text{y}}\right)}^2}$

Substitute $3\text{m}$ for $A_{\text{x}}$, $-3\text{m}$ for $A_{\text{y}}$, $1\text{m}$ for $B_{\text{x}}$, $-4\text{m}$ for $B_{\text{y}}$, $-2\text{m}$ for $C_{\text{x}}$ and $5\text{m}$ for $C_{\text{y}}$ in above expression.

$\begin{array}{c}E=\sqrt{{\left(\left(-3\text{m}\right)+\left(-1\text{m}\right)+\left(-2\text{m}\right)\right)}^2+{\left(3\text{m}+4\text{m}+5\text{m}\right)}^2}\\ =\sqrt{180\text{m}}\\ =13.42\text{m}\end{array}$

Conclusion:

Therefore, the magnitude of $\overrightarrow{\text{E}}=-\overrightarrow{\text{A}}-\overrightarrow{\text{B}}+\overrightarrow{\text{C}}$ is $13.42\text{m}$.

Section 2:

To determine: The direction of $\overrightarrow{\text{E}}=-\overrightarrow{\text{A}}-\overrightarrow{\text{B}}+\overrightarrow{\text{C}}$.

Answer: The direction of $\overrightarrow{\text{E}}=-\overrightarrow{\text{A}}-\overrightarrow{\text{B}}+\overrightarrow{\text{C}}$ is $296.57°$ counterclockwise from the positive $x$-axis

Given information:

The three displacement vectors are $\overrightarrow{\text{A}}=\left(3\widehat{i}-3\widehat{j}\right)\text{m}$, $\overrightarrow{\text{B}}=\left(\widehat{i}-4\widehat{j}\right)\text{m}$ and $\overrightarrow{C}=\left(-2\widehat{i}+5\widehat{j}\right)\text{m}$.

Formula to determine the direction of $\overrightarrow{\text{E}}=-\overrightarrow{\text{A}}-\overrightarrow{\text{B}}+\overrightarrow{\text{C}}$ is,

$\theta ={\tan }^{-1}\left(\frac{\left(-A_{\text{y}}\right)+\left(-B_{\text{y}}\right)+C_{\text{y}}}{\left(-A_{\text{x}}\right)+\left(-B_{\text{x}}\right)+C_{\text{x}}}\right)$

Substitute $3\text{m}$ for $A_{\text{x}}$, $-3\text{m}$ for $A_{\text{y}}$, $1\text{m}$ for $B_{\text{x}}$, $-4\text{m}$ for $B_{\text{y}}$, $-2\text{m}$ for $C_{\text{x}}$ and $5\text{m}$ for $C_{\text{y}}$ in above expression.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{\left(3\text{m}+\left(4\text{m}\right)+5\text{m}\right)}{\left(-3\text{m}+\left(-1\text{m}\right)+\left(-2\text{m}\right)\right)}\right)\\ ={\tan }^{-1}\left(\frac{12\text{m}}{-6\text{m}}\right)\\ =-63.43°\end{array}$

Thus, the angle measured from counterclockwise is $360°-63.43°=296.57°$

Conclusion:

Therefore, the direction of $\overrightarrow{\text{E}}=-\overrightarrow{\text{A}}-\overrightarrow{\text{B}}+\overrightarrow{\text{C}}$ is $296.57°$ counterclockwise from the positive $x$-axis.