Chapter 2, Problem 26P

(a)

To determine

The $x$ and $y$ components of $\overrightarrow{b}$.

Answer to Problem 26P

The $x$ and $y$ components of $\overrightarrow{b}$ are $7$ and $2$ respectively.

Explanation of Solution

The magnitude of $\overrightarrow{b}$ is $7.1$ and it is in direction $14°$ the $+x$ axis.

Write the expression for $x$ component of $\overrightarrow{b}$.

  $b_x=b\cos \theta$        (I)

Here, $b$ is the magnitude of $\overrightarrow{b}$, and $\theta$ is the angle inclined by $\overrightarrow{b}$ with $x$ axis.

Write the expression for $y$ component of $\overrightarrow{b}$.

  $b_y=b\sin \theta$        (II)

Conclusion:

Substitute, $7.1$ for $b$, $14°$ for $\theta$ in equation (I) to find the $x$ component of $\overrightarrow{b}$.

  $\begin{array}{c}{b}_x=\left(7.1\right)\cos \left(14°\right)\\ =7.1\times 0.970\\ =6.9\end{array}$

Substitute, $7.1$ for $b$, $14°$ for $\theta$ in equation (II) to find the $y$ component of $\overrightarrow{b}$.

  $\begin{array}{c}{b}_y=\left(7.1\right)\sin \left(14°\right)\\ =7.1\times 0.24\\ =1.7\end{array}$

Hence the vector representation of $\overrightarrow{b}$ is.

  $\begin{array}{c}\overrightarrow{b}={\text{b}}_x\widehat{i}+{\text{b}}_y\widehat{j}\\ =6.9\widehat{i}+1.7\widehat{j}\end{array}$

Conclusion:

Therefore, the $x$ and $y$ components of $\overrightarrow{b}$ are $6.9$ and $1.7$ respectively.

(b)

To determine

The magnitude and direction of $\overrightarrow{c}$.

Answer to Problem 26P

The magnitude of $\overrightarrow{c}$ is $\underset{\_}{6.94}$ and direction of $\overrightarrow{c}$ $\underset{\_}{74.9°}$.

Explanation of Solution

The $x$ component of $\overrightarrow{c}$ is $-1.8$, and $y$ component of $\overrightarrow{c}$ is $-6.7$.

Write the vector expression for $\overrightarrow{c}$.

  $\overrightarrow{c}=c_x\widehat{i}+c_y\widehat{j}$        (III)

Here, $c_x$ is the $x$ component of $\overrightarrow{c}$, and $c_y$ is the $y$ component of $\overrightarrow{c}$.

Write the equation to find the magnitude of $\overrightarrow{c}$.

  $c=\sqrt{c_x^2+c_y^2}$        (IV)

Write the expression for the direction of $\overrightarrow{c}$.

  $\theta ={\tan }^{-1}\left(\frac{c_y}{c_x}\right)$        (V)

Conclusion:

Substitute, $-1.8$ for $c_x$, and $-6.7$ for $c_y$ in equation (IV) to find the magnitude of $\overrightarrow{c}$.

  $\begin{array}{c}c=\sqrt{{\left(-1.8\right)}^2+{\left(-6.7\right)}^2}\\ =6.94\end{array}$

Substitute, $-1.8$ for $c_x$, and $-6.7$ for $c_y$ in equation (V) to find the direction of $\overrightarrow{c}$.

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-6.7}{-1.8}\right)\\ =74.9°\end{array}$

Therefore, the magnitude of $\overrightarrow{c}$ is $\underset{\_}{6.94}$ and direction of $\overrightarrow{c}$ $\underset{\_}{74.9°}$.

(c)

To determine

The magnitude and direction of $\overrightarrow{c}+\overrightarrow{b}$.

Answer to Problem 26P

The magnitude of $\overrightarrow{c}+\overrightarrow{b}$ is $\underset{\_}{7.14}$, and direction of $\overrightarrow{c}+\overrightarrow{b}$ is .

Explanation of Solution

Let the sum of vector $c$ and $b$ be vector $a$.

Write the vector expression for $\overrightarrow{a}$.

  $\begin{array}{c}\overrightarrow{a}=\overrightarrow{c}+\overrightarrow{b}\\ a_x\widehat{i}+a_y\widehat{j}=\left(c_x\widehat{i}+c_y\widehat{j}\right)+\left(b_x\widehat{i}+b_y\widehat{j}\right)\end{array}$        (VI)

Write the expression for finding the magnitude of $\overrightarrow{a}$.

  $a=\sqrt{a_x^2+a_y^2}$        (VII)

Write the expression for finding the direction of $\overrightarrow{a}$.

  $\theta ={\tan }^{-1}\left(\frac{a_y}{a_x}\right)$        (VIII)

Conclusion:

Substitute, $6.9$ for $b_x$, $1.7$ for $b_y$, $-1.8$ for $c_x$, and $-6.7$ for $c_y$ in equation (VI).

  $\begin{array}{c}{a}_x\widehat{i}+a_y\widehat{j}=\left(-1.8\widehat{i}+-{6.7}_y\widehat{j}\right)+\left(6.9\widehat{i}+1.7\widehat{j}\right)\\ =5.1\widehat{i}-5\widehat{j}\end{array}$        (IX)

Substitute, $5.1$ for $a_x$, and $-5$ for $a_y$ in equation (VII) to find the magnitude of $\overrightarrow{a}$.

  $\begin{array}{c}a=\sqrt{{\left(5.1\right)}^2+{\left(-5\right)}^2}\\ =7.14\end{array}$

Substitute, $5.1$ for $a_x$, and $-5$ for $a_y$ in equation (VIII) to find the direction of $\overrightarrow{a}$.

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-5}{5.1}\right)\\ =\end{array}$

Therefore, the magnitude of $\overrightarrow{c}+\overrightarrow{b}$ is $\underset{\_}{7.14}$, and direction of $\overrightarrow{c}+\overrightarrow{b}$ is .