Chapter 1, Problem 57P

(a)

To determine

The components of $\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}$.

Answer to Problem 57P

The components of $\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}$, are $\underset{\_}{C_x=22}$, and $\underset{\_}{C_y=30}$, magnitude of C is $\underset{\_}{37}$, and direction is $\underset{\_}{53°}$.

Explanation of Solution

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

  $C_x=A_x+B_x$        (I)

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

  $C_y=A_y+B_y$        (II)

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

  $C=\sqrt{C_x^2+C_y^2}$        (III)

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

  $\theta ={\tan }^{-1}\left(\frac{C_y}{C_x}\right)$        (IV)

Conclusion:

Substitute, $13.6$ for $A_x$, and $8.55$ for $B_x$ in equation (I) to find $C_x$.

  $\begin{array}{c}{C}_x=13.6+8.55\\ =22.2\end{array}$

Substitute, $6.34$ for $A_y$, and $23.5$ for $B_y$ in equation (I) to find $C_y$.

  $\begin{array}{c}{C}_y=6.34+23.5\\ =29.8\end{array}$

Substitute, $22.2$ for $C_x$, and $29.8$ for $C_y$ in equation (III) to find the magnitude of vector C.

  $\begin{array}{c}C=\sqrt{{\left(22.2\right)}^2+{\left(29.8\right)}^2}\\ =37\end{array}$

Substitute, $22.2$ for $C_x$, and $29.8$ for $C_y$ in equation (IV) to find the direction of vector C.

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{29.8}{22.2}\right)\\ =53°\end{array}$

Therefore, the components of $\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}$, are $\underset{\_}{C_x=22}$, and $\underset{\_}{C_y=30}$, magnitude of C is $\underset{\_}{37}$, and direction is $\underset{\_}{53°}$.

(b)

To determine

The components of $\overrightarrow{C}=\overrightarrow{A}-\overrightarrow{B}$.

Answer to Problem 57P

The components of $\overrightarrow{C}=\overrightarrow{A}-\overrightarrow{B}$, are $\underset{\_}{C_x=5.05}$, and $\underset{\_}{C_y=-17.6}$, magnitude of C is $\underset{\_}{18}$, and direction is $\underset{\_}{-74°}$.

Explanation of Solution

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

  $C_x=A_x-B_x$        (V)

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

  $C_y=A_y-B_y$        (VI)

Conclusion:

Substitute, $13.6$ for $A_x$, and $8.55$ for $B_x$ in equation (V) to find $C_x$.

  $\begin{array}{c}{C}_x=13.6-8.55\\ =5.05\end{array}$

Substitute, $6.34$ for $A_y$, and $23.5$ for $B_y$ in equation (VI) to find $C_y$.

  $\begin{array}{c}{C}_y=6.34-23.5\\ =-17.16\end{array}$

Substitute, $5.05$ for $C_x$, and $-17.2$ for $C_y$ in equation (III) to find the magnitude of vector C.

  $\begin{array}{c}C=\sqrt{{\left(5.05\right)}^2+{\left(-17.2\right)}^2}\\ =18\end{array}$

Substitute, $5.05$ for $C_x$, and $-17.2$ for $C_y$  in equation (IV) to find the direction of vector C.

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-17.16}{5.05}\right)\\ =-74°\end{array}$

Therefore, the components of $\overrightarrow{C}=\overrightarrow{A}-\overrightarrow{B}$, are $\underset{\_}{C_x=5.05}$, and $\underset{\_}{C_y=-17.6}$, magnitude of C is $\underset{\_}{18}$, and direction is $\underset{\_}{-74°}$.

(c)

To determine

The components of $\overrightarrow{C}=\overrightarrow{A}+4\overrightarrow{B}$.

Answer to Problem 57P

The components of $\overrightarrow{C}=\overrightarrow{A}+4\overrightarrow{B}$, are $\underset{\_}{C_x=47.8}$, and $\underset{\_}{C_y=100}$, magnitude of C is $\underset{\_}{110}$, and direction is $\underset{\_}{64°}$.

Explanation of Solution

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

  $C_x=A_x+4B_x$        (VII)

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

  $C_y=A_y+4B_y$        (VIII)

Conclusion:

Substitute, $13.6$ for $A_x$, and $8.55$ for $B_x$ in equation (VII) to find $C_x$.

  $\begin{array}{c}{C}_x=13.6+4\left(8.55\right)\\ =47.8\end{array}$

Substitute, $6.34$ for $A_y$, and $23.5$ for $B_y$ in equation (VIII) to find $C_y$.

  $\begin{array}{c}{C}_y=6.34-4\left(23.5\right)\\ =100\end{array}$

Substitute, $47.8$ for $C_x$, and $100$ for $C_y$ in equation (III) to find the magnitude of vector C.

  $\begin{array}{c}C=\sqrt{{\left(47.8\right)}^2+{\left(100\right)}^2}\\ =110\end{array}$

Substitute, $47.8$ for $C_x$, and $100$ for $C_y$ in equation (IV) to find the direction of vector C.

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{100}{47.8}\right)\\ =64°\end{array}$

Therefore, the components of $\overrightarrow{C}=\overrightarrow{A}+4\overrightarrow{B}$, are $\underset{\_}{C_x=47.8}$, and $\underset{\_}{C_y=100}$, magnitude of C is $\underset{\_}{110}$, and direction is $\underset{\_}{64°}$.

(b)

To determine

The components of $\overrightarrow{C}=-\overrightarrow{A}-7\overrightarrow{B}$.

Answer to Problem 57P

The components of $\overrightarrow{C}=-\overrightarrow{A}-7\overrightarrow{B}$, are $\underset{\_}{C_x=-73.5}$, and $\underset{\_}{C_y=-171}$, magnitude of C is $\underset{\_}{190}$, and direction is $\underset{\_}{67°}$.

Explanation of Solution

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

  $C_x=-A_x-7B_x$        (IX)

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

  $C_y=-A_y-7B_y$        (X)

Conclusion:

Substitute, $13.6$ for $A_x$, and $8.55$ for $B_x$ in equation (IX) to find $C_x$.

  $\begin{array}{c}{C}_x=-13.6-7\left(8.55\right)\\ =-73.5\end{array}$

Substitute, $6.34$ for $A_y$, and $23.5$ for $B_y$ in equation (X) to find $C_y$.

  $\begin{array}{c}{C}_y=-6.34-7\left(23.5\right)\\ =-171\end{array}$

Substitute, $-73.5$ for $C_x$, and $-171$ for $C_y$ in equation (III) to find the magnitude of vector C.

  $\begin{array}{c}C=\sqrt{{\left(-73.5\right)}^2+{\left(-171\right)}^2}\\ =190\end{array}$

Substitute, $-73.5$ for $C_x$, and $-171$ for $C_y$ in equation (IV) to find the direction of vector C.

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-171}{-73.5}\right)\\ =67°\end{array}$

Therefore, components of $\overrightarrow{C}=-\overrightarrow{A}-7\overrightarrow{B}$, are $\underset{\_}{C_x=-73.5}$, and $\underset{\_}{C_y=-171}$, magnitude of C is $\underset{\_}{190}$, and direction is $\underset{\_}{67°}$.