Chapter 1, Problem 56P

(a)

To determine

The magnitude and direction of the vector.

Answer to Problem 56P

The magnitude and direction of the vector is $37$ and $53°$ respectively.

Explanation of Solution

Write the expression to determine the magnitude of the vector.

  $C=\sqrt{C_x^2+C_y^2}$        (I)

Write the expression to determine the direction of the vector.

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

Calculate the components of each given vectors by substituting $15$ for $A$ , $25$ for $B$ and $\theta =2\text{5°and 70°}$

  $\begin{array}{l}{\text{A}}_{\text{x}}\text{=Acosq = }\left(\text{15}\right)\text{cos}\left(\text{25º}\right)\text{ = 13}\text{.6}\\ \begin{array}{c}{\text{A}}_{\text{y}}\text{ =Asinq = }\left(\text{15}\right)\text{sin}\left(\text{25º}\right)\text{ = 6}\text{.34}\\ {\text{B}}_{\text{x}}\text{ =Bcosq = }\left(\text{25}\right)\text{cos}\left(\text{70º}\right)\text{ = 8}\text{.55}\\ {\text{B}}_{\text{y}}\text{ =Bsinq = }\left(\text{25}\right)\text{sin}\left(\text{70º}\right)\text{ = 23}\text{.5}\end{array}\end{array}$

Solve for $\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}$ by substituting the values from above expression,

  $\begin{array}{l}\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}\\ C_x=A_x+B_x=22.2\\ C_x=13.6+8.55=22.2\end{array}$

  $\begin{array}{l}{C}_y=A_y+B_y=29.8\\ C_y=6.34+23.5=29.8\end{array}$

The magnitude of the vector from expression (I)

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

The direction of the vector from expression (II)

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

Conclusion:

The magnitude and direction of the vector is $37$ and $53°$ respectively.

(b)

To determine

The magnitude and direction of the vector.

Answer to Problem 56P

The magnitude and direction of the vector is $18$ and $-74°$ respectively.

Explanation of Solution

Write the expression to determine the magnitude of the vector.

  $C=\sqrt{C_x^2+C_y^2}$        (I)

Write the expression to determine the direction of the vector.

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

Calculate the components of each given vectors by substituting $15$ for $A$ , $25$ for $B$ and $\theta =2\text{5°and 70°}$

  $\begin{array}{l}{\text{A}}_{\text{x}}\text{=Acosq = }\left(\text{15}\right)\text{cos}\left(\text{25º}\right)\text{ = 13}\text{.6}\\ \begin{array}{c}{\text{A}}_{\text{y}}\text{ =Asinq = }\left(\text{15}\right)\text{sin}\left(\text{25º}\right)\text{ = 6}\text{.34}\\ {\text{B}}_{\text{x}}\text{ =Bcosq = }\left(\text{25}\right)\text{cos}\left(\text{70º}\right)\text{ = 8}\text{.55}\\ {\text{B}}_{\text{y}}\text{ =Bsinq = }\left(\text{25}\right)\text{sin}\left(\text{70º}\right)\text{ = 23}\text{.5}\end{array}\end{array}$

Solve for $\overrightarrow{C}=\overrightarrow{A}-\overrightarrow{B}$ by substituting the values from above expression,

  $\begin{array}{c}\overrightarrow{C}=\overrightarrow{A}-\overrightarrow{B}\\ C_x=A_x-B_x\\ C_x=13.6-8.55\\ =5.04\end{array}$

  $\begin{array}{c}{C}_y=A_y-B_y\\ C_y=6.34-23.5\\ =-17.2\end{array}$

The magnitude of the vector from expression (I)

  $\begin{array}{c}C=\sqrt{{\left(5.04\right)}^2+{\left(-17.2\right)}^2}\\ =18\end{array}$

The direction of the vector from expression (II)

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-17.2}{5.04}\right)\\ =-74°\end{array}$

Conclusion:

The magnitude and direction of the vector is $18$ and $-74°$ respectively.

(c)

To determine

The magnitude and direction of the vector.

Answer to Problem 56P

The magnitude and direction of the vector is $110$ and $64°$ respectively.

Explanation of Solution

Write the expression to determine the magnitude of the vector.

  $C=\sqrt{C_x^2+C_y^2}$        (I)

Write the expression to determine the direction of the vector.

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

Calculate the components of each given vectors by substituting $15$ for $A$ , $25$ for $B$ and $\theta =2\text{5°and 70°}$

  $\begin{array}{l}{\text{A}}_{\text{x}}\text{=Acosq = }\left(\text{15}\right)\text{cos}\left(\text{25º}\right)\text{ = 13}\text{.6}\\ \begin{array}{c}{\text{A}}_{\text{y}}\text{ =Asinq = }\left(\text{15}\right)\text{sin}\left(\text{25º}\right)\text{ = 6}\text{.34}\\ {\text{B}}_{\text{x}}\text{ =Bcosq = }\left(\text{25}\right)\text{cos}\left(\text{70º}\right)\text{ = 8}\text{.55}\\ {\text{B}}_{\text{y}}\text{ =Bsinq = }\left(\text{25}\right)\text{sin}\left(\text{70º}\right)\text{ = 23}\text{.5}\end{array}\end{array}$

Solve for $\overrightarrow{C}=\overrightarrow{A}+4\overrightarrow{B}$ by substituting the values from above expression,

  $\begin{array}{c}\overrightarrow{C}=\overrightarrow{A}+4\overrightarrow{B}\\ C_x=A_x+4B_x\\ C_x=13.6+4\times 8.55\\ =47.8\end{array}$

  $\begin{array}{c}{C}_y=A_y+4B_y\\ C_y=6.34+4\times 23.5\\ =100\end{array}$

The magnitude of the vector from expression (I)

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

The direction of the vector from expression (II)

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

Conclusion:

The magnitude and direction of the vector is $110$ and $64°$ respectively.

(d)

To determine

The magnitude and direction of the vector.

Answer to Problem 56P

The magnitude and direction of the vector is $190$ and $180+\theta$ that is $250°$ respectively.

Explanation of Solution

Write the expression to determine the magnitude of the vector.

  $C=\sqrt{C_x^2+C_y^2}$        (I)

Write the expression to determine the direction of the vector.

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

Calculate the components of each given vectors by substituting $15$ for $A$ , $25$ for $B$ and $\theta =2\text{5°and 70°}$

  $\begin{array}{l}{\text{A}}_{\text{x}}\text{=Acosq = }\left(\text{15}\right)\text{cos}\left(\text{25º}\right)\text{ = 13}\text{.6}\\ \begin{array}{c}{\text{A}}_{\text{y}}\text{ =Asinq = }\left(\text{15}\right)\text{sin}\left(\text{25º}\right)\text{ = 6}\text{.34}\\ {\text{B}}_{\text{x}}\text{ =Bcosq = }\left(\text{25}\right)\text{cos}\left(\text{70º}\right)\text{ = 8}\text{.55}\\ {\text{B}}_{\text{y}}\text{ =Bsinq = }\left(\text{25}\right)\text{sin}\left(\text{70º}\right)\text{ = 23}\text{.5}\end{array}\end{array}$

Solve for $\overrightarrow{C}=-\overrightarrow{A}-7\overrightarrow{B}$ by substituting the values from above expression,

  $\begin{array}{c}\overrightarrow{C}=-\overrightarrow{A}-7\overrightarrow{B}\\ C_x=-A_x-7B_x\\ C_x=-13.6-7\times 8.55\\ =-73.5\end{array}$

  $\begin{array}{c}{C}_y=-A_y-7B_y\\ C_y=-6.34-7\times 23.5\\ =-171\end{array}$

The magnitude of the vector from expression (I)

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

The direction of the vector from expression (II)

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-171}{-73.8}\right)\\ =67°\end{array}$

Conclusion:

The magnitude and direction of the vector is $190$ and $67°$ respectively. Here as both components are negative, the vector $\overrightarrow{C}$ will be in third quadrant. Thus the angle will be $180+\theta$ that is $250°$