Chapter 3, Problem 13P

Solution

1. $\overrightarrow{A}-\overrightarrow{B}+\overrightarrow{C}$
2. $\overrightarrow{A}+\overrightarrow{B}-\overrightarrow{C}$
3. $\overrightarrow{C}-\overrightarrow{A}-\overrightarrow{B}$

Solution:

1. The magnitude $\overrightarrow{A}-\overrightarrow{B}+\overrightarrow{C}$ is 62.6 with a direction of ${328.9}^{\text{o}}$ counterclockwise from + x -axis.
2. The magnitude $\overrightarrow{A}+\overrightarrow{B}-\overrightarrow{C}$ is 77.4 with a direction of ${71.9}^{\text{o}}$ counterclockwise from +x-axis.
3. The magnitude $\overrightarrow{C}-\overrightarrow{A}-\overrightarrow{B}$ of resultant vector is 77.4 with a direction of ${251.9}^{\text{o}}$ counterclockwise from +x-axis.

Explanation:

Formula used:

Given: The magnitude $\left|\overrightarrow{A}\right|$ is 44.0 with a direction $\theta {\text{'}}_A$ of $28.0˚$ counterclockwise from the positive x -axis. For $\left|\overrightarrow{B}\right|$ the magnitude is 26.5with a direction $\theta {\text{'}}_B$ of $56.0˚$ clockwise from the negative x -axis and the magnitude of $\left|\overrightarrow{C}\right|$ is 31.0 with a direction of $\theta {\text{'}}_C$ of $270˚$ on the negative y-axis.

Formula used:

The resultant $\overrightarrow{R}$ in terms of its components can be defined as:

The x-component of the resultant $R_x$ can be obtained using:

  $R_x=A_x+B_x+C_x=\left|\overrightarrow{A}\right|\cos ({\theta }_A)+\left|\overrightarrow{B}\right|\cos ({\theta }_B)+\left|\overrightarrow{C}\right|\cos ({\theta }_C)$

The y-component of the resultant $R_y$ can be obtained using:

  $R_y=A_y+B_y+C_y=\left|\overrightarrow{A}\right|\sin ({\theta }_A)+\left|\overrightarrow{B}\right|\sin ({\theta }_B)+\left|\overrightarrow{C}\right|\sin ({\theta }_C)$

Where:

• ${\theta }_A$ , ${\theta }_B$ and ${\theta }_C$ are the directions for vectors $\overrightarrow{A},\overrightarrow{B}\text{and}\overrightarrow{C}$ respectively measured counterclockwise from the + x -axis.
• $A_y$ , $B_y$ and $C_y$ are the y-components for vectors $\overrightarrow{A},\overrightarrow{B}\text{and}\overrightarrow{C}$ respectively.
• $A_x$ , $B_x$ and $C_x$ are the x-components for vectors $\overrightarrow{A},\overrightarrow{B}\text{and}\overrightarrow{C}$ respectively.
• $\left|\overrightarrow{A}\right|$ , $$ \overrightarrow{A},\overrightarrow{B}\text{and}\overrightarrow{C}$\left|\overrightarrow{B}\right|$ and $\left|\overrightarrow{C}\right|$ are the magnitudes for vectors respectively.

The resultant $\left|\overrightarrow{R}\right|$ of the sum of three vectors $\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}$ in terms of its magnitude $\left|\overrightarrow{R}\right|$ can be obtained using:

  $\left|\overrightarrow{R}\right|=\sqrt{R_x{}^2+R_y{}^2}$

Similarly, the direction (angle) ${\theta }_R$ is defined as:

  ${\theta }_R={\tan }^{-1}\left(\frac{R_y}{R_x}\right)$

Where:

• $R_x$ is the x -component of the resultant vector $\overrightarrow{R}$ .
• $R_y$ is the y -component of the resultant vector $\overrightarrow{R}$ .

Calculation:

x -component of $\overrightarrow{A}$

  $\begin{array}{c}{A}_x=44\cos {28}^o\\ =38.8\end{array}$

y -component of $\overrightarrow{A}$

  $\begin{array}{c}{A}_y=44\sin {28}^o\\ =20.6\end{array}$

x -component of $\overrightarrow{B}$

  $\begin{array}{c}{B}_x=26.5\cos \left({180}^o-{56}^o\right)\\ =-14.8\end{array}$

y -component of $\overrightarrow{B}$

  $\begin{array}{c}{B}_y=26.5\sin \left({180}^o-{56}^o\right)\\ =22\end{array}$

x -component of $\overrightarrow{C}$

  $\begin{array}{c}{C}_x=31\cos \left({270}^o\right)\\ =0\end{array}$

y -component of $\overrightarrow{C}$

  $\begin{array}{c}{C}_y=31\sin \left({270}^o\right)\\ =-31\end{array}$

a. $\overrightarrow{R}=\overrightarrow{A}-\overrightarrow{B}+\overrightarrow{C}$

  $\begin{array}{l}{R}_x=A_x-B_x+C_x\\ R_x=\left|\overrightarrow{A}\right|\cos ({\theta }_A)-\left|\overrightarrow{B}\right|\cos ({\theta }_B)+\left|\overrightarrow{C}\right|\cos ({\theta }_C)\\ R_x=\left(44.0\right)\cos ({28}^o)-\left(26.5\right)\cos \left({180}^o-{56}^o\right)+\left(31.0\right)\cos ({270}^o)\\ R_x=53.6\end{array}$

  $\begin{array}{l}{R}_y=A_y-B_y+C_y\\ R_y=\left|\overrightarrow{A}\right|\sin ({\theta }_A)-\left|\overrightarrow{B}\right|\sin ({\theta }_B)+\left|\overrightarrow{C}\right|\sin ({\theta }_C)\\ R_y=44.0\sin ({28.0}^{\circ })-\left(26.5\right)\sin ({180}^{\circ }-{56.0}^{\circ })+\left(31.0\right)\sin ({270}^{\circ })\\ R_y=-32.3\end{array}$

The magnitude can be calculated as

  $\begin{array}{l}\left|\overrightarrow{R}\right|=\sqrt{R_x{}^2+R_y{}^2}\\ =\sqrt{{\left(53.6\right)}^2+{\left(-32.3\right)}^2}\\ =62.6\end{array}$

  $\begin{array}{l}{\theta }_R={\tan }^{-1}\left(\frac{-32.3}{53.6}\right)\\ {\theta }_R=-31.{\text{1}}^{\text{o}}\\ {\theta }_R={360}^{\text{o}}-31.{\text{1}}^{\text{o}}\\ {\theta }_R={328.9}^{\text{o}}\end{array}$

b. $\overrightarrow{A}+\overrightarrow{B}-\overrightarrow{C}$

  $\begin{array}{l}{R}_x=A_x+B_x-C_x\\ R_x=\left|\overrightarrow{A}\right|\cos ({\theta }_A)+\left|\overrightarrow{B}\right|\cos ({\theta }_B)-\left|\overrightarrow{C}\right|\cos ({\theta }_C)\\ R_x=\left(44.0\right)\cos ({28}^o)+\left(26.5\right)\cos \left({180}^o-{56}^o\right)-\left(31.0\right)\cos ({270}^o)\\ R_x=24\end{array}$

  $\begin{array}{l}{R}_y=A_y+B_y-C_y\\ R_y=\left|\overrightarrow{A}\right|\sin ({\theta }_A)+\left|\overrightarrow{B}\right|\sin ({\theta }_B)-\left|\overrightarrow{C}\right|\sin ({\theta }_C)\\ R_y=44.0\sin ({28.0}^{\circ })+\left(26.5\right)\sin ({180}^{\circ }-{56.0}^{\circ })-\left(31.0\right)\sin ({270}^{\circ })\\ R_y=73.6\end{array}$

The magnitude can be calculated as

  $\begin{array}{c}\left|\overrightarrow{R}\right|=\sqrt{R_x{}^2+R_y{}^2}\\ =\sqrt{{\left(24\right)}^2+{\left(73.6\right)}^2}\\ =77.4\end{array}$

  $\begin{array}{l}{\theta }_R={\tan }^{-1}\left(\frac{73.6}{24}\right)\\ {\theta }_R={71.9}^{\text{o}}\end{array}$

c. $\overrightarrow{C}-\overrightarrow{A}-\overrightarrow{B}$

This is equal to the negative of previous part.

  $\overrightarrow{C}-\overrightarrow{A}-\overrightarrow{B}=-\left(\overrightarrow{A}+\overrightarrow{B}-\overrightarrow{C}\right)$

Thus, there will be no change in the resultant but its direction would be opposite.

  $\begin{array}{l}\left|\overrightarrow{R}\right|=\left|\overrightarrow{C}-\overrightarrow{A}-\overrightarrow{B}\right|=77.4\\ \theta ={180}^{\text{o}}+{71.9}^{\text{o}}={251.9}^{\text{o}}\end{array}$

Conclusion:

1. The magnitude $\overrightarrow{A}-\overrightarrow{B}+\overrightarrow{C}$ is 62.6 with a direction of ${328.9}^{\text{o}}$ counterclockwise from + x -axis.
2. The magnitude $\overrightarrow{A}+\overrightarrow{B}-\overrightarrow{C}$ is 77.4 with a direction of ${71.9}^{\text{o}}$ counterclockwise from +x-axis.
3. The magnitude $\overrightarrow{C}-\overrightarrow{A}-\overrightarrow{B}$ of resultant vector is 77.4 with a direction of ${251.9}^{\text{o}}$ counterclockwise from +x-axis.