Chapter 3, Problem 12P

For the vectors shown in Fig. 3—35, determine (a) $\stackrel{\to }{\text{B}}-\text{3}\stackrel{\to }{\text{A}}$ , (b) $2\stackrel{\to }{\text{A}}-3\stackrel{\to }{\text{B}}\text{ + 3}\stackrel{\to }{\text{C}}$

Solution

To Determine:

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

Solution:

1. The magnitude of resultant vector $\overrightarrow{B}-3\overrightarrow{A}$ is 137.1 with a direction of $196.9˚$ counterclockwise from +x-axis.
2. The magnitude of resultant vector $\overrightarrow{R}=2\overrightarrow{A}-3\overrightarrow{B}+2\overrightarrow{C}$ is 149.7 with a direction of $324.6˚$ counterclockwise from +x-axis.

Explanation:

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 x -component $V_x$ of vector $\overrightarrow{V}$ is defined as:

  $V_x=\left|\overrightarrow{V}\right|\cos \theta$

The y-component of $V_y$ vector $\overrightarrow{V}$ is defined as:

  $V_y=\left|\overrightarrow{V}\right|\sin \theta$

Where:

• $\left|\overrightarrow{V}\right|$ is the magnitude of the vector $\overrightarrow{V}$
• $\theta$ is the direction of the vector $\overrightarrow{V}$ measured respect to the positive x -axis counter-clockwise.

The magnitude of vector can be obtained by:

  $\left|\overrightarrow{V}\right|=\sqrt{\left(V_x{}^2\right)+{\left(V_y\right)}^2}$

And, the direction θ is defined as:

  $\theta ={\tan }^{-1}\left(\frac{V_y}{V_x}\right)$

Calculation:Section 1: 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 $\overrightarrow{B}$

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

Therefore, the x-component of $\overrightarrow{B}-3\overrightarrow{A}$ is:

  $\begin{array}{c}{B}_x-3A_x=-14.8-3\left(38.8\right)\\ =-131.2\end{array}$

Similarly, the y-component of $\overrightarrow{B}-3\overrightarrow{A}$ is:

  $\begin{array}{c}{B}_y-3A_y=22-3\left(20.6\right)\\ =-39.8\end{array}$

Therefore, the magnitude of $\overrightarrow{B}-3\overrightarrow{A}$ is:

  $\begin{array}{c}\left|\overrightarrow{B}-3\overrightarrow{A}\right|=\sqrt{{\left(-131.2\right)}^2+{\left(-39.8\right)}^2}\\ =137.1\end{array}$

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-39.8}{-131.2}\right)\\ =16.9\\ ={180}^o+{16.9}^o\\ ={196.9}^o\end{array}$

Therefore, the vector $\overrightarrow{B}-3\overrightarrow{A}$ has a magnitude of 137.1 with a direction of $196.9˚$ counterclockwise from + x -axis.

Section 2: x -component of $\overrightarrow{C}$

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

y -component of $\overrightarrow{C}$

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

  $\overrightarrow{R}=2\overrightarrow{A}-3\overrightarrow{B}+2\overrightarrow{C}$

  $\begin{array}{c}{R}_x=2A_x-3B_x+2C_x\\ =2\left(38.8\right)-3\left(-14.8\right)+2\left(0\right)\\ =122\\ R_y=2A_y-3B_y+2C_y\\ =2\left(20.6\right)-3\left(22\right)+2\left(-31\right)\\ =-86.8\end{array}$

  $\begin{array}{c}\left|\overrightarrow{R}\right|=\sqrt{{\left(122\right)}^2+{\left(-86.8\right)}^2}\\ =149.7\end{array}$

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-86.8}{122}\right)\\ =-35.4\\ ={360}^o-{35.4}^o\\ ={324.6}^o\end{array}$

Conclusion:

The magnitude of resultant vector $\overrightarrow{B}-3\overrightarrow{A}$ is 137.1 with a direction of $196.9˚$ counterclockwise from +x-axis.

The magnitude of resultant vector $\overrightarrow{R}=2\overrightarrow{A}-3\overrightarrow{B}+2\overrightarrow{C}$ is 149.7 with a direction of $324.6˚$ counterclockwise from +x-axis.