Chapter 3, Problem 10P

(a) given the vectors $\stackrel{\to }{\text{A}}$ and $\stackrel{\to }{\text{B}}$ shown in Fig. 3-35, determine $\stackrel{\to }{\text{B}}-\stackrel{\to }{\text{A}}$ . (b) Determine $\stackrel{\to }{\text{A}}-\stackrel{\to }{\text{B}}$ without using your answer in (a) then compare your results and see if they are opposite.

Solution

To Determine:

1. $\overrightarrow{B}-\overrightarrow{A}$
2. $\overrightarrow{A}-\overrightarrow{B}$ without using results obtained in (a)comparingresults.

Solution:

1. The vector $\overrightarrow{B}-\overrightarrow{A}$ has a magnitude of 53.6 with a direction of ${178.5}^o$ counterclockwise from +x-axis.
2. The vector $\overrightarrow{A}-\overrightarrow{B}$ has a magnitude of 53.6 with a direction of ${358.5}^o$ counterclockwise from +x-axis. Comparing vector $\overrightarrow{B}-\overrightarrow{A}$ with vector $\overrightarrow{A}-\overrightarrow{B}$ we can observe that both vectors have the same magnitude both points in opposite directions.

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

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 $\overrightarrow{B}-\overrightarrow{A}$ is:

  $\begin{array}{c}{B}_x-A_x=-14.8-38.8\\ =-53.6\end{array}$

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

  $\begin{array}{c}{B}_y-A_y=22-20.6\\ =1.4\end{array}$

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

  $\begin{array}{c}\left|\overrightarrow{B}-\overrightarrow{A}\right|=\sqrt{{\left(-53.6\right)}^2+{\left(1.4\right)}^2}\\ =53.6\end{array}$

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{1.4}{-53.6}\right)\\ =-{1.5}^o\\ ={180}^o-{1.5}^o\\ ={178.5}^o\end{array}$

Therefore, the vector $\overrightarrow{B}-\overrightarrow{A}$ has a magnitude of 53.6 with a direction of $178.5˚$ counterclockwise from + x -axis.

Section 2: The x-component of $\overrightarrow{A}-\overrightarrow{B}$ is:

  $\begin{array}{c}{A}_x-B_x=38.8-\left(-14.8\right)\\ =53.6\end{array}$

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

  $\begin{array}{c}{A}_y-B_y=20.6-22\\ =-1.4\end{array}$

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

  $\begin{array}{c}\left|\overrightarrow{A}-\overrightarrow{B}\right|=\sqrt{{\left(53.6\right)}^2+{\left(-1.4\right)}^2}\\ =53.6\end{array}$

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-1.4}{53.6}\right)\\ =-{1.5}^o\\ ={360}^o-{1.5}^o\\ ={358.5}^o\end{array}$

Conclusion:

The vector $\overrightarrow{B}-\overrightarrow{A}$ has a magnitude of 53.6 with a direction of ${178.5}^o$ counterclockwise from +x-axis.The vector $\overrightarrow{A}-\overrightarrow{B}$ has a magnitude of 53.6 with a direction of ${358.5}^o$ counterclockwise from +x-axis. Comparing vector $\overrightarrow{B}-\overrightarrow{A}$ with vector $\overrightarrow{A}-\overrightarrow{B}$ we can observe that both vectors have the same magnitude but point in opposite directions.