Chapter 3, Problem 6P

Vector is 6.6 using long and points along the negative x axis. Vector is 8.5 units long and points at ${\text{+55}}^{\circ }$ to the positive x axis. (a) What are the x any y components of each vector? (b) Determine the sum (magnitude and angle.)

Solution

(a)

To Determine:

x and y components of each vector.

Solution:

The x and y components of ${\overrightarrow{V}}_1$ are $-6.6$ and 0 respectively.

The x and y components of ${\overrightarrow{V}}_2$ are 4.87 and 6.96 respectively.

Explanation:

Given:

  ${\overrightarrow{V}}_1=6.6\text{units}\left(-\widehat{i}\right)$ (because it is along the negative x -axis)

  ${\overrightarrow{V}}_2=8.5\text{units}$ and points ${55}^o$ to the positive 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.

Calculations:

The x and y components of first vector are:

  $\begin{array}{l}{V}_{1x}=\left|{\overrightarrow{V}}_1\right|\cos \theta \\ V_{1x}=6.6\cos \left({180}^{°}\right)=-6.6\end{array}$

  $\begin{array}{l}{V}_{1y}=\left|{\overrightarrow{V}}_1\right|\sin \theta \\ V_{1y}=6.6\sin \left({180}^o\right)=0\end{array}$

The x and y components of second vector are:

  $\begin{array}{l}{V}_{2x}=\left|{\overrightarrow{V}}_2\right|\cos \theta \\ V_{2x}=8.5\cos \left({55}^{°}\right)=4.87\end{array}$

  $\begin{array}{l}{V}_{1y}=\left|{\overrightarrow{V}}_1\right|\sin \theta \\ V_{1y}=8.5\sin \left({55}^o\right)=6.96\end{array}$

(b)

To Determine:

The magnitude and angle of $\overrightarrow{V}={\overrightarrow{V}}_1+{\overrightarrow{V}}_2$ .

Solution:

The magnitude and angle of $\overrightarrow{V}={\overrightarrow{V}}_1+{\overrightarrow{V}}_2$ is 71.7 and ${104}^o$ respectively.

Explanation:

Given:

  ${\overrightarrow{V}}_1=6.6\text{units}\left(-\widehat{i}\right)$ (because it is along the negative x -axis)

  ${\overrightarrow{V}}_2=8.5\text{units}$ and points ${55}^o$ to the positive x -axis.

Formula Used:

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:

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

Calculations:${\overrightarrow{V}}_1=-6.6\widehat{i}$

  $\begin{array}{l}{\overrightarrow{V}}_2=4.87\widehat{i}+6.96\widehat{j}\\ \overrightarrow{V}={\overrightarrow{V}}_1+{\overrightarrow{V}}_2\\ \overrightarrow{V}=\left(-6.6+4.87\right)\widehat{i}+6.96\widehat{j}\\ \overrightarrow{V}=-1.73\widehat{i}+6.96\widehat{j}\end{array}$

Find the resultant magnitude:

  $\begin{array}{l}\left|\overrightarrow{V}\right|=\sqrt{\left(-{1.73}^2\right)+{\left(6.96\right)}^2}\\ \left|\overrightarrow{V}\right|=7.17\end{array}$

Find the angle between the vectors:

  $\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{6.96}{-1.73}\right)\\ \theta =-{76}^o\\ \theta ={180}^o-{76}^o={104}^o\end{array}$