Q35E - University Physics with Modern Physics

Question

Vector $\overrightarrow{A} \mathrm{S}$ is $2.80 \mathrm{cm}$ long and is $60°$ above the x-axis in the first quadrant. Vector ${\overrightarrow{B}}_{\text{is }}1.90 \mathrm{cm}$ long and is $60°$ below the $x$-axis in the fourth quadrant (Fig. E1.35). Use components to find the magnitude and direction of (a) $\overrightarrow{A}+\overrightarrow{B}$; (b) $\overrightarrow{A}-\overrightarrow{B}$; (c) $\overrightarrow{B}-\overrightarrow{A}$. In each case, sketch the vector addition or subtraction and show that your numerical answers are in qualitative agreement with your sketch.

Step-by-Step Solution

Answer

a) The magnitude of $\overrightarrow{A}+\overrightarrow{B}$ is $2.48 \mathrm{cm}$and its angle with x-axis is $18.4°$and it is matches vith graphical diagram

b) the magnitude of $\overrightarrow{A}-\overrightarrow{B}$ is $4.09 \mathrm{cm}$ and its angle with x-axis is $83.7°$ and it is matches with graphical diagram

c) The magnitude of $\overrightarrow{B}-\overrightarrow{A}$ is $4.09 \mathrm{cm}$ and its angle with x-axis is $264°$ and it is matches with graphical diagram

Step 1: Identification of given data

•  The vector $\overrightarrow{A}$ is $2.80 \mathrm{cm}$ long and makes an angle of $60°$ with x-axis in the first quadrant
• The vector $\overrightarrow{B}$ is $1.90 \mathrm{cm}$ long and makes an angle of $60°$ with x-axis in the fourth quadrant

Step 2: Magnitude and direction of a vector

Consider a vector quantity $\overrightarrow{V}=V_x\hat{i}+V_y\hat{j}$, Here $V_x$ and $V_y$ are the components along x, and y directions respectively and $\hat{i},\hat{j}$are the unit vectors along x, and y directions respectively.

The magnitude of $\overrightarrow{V}$ can be expressed as,

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

The direction of this vector is expressed as,

$\tan \theta =\frac{V_y}{V_x}$

Step 3: a) Determination of magnitude and direction of A → + B →

The Representation of in terms of unit vectors is,

$\overrightarrow{A}=A_x\hat{i}+A_y\hat{j}\text{. }$

From the given diagram angle between $\overrightarrow{A}$ and x is,

$\theta =60°$

Thus, the components of vector $\overrightarrow{A}$ is

$$\begin{gathered} A_x=A\cos \theta \\ {A}_y=A\sin \theta \end{gathered}$$

Substitute $2.80 \mathrm{cm}$ for $\mathrm{A}$ and $60°$ for $\theta$ in the above equations,

$$\begin{gathered} A_x=(2.80 \mathrm{cm})\cos \left(60°\right) \\ =1.40 \mathrm{cm} \\ {A}_y=(2.80 \mathrm{cm})\sin \left(60°\right) \\ =2.42 \mathrm{cm} \end{gathered}$$

Thus, the representation of vector $\overrightarrow{A}$ in terms of unit vectors is $\overrightarrow{A}=(1.40 \mathrm{cm})\hat{i}+(2.42 \mathrm{cm})\hat{j}$

The Representation of $\overrightarrow{B}$ in terms of unit vectors is,

$\overrightarrow{B}=B_x\hat{i}+B_y\hat{j}\text{. }$

From the given diagram angle between $\overrightarrow{B}$ and x is,

$\theta =-60°$

Thus, the components of vector $\overrightarrow{B}$ is

$$\begin{gathered} B_x=B\cos \theta \\ {B}_y=B\sin \theta \end{gathered}$$

Substitute $1.90 \mathrm{cm}$ for B and $60°$ for $\theta$ in the above equations.

$$\begin{gathered} B_x=(1.90 \mathrm{cm})\cos \left(-60°\right) \\ =0.95 \mathrm{cm} \\ {B}_y=(1.90 \mathrm{mm})\sin \left(-60°\right) \\ =-1.64 \mathrm{cm} \end{gathered}$$

Thus, the representation of vector $\overrightarrow{B}$ in terms of unit vectors is $\overrightarrow{B}=(0.95 \mathrm{cm})\hat{i}+(-1.64 \mathrm{cm})\hat{j}$

Thus, the vector sum of $\overrightarrow{A}+\overrightarrow{B}$ can be expressed as,

$\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}$

Substitute for $\overrightarrow{A}$ and $\overrightarrow{B}$,

$$\begin{gathered} \overrightarrow{R}=\left(\right(1.40 \mathrm{cm})\hat{i}+(2.42 \mathrm{cm}\left)\hat{j}\right)+\left(\right(0.95 \mathrm{cm})\hat{i}+(-1.64 \mathrm{cm}\left)\hat{j}\right) \\ =(1.40 \mathrm{cm}+0.95 \mathrm{cm})\hat{i}+(2.42 \mathrm{cm}+(-1.64 \mathrm{cm}\left)\right)\hat{j} \\ =(2.35 \mathrm{cm})\hat{i}+(0.78 \mathrm{cm})\hat{j} \end{gathered}$$

The vector diagram representation of $\overrightarrow{R}$ is shown below,

The magnitude of can be expressed as,

$R=\sqrt{R_x^2+R_y^2}$

Substitute $2.35 \mathrm{cm}$ for $R_x$ and $0.78 \mathrm{cm}{R}_y$ in the above equation

$$\begin{gathered} R=\sqrt{(2.35 \mathrm{cm}{)}^2+(0.78 \mathrm{cm}{)}^2} \\ =2.48 \mathrm{cm} \end{gathered}$$

The direction of $\overrightarrow{R}$can be expressed as,

$\tan \theta =\frac{R_y}{R_x}$

Substitute $2.35 \mathrm{cm}$ for $R_x$ and $0.78 \mathrm{cm}{R}_y$ in the above equation

$$\begin{gathered} \tan \theta =\frac{0.78 \mathrm{cm}}{2.35 \mathrm{cm}} \\ =0.3319 \\ \theta ={\tan }^{-1}(0.3319) \\ =18.4° \end{gathered}$$

Thus, the magnitude of$\overrightarrow{A}+\overrightarrow{B}$ is $2.48 \mathrm{cm}$and its angle with x-axis is $18.4°$ and it is matches with graphical diagram.

Step 4: b) Determination of magnitude and direction of A → - B →

Part(b)

The vector difference of $\vec{A}-\vec{B}$can be expressed as,

$\vec{R}=\vec{A}-\vec{B}$

Substitute for \[\vec{A}\]and \[\vec{B}\],

$\begin{align}

  & \vec{R}=\left( \left( 1.40\ \text{cm} \right)\hat{i}+\left( 2.42\ \text{cm} \right)\hat{j} \right)-\left( \left( 0.95\ \text{cm} \right)\hat{i}+\left( -1.64\ \text{cm} \right)\hat{j} \right) \\ 

 & =\left( 1.40\ \text{cm}-0.95\ \text{cm} \right)\hat{i}+\left( 2.42\ \text{cm}-\left( -1.64\ \text{cm} \right) \right)\hat{j} \\ 

 & =\left( 0.45\ \text{cm} \right)\hat{i}+\left( 4.06\ \text{cm} \right)\hat{j}  

\end{align}$

The vector diagram representation of $\vec{R}$is shown below,

The magnitude of $\vec{R}$can be expressed as,

$R=\sqrt{R_{x}^{2}+R_{y}^{2}}$

Substitute 0.45 cm for ${{R}_{x}}$and 4.06 cm ${{R}_{y}}$in the above equation

$\begin{align}

  & R=\sqrt{{{\left( 0.45\ \text{cm} \right)}^{2}}+{{\left( 4.06\ \text{cm} \right)}^{2}}} \\ 

 & =4.09\ \text{cm}  

\end{align}$

The direction of $\vec{R}$can be expressed as,

\[\tan \theta =\frac{{{R}_{y}}}{{{R}_{x}}}\]

Substitute 0.45 cm for ${{R}_{x}}$and 4.06 cm ${{R}_{y}}$in the above equation

\[\begin{align}

  & \tan \theta =\frac{4.06\ \text{cm}}{0.45\ \text{cm}} \\ 

 & =9.02 \\ 

 & \theta ={{\tan }^{-1}}\left( 9.02 \right) \\ 

 & =83.7{}^\circ   

\end{align}\]

Thus, the magnitude of \[\vec{A}-\vec{B}\] is 4.09 cm and its angle with x-axis is \[83.7{}^\circ \]and it is matches with graphical diagram

Step 5: c) Determination of magnitude and direction of B → - A → ∣

Part(b)

 

The vector difference of $\vec{B}-\vec{A}$can be expressed as,

$\vec{R}=\vec{B}-\vec{A}$

Substitute for \[\vec{B}\]and \[\vec{A}\],

$\begin{align}

  & \vec{R}=\left( \left( 0.95\ \text{cm} \right)\hat{i}+\left( -1.64\ \text{cm} \right)\hat{j} \right)-\left( \left( 1.40\ \text{cm} \right)\hat{i}+\left( 2.42\ \text{cm} \right)\hat{j} \right) \\ 

 & =\left( 0.95\ \text{cm}-1.40\ \text{cm} \right)\hat{i}+\left( -1.64\ \text{cm}-2.42\ \text{cm} \right)\hat{j} \\ 

 & =-0.45\hat{i}-4.06\hat{j}  

\end{align}$

The vector diagram representation of $\vec{R}$is shown below,

 

The magnitude of $\vec{R}$can be expressed as,

$R=\sqrt{R_{x}^{2}+R_{y}^{2}}$

Substitute -0.45 cm for ${{R}_{x}}$and -4.06 cm ${{R}_{y}}$in the above equation

$\begin{align}

  & R=\sqrt{{{\left( -0.45\ \text{cm} \right)}^{2}}+{{\left( -4.06\ \text{cm} \right)}^{2}}} \\ 

 & =4.09\ \text{cm}  

\end{align}$

The direction of $\vec{R}$can be expressed as,

\[\tan \theta =\frac{{{R}_{y}}}{{{R}_{x}}}\]

Substitute 0.45 cm for ${{R}_{x}}$and 4.06 cm ${{R}_{y}}$in the above equation

\[\begin{align}

  & \tan \theta =\frac{4.06\ \text{cm}}{0.45\ \text{cm}} \\ 

 & =9.02 \\ 

 & \theta ={{\tan }^{-1}}\left( 9.02 \right) \\ 

 & =83.7{}^\circ +180{}^\circ  \\ 

 & =264{}^\circ   

\end{align}\]

Thus, the magnitude of \[\vec{B}-\vec{A}\] is 4.09 cm and its angle with x-axis is \[264{}^\circ \]and it is matches with graphical diagram

 

 

$$