Q35 E - University Physics with Modern Physics

Question

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

Step-by-Step Solution

Answer

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

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

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

Step-1: Identification of given data

• The vector $\overrightarrow{\mathrm{A}}$ is 2.80 cm long and makes an angle of $60°$ with x-axis in the first quadrant 
• The vector $\overrightarrow{\mathrm{B}}$ is 1.90 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 $\overrightarrow{V}=V_x\stackrel{⏜}{i}+V_y\stackrel{⏜}{j}$, Here $V_x$ and $V_y$ are the components along x, and y directions respectively and $\stackrel{⏜}{i},\stackrel{⏜}{j}$ 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 →

Part(a)

 

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

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

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 cm for 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,

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

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 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 $\overrightarrow{R}$ can be expressed as,

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

Substitute 2.35 cm for $R_x$ and 0.78 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 cm for $R_x$ and 0.78 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 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 $\overrightarrow{A}-\overrightarrow{B}$ can be expressed as,

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

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

$$\begin{gathered} \overrightarrow{R}=\left(\right(1.40 \mathrm{cm})\hat{i}+(2.42 \mathrm{cm}\left)\hat{j}\right)-\left(\right(0.95 \mathrm{cm})\hat{i}+(-1.64 \mathrm{cm}\left)\hat{j}\right) \\ =(1.40 \mathrm{cm}-0.95 \mathrm{cm})\hat{i}+(2.42 \mathrm{cm}-(-1.64 \mathrm{cm}\left)\right)\hat{j} \\ =(0.45 \mathrm{cm})\hat{i}+(4.06 \mathrm{cm})\hat{j} \end{gathered}$$

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

The magnitude of $\overrightarrow{R}$ 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

$$\begin{gathered} R=\sqrt{(0.45 \mathrm{cm}{)}^2+(4.06 \mathrm{cm}{)}^2} \\ =4.09 \mathrm{cm} \end{gathered}$$

The direction of $\overrightarrow{R}$ 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

$$\begin{gathered} \tan \theta =\frac{4.06 \mathrm{cm}}{0.45 \mathrm{cm}} \\ =9.02 \\ \theta ={\tan }^{-1}(9.02) \\ =83.7° \end{gathered}$$

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

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

Part(b)

 

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

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

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

$$\begin{gathered} \overrightarrow{R}=\left(\right(0.95 \mathrm{cm})\hat{i}+(-1.64 \mathrm{cm}\left)\hat{j}\right)-\left(\right(1.40 \mathrm{cm})\hat{i}+(2.42 \mathrm{cm}\left)\hat{j}\right) \\ =(0.95 \mathrm{cm}-1.40 \mathrm{cm})\hat{i}+(-1.64 \mathrm{cm}-2.42 \mathrm{cm})\hat{j} \\ =-0.45\hat{i}-4.06\hat{j} \end{gathered}$$

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

The magnitude of $\overrightarrow{R}$ 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

$$\begin{gathered} R=\sqrt{{\left(-0.45 cm\right)}^2+{\left(-4.06 cm\right)}^2} \\ =4.09 cm \end{gathered}$$

The direction of $\overrightarrow{R}$ 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

$$\begin{gathered} \tan \theta =\frac{4.06 cm}{0.45 cm} \\ =9.02 \\ \theta ={\tan }^{-1}\left(9.02\right) \\ =83.7°+180° \\ =264° \end{gathered}$$

Thus, the magnitude of $\overrightarrow{B}-\overrightarrow{A}$ is 4.09 cm and its angle with x-axis is $264°$ and it is matches with graphical diagram