Chapter 3, Problem 61AP

(a)

To determine

The magnitude of $\overrightarrow{\text{A}}+\overrightarrow{\text{B}}$ as a function of angle $\theta$.

Answer to Problem 61AP

The magnitude of $\overrightarrow{\text{A}}+\overrightarrow{\text{B}}$ as a function of angle $\theta$ is $\underset{\_}{{\left[10,000-\left(9600\right)\sin \theta \right]}^{\frac{1}{2}}\text{cm}}$.

Explanation of Solution

Write the expression for $\overrightarrow{\text{A}}$ in terms of cos and sine of angles.

    $\overrightarrow{\text{A}}=\left(x_1\cos {\theta }_1\widehat{i}+y_1\sin {\theta }_1\widehat{j}\right)$                                                                                        (I)

Here, $x$ and $y$ are the position coordinates in $x$ and $y$ direction, and ${\theta }_1$ is the angle between $\overrightarrow{\text{A}}$ and horizontal.

Write the vector representation for $\overrightarrow{\text{B}}$ in terms of cos and sine of angle.

    $\overrightarrow{\text{B}}=\left(x_2\text{cos}\theta \right)\widehat{i}+\left(y_2\text{sin}\theta \right)\widehat{j}$                                                                                    (II)

Here, $x_2$ is the position coordinate in the $x$ direction, $y_2$ is the position coordinate in the $y$ direction, and $\theta$ is the angle between $\overrightarrow{\text{B}}$ and horizontal.

Add expressions (I) and (II).

    $\overrightarrow{\text{R}}=\overrightarrow{\text{A}}+\overrightarrow{\text{B}}$                                                                                                             (III)

Here, $\overrightarrow{\text{R}}$ is the sum of vectors.

Write the expression for the magnitude of sum of vectors.

    $\left|\overrightarrow{\text{R}}\right|=\sqrt{A^2+B^2}$                                                                                                       (IV)

Here, $\left|\overrightarrow{\text{R}}\right|$ is the magnitude of vector $\overrightarrow{\text{R}}$, $A$ and $B$ are the magnitude part of vectors $\overrightarrow{\text{A}}$ and $\overrightarrow{\text{B}}$.

Conclusion:

Given that $\overrightarrow{\text{A}}=60.0\text{cm at 270}°\text{ from the horizontal}$ and $\overrightarrow{\text{B}}=80.0\text{cm at an angle }\theta$. That is $\overrightarrow{\text{A}}$ is oriented in negative $y$ direction.

Substitute $60\text{cm}$ for $x_1$, and $270°$ for ${\theta }_1$ in equation (I) to find $\overrightarrow{\text{A}}$.

    $\begin{array}{c}\overrightarrow{\text{A}}=\left(60\text{cm}\cos 270\widehat{i}+60\text{cm}\sin 270\widehat{j}\right)\\ =-60\text{cm }\widehat{j}\end{array}$

Substitute $80\text{cm}$ for $x_2$ in equation (II) to find $\overrightarrow{\text{B}}$.

    $\overrightarrow{\text{B}}=\left(80\text{cm}\cos \theta \widehat{i}+80\text{cm}\sin \theta \widehat{j}\right)$

Substitute $-60\text{cm }\widehat{j}$ for $\overrightarrow{\text{A}}$, and $\left(80\text{cm}\cos \theta \widehat{i}+80\text{cm}\sin \theta \widehat{j}\right)$ for $\overrightarrow{\text{B}}$ in equation (III).

    $\begin{array}{c}\overrightarrow{\text{A}}+\overrightarrow{\text{B}}=\left(-60\text{cm}\widehat{j}\right)+\left(80\text{cm}\text{cos}\theta \right)\widehat{i}+\left(80\text{cm}\text{sin}\theta \right)\widehat{j}\\ =\left(80\text{cm}\text{cos}\theta \right)\widehat{i}+\left(80\text{cm}\text{sin}\theta -60\text{cm}\right)\widehat{j}\end{array}$

Substitute $80\text{cos}\theta$ for $A$ and $80\text{sin}\theta -60$ for $B$ in equation (IV) to find $\left|\overrightarrow{\text{R}}\right|$.

    $\begin{array}{c}\left|\overrightarrow{\text{R}}\right|=\sqrt{{\left(80\text{cos}\theta \right)}^2+{\left(80\text{sin}\theta -60\right)}^2}\\ =\sqrt{\left[{\left(80\right)}^2\left({\cos }^2\theta +{\sin }^2\theta \right)-2\left(80\right)\left(60\right)\sin \theta +{\left(60\right)}^2\right]}\\ ={\left[10,000-\left(9600\right)\sin \theta \right]}^{\frac{1}{2}}\text{cm}\end{array}$

Therefore, the magnitude of $\overrightarrow{\text{A}}+\overrightarrow{\text{B}}$ as a function of angle $\theta$ is $\underset{\_}{{\left[10,000-\left(9600\right)\sin \theta \right]}^{\frac{1}{2}}\text{cm}}$.

(b)

To determine

The value of $\theta$ for which the magnitude of $\overrightarrow{\text{A}}+\overrightarrow{\text{B}}$ becomes maximum and also determine the maximum value.

Answer to Problem 61AP

The maximum value of $\overrightarrow{\text{A}}+\overrightarrow{\text{B}}$ is $\underset{\_}{140\text{cm}}$ and it occurs at an angle of $\underset{\_}{270°}$.

Explanation of Solution

Write the expression for the magnitude of $\overrightarrow{\text{A}}+\overrightarrow{\text{B}}$.

    $\left|\overrightarrow{\text{A}}+\overrightarrow{\text{B}}\right|={\left[10,000-\left(9,600\right)\sin \theta \right]}^{\frac{1}{2}}\text{cm}$

The magnitude depends on the sin of angle $\theta$. The magnitude is maximum when $\sin \theta$ is maximum. When $\theta =270°$, the magnitude takes the maximum value.

Conclusion:

Substitute $270°$ for $\theta$ in above expression to find the maximum value of $\overrightarrow{\text{A}}+\overrightarrow{\text{B}}$.

    $\begin{array}{c}\left|\overrightarrow{\text{A}}+\overrightarrow{\text{B}}\right|={\left[10,000-\left(9,600\right)\sin 270\right]}^{\frac{1}{2}}\text{cm}\\ \text{=}{\left[10,000-\left(9,600\right)\left(-1\right)\right]}^{\frac{1}{2}}\text{cm}\\ \text{=140}\text{cm}\end{array}$

Therefore, the maximum value of $\overrightarrow{\text{A}}+\overrightarrow{\text{B}}$ is $\underset{\_}{140\text{cm}}$ and it occurs at an angle of $\underset{\_}{270°}$.

(c)

To determine

The value of $\theta$ for which the magnitude of $\overrightarrow{\text{A}}+\overrightarrow{\text{B}}$ becomes minimum and also determine the minimum value.

Answer to Problem 61AP

The maximum value of $\overrightarrow{\text{A}}+\overrightarrow{\text{B}}$ is $\underset{\_}{20.0\text{cm}}$ and it occurs at an angle of $\underset{\_}{90°}$.

Explanation of Solution

Write the expression for the magnitude of $\overrightarrow{\text{A}}+\overrightarrow{\text{B}}$.

    $\left|\overrightarrow{\text{A}}+\overrightarrow{\text{B}}\right|={\left[10,000-\left(9,600\right)\sin \theta \right]}^{\frac{1}{2}}\text{cm}$

The magnitude depends on the sin of angle $\theta$. When $\theta =90°$, the magnitude takes the minimum value.

Conclusion:

Substitute $90°$ for $\theta$ in above expression to find the minimum value of $\overrightarrow{\text{A}}+\overrightarrow{\text{B}}$.

    $\begin{array}{c}\left|\overrightarrow{\text{A}}+\overrightarrow{\text{B}}\right|={\left[10,000-\left(9,600\right)\sin 90\right]}^{\frac{1}{2}}\text{cm}\\ \text{=}{\left[10,000-\left(9,600\right)\left(1\right)\right]}^{\frac{1}{2}}\text{cm}\\ \text{=20}\text{.0}\text{cm}\end{array}$

Therefore, the minimum value of $\overrightarrow{\text{A}}+\overrightarrow{\text{B}}$ is $\underset{\_}{20.0\text{cm}}$ and it occurs at an angle of $\underset{\_}{90°}$.