Chapter 3, Problem 3.61AP

Lei $\stackrel{\to }{A}$ = 60.0 cm at 270º measured from the horizontal. Let $\stackrel{\to }{B}$ = 80.0 cm at some angle θ. (a) Find the magnitude of $\stackrel{\to }{A}$ + $\stackrel{\to }{B}$ as a function of θ. (b) From the answer to part (a), for what value of θ does | $\stackrel{\to }{A}$ + $\stackrel{\to }{B}$| take on its maximum value? What is this maximum value? (c) From the answer to pan (a), for what value of θ does | $\stackrel{\to }{A}$ + $\stackrel{\to }{B}$| take on its minimum value? What is this minimum value? (d) Without reference to the answer to part (a), argue that the answers to each of parts (b) and (c) do or do not make sense.

To determine

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

Answer to Problem 3.61AP

The magnitude of $\overrightarrow{\text{A}}+\overrightarrow{\text{B}}$ as a function of $\theta$ is $\sqrt{10000{\text{cm}}^2-9600{\text{cm}}^2\sin \theta }$.

Explanation of Solution

Given info: The vector $\overrightarrow{\text{A}}$ is $60.0\text{cm}$ at $270°$ direction from the horizontal. The vector $\overrightarrow{\text{B}}$ is $80.0\text{cm}$ at $\theta$.

The component of vector $\overrightarrow{\text{A}}$ in the $x$ direction is,

${\overrightarrow{\text{A}}}_{\text{x}}=A\left(\text{cos}{\theta }_1\right)\widehat{\text{i}}$

Here,

${\overrightarrow{\text{A}}}_{\text{x}}$ is the component of vector $\overrightarrow{\text{A}}$ in the $x$ direction.

$A$ is the magnitude of vector $\overrightarrow{\text{A}}$.

$\widehat{\text{i}}$ is the unit vector component in x direction.

${\theta }_1$ is the angle of the vector $\overrightarrow{\text{A}}$.

Substitute $60.0\text{cm}$ for $A$ and $270°$ for ${\theta }_1$ in the above equation.

$\begin{array}{c}{\overrightarrow{\text{A}}}_{\text{x}}=60.0\text{cm}\left(\text{cos}\left(270°\right)\right)\widehat{\text{i}}\\ =\left(0\text{cm}\right)\widehat{\text{i}}\end{array}$

Thus, the component of vector $\overrightarrow{\text{A}}$ in the $x$ direction is $\left(0\text{cm}\right)\widehat{\text{i}}$.

The component of vector $\overrightarrow{\text{A}}$ in the $y$ direction is,

${\overrightarrow{\text{A}}}_{\text{y}}=A\left(\text{sin}{\theta }_1\right)\widehat{\text{j}}$

Here,

${\overrightarrow{\text{A}}}_{\text{y}}$ is the component of vector $\overrightarrow{\text{A}}$ in the $y$ direction.

$\widehat{\text{j}}$ is the unit vector component in $y$ direction.

Substitute $60.0\text{cm}$ for $A$ and $270°$ for ${\theta }_1$ in the above equation.

$\begin{array}{c}{\overrightarrow{\text{A}}}_{\text{y}}=60.0\text{cm}\left(\text{sin}270°\right)\widehat{\text{j}}\\ =\left(-\text{60}\text{cm}\right)\widehat{\text{j}}\end{array}$

Thus, the component of vector $\overrightarrow{\text{A}}$ in the $y$ direction is $\left(-\text{60}\text{cm}\right)\widehat{\text{j}}$.

The position vector of the $\overrightarrow{\text{A}}\text{'}$ is,

$\overrightarrow{\text{A}}\text{'}={\overrightarrow{\text{A}}}_{\text{x}}+{\overrightarrow{\text{A}}}_{\text{y}}$

Here,

$\overrightarrow{\text{A}}\text{'}$ is the position vector of the vector $\overrightarrow{\text{A}}$.

Substitute $\left(0\text{cm}\right)\widehat{\text{i}}$ for ${\overrightarrow{\text{A}}}_{\text{x}}$, $\left(-\text{60}\text{cm}\right)\widehat{\text{j}}$ for ${\overrightarrow{\text{A}}}_{\text{y}}$ in the above equation.

$\overrightarrow{\text{A}}\text{'}=\left(0\text{cm}\right)\widehat{\text{i}}+\left(-\text{60}\text{cm}\right)\widehat{\text{j}}$

Thus, the position vector of the vector $\overrightarrow{\text{A}}$ is $\left(0\text{cm}\right)\widehat{\text{i}}+\left(-\text{60}\text{cm}\right)\widehat{\text{j}}$.

The component of vector $\overrightarrow{\text{B}}$ in the $x$ direction is,

${\overrightarrow{\text{B}}}_{\text{x}}=B\left(\text{cos}\theta \right)\widehat{\text{i}}$

Here,

${\overrightarrow{\text{B}}}_{\text{x}}$ is the component of vector $\overrightarrow{\text{B}}$ in the $x$ direction.

$B$ is the magnitude of vector $\overrightarrow{\text{B}}$.

$\widehat{\text{i}}$ is the unit vector component in x direction.

$\theta$ is the angle of the vector $\overrightarrow{\text{B}}$.

Substitute $80.0\text{cm}$ for $B$ in the above equation.

${\overrightarrow{\text{B}}}_{\text{x}}=80.0\text{cm}\left(\text{cos}\theta \right)\widehat{\text{i}}$

Thus, the component of vector $\overrightarrow{\text{B}}$ in the $x$ direction is $80.0\text{cm}\left(\text{cos}\theta \right)\widehat{\text{i}}$.

The component of vector $\overrightarrow{\text{B}}$ in the $y$ direction is,

${\overrightarrow{\text{B}}}_{\text{y}}=B\left(\text{sin}\theta \right)\widehat{\text{j}}$

Here,

${\overrightarrow{\text{B}}}_{\text{y}}$ is the component of vector $\overrightarrow{\text{B}}$ in the $y$ direction.

$\widehat{\text{j}}$ is the unit vector component in $y$ direction.

Substitute $80.0\text{cm}$ for $B$ in the above equation.

${\overrightarrow{\text{B}}}_{\text{y}}=80.0\text{cm}\left(\text{sin}\theta \right)\widehat{\text{j}}$

Thus, the component of vector $\overrightarrow{\text{B}}$ in the $y$ direction is $80.0\text{cm}\left(\text{sin}\theta \right)\widehat{\text{j}}$.

The position vector of the $\overrightarrow{\text{B}}\text{'}$ is,

$\overrightarrow{\text{B}}\text{'}={\overrightarrow{\text{B}}}_{\text{x}}+{\overrightarrow{\text{B}}}_{\text{y}}$

Here,

$\overrightarrow{\text{B}}\text{'}$ is the position vector of the vector $\overrightarrow{\text{B}}$.

Substitute $80.0\text{cm}\left(\text{cos}\theta \right)\widehat{\text{i}}$ for ${\overrightarrow{\text{B}}}_{\text{x}}$, $80.0\text{cm}\left(\text{sin}\theta \right)\widehat{\text{j}}$ for ${\overrightarrow{\text{B}}}_{\text{y}}$ in the above equation.

$\overrightarrow{\text{B}}\text{'}=80.0\text{cm}\left(\text{cos}\theta \right)\widehat{\text{i}}+80.0\text{cm}\left(\text{sin}\theta \right)\widehat{\text{j}}$

Thus, the position vector of the vector $\overrightarrow{\text{B}}$ is $80.0\text{cm}\left(\text{cos}\theta \right)\widehat{\text{i}}+80.0\text{cm}\left(\text{sin}\theta \right)\widehat{\text{j}}$.

The vector $\overrightarrow{\text{A}}+\overrightarrow{\text{B}}$ is,

$\overrightarrow{\text{C}}=\overrightarrow{\text{A}}+\overrightarrow{\text{B}}$

The vector $\overrightarrow{\text{C}}$ can be written as,

$\overrightarrow{\text{C}}=\overrightarrow{\text{A}}\text{'}+\overrightarrow{\text{B}}\text{'}$

Substitute $\left(0\text{cm}\right)\widehat{\text{i}}+\left(-\text{60}\text{cm}\right)\widehat{\text{j}}$ for $\overrightarrow{\text{A}}\text{'}$ and $80.0\text{cm}\left(\text{cos}\theta \right)\widehat{\text{i}}+80.0\text{cm}\left(\text{sin}\theta \right)\widehat{\text{j}}$ for $\overrightarrow{\text{B}}\text{'}$ in the above equation.

$\begin{array}{c}\overrightarrow{\text{C}}=\left(0\text{cm}\right)\widehat{\text{i}}+\left(-\text{60}\text{cm}\right)\widehat{\text{j}}+80.0\text{cm}\left(\text{cos}\theta \right)\widehat{\text{i}}+80.0\text{cm}\left(\text{sin}\theta \right)\widehat{\text{j}}\\ \text{=}\left[80.0\text{cm}\left(\text{cos}\theta \right)\right]\widehat{\text{i}}+\left[-\text{60}\text{cm}+80.0\text{cm}\left(\text{sin}\theta \right)\right]\widehat{\text{j}}\end{array}$

Thus, the vector $\overrightarrow{\text{C}}$ is $\left[80.0\text{cm}\left(\text{cos}\theta \right)\right]\widehat{\text{i}}+\left[-\text{60}\text{cm}+80.0\text{cm}\left(\text{sin}\theta \right)\right]\widehat{\text{j}}$.

The magnitude for the vector $\overrightarrow{\text{C}}$ is,

$C=\sqrt{{\left({\overrightarrow{\text{C}}}_{\text{x}}\right)}^2+{\left({\overrightarrow{\text{C}}}_{\text{y}}\right)}^2}$

Here,

${\overrightarrow{\text{C}}}_{\text{x}}$ is the component of vector $\overrightarrow{\text{C}}$ in $x$ direction.

${\overrightarrow{\text{C}}}_{\text{y}}$ is the component of vector $\overrightarrow{\text{C}}$ in $y$ direction.

Substitute $80.0\text{cm}\left(\text{cos}\theta \right)$ for ${\overrightarrow{\text{C}}}_{\text{x}}$, $\left[-\text{60}\text{cm}+80.0\text{cm}\left(\text{sin}\theta \right)\right]$ for ${\overrightarrow{\text{C}}}_{\text{y}}$ in the above equation.

$\begin{array}{c}C=\sqrt{{\left(80.0\text{cm}\left(\text{cos}\theta \right)\right)}^2+{\left(80.0\text{cm}\left(\text{sin}\theta \right)-\text{60}\text{cm}\right)}^2}\\ =\sqrt{6400{\text{cm}}^2\left({\text{cos}}^2\theta \right)+360\text{0}{\text{cm}}^2+6400{\text{cm}}^2\left({\text{sin}}^2\theta \right)-\left(9600{\text{cm}}^2\right)\sin \theta }\\ =\sqrt{6400{\text{cm}}^2+360\text{0}{\text{cm}}^2-\left(9600{\text{cm}}^2\right)\sin \theta }\\ =\sqrt{10000{\text{cm}}^2-9600{\text{cm}}^2\sin \theta }\end{array}$

Conclusion:

Therefore, the magnitude of $\overrightarrow{\text{A}}+\overrightarrow{\text{B}}$ as a function of $\theta$ is $\sqrt{10000{\text{cm}}^2-9600{\text{cm}}^2\sin \theta }$.

To determine

The value of $\theta$ for which the magnitude of $\overrightarrow{\text{A}}+\overrightarrow{\text{B}}$ is maximum.

Answer to Problem 3.61AP

The value of $\theta$ for which the magnitude of $\overrightarrow{\text{A}}+\overrightarrow{\text{B}}$ is maximum is $270°$.

Explanation of Solution

Given info: The vector $\overrightarrow{\text{A}}$ is $60.0\text{cm}$ at $270°$ direction from the horizontal. The vector $\overrightarrow{\text{B}}$ is $80.0\text{cm}$ at $\theta$.

The magnitude of $\overrightarrow{\text{A}}+\overrightarrow{\text{B}}$ is maximum when both the vector $\overrightarrow{\text{A}}$ and $\overrightarrow{\text{B}}$ move in same direction. Thus, the $\overrightarrow{\text{A}}+\overrightarrow{\text{B}}$ is maximum at $270°$.

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

$\overrightarrow{\text{A}}+\overrightarrow{\text{B}}=\sqrt{10000{\text{cm}}^2-9600{\text{cm}}^2\sin \theta }$

Substitute $270°$ for $\theta$ in the above equation.

$\begin{array}{c}\overrightarrow{\text{A}}+\overrightarrow{\text{B}}=\sqrt{10000{\text{cm}}^2-9600{\text{cm}}^2\sin 270°}\\ =140\text{cm}\end{array}$

Conclusion:

Therefore, the value of $\theta$ for which the magnitude of $\overrightarrow{\text{A}}+\overrightarrow{\text{B}}$ is maximum is $270°$.

To determine

The value of $\theta$ for which the magnitude of $\overrightarrow{\text{A}}+\overrightarrow{\text{B}}$ is minimum.

Answer to Problem 3.61AP

The value of $\theta$ for which the magnitude of $\overrightarrow{\text{A}}+\overrightarrow{\text{B}}$ is minimum is $90°$.

Explanation of Solution

Given info: The vector $\overrightarrow{\text{A}}$ is $60.0\text{cm}$ at $270°$ direction from the horizontal. The vector $\overrightarrow{\text{B}}$ is $80.0\text{cm}$ at $\theta$.

The magnitude of $\overrightarrow{\text{A}}+\overrightarrow{\text{B}}$ is minimum when both the vector $\overrightarrow{\text{A}}$ and $\overrightarrow{\text{B}}$ move in opposite direction. Thus, the $\overrightarrow{\text{A}}+\overrightarrow{\text{B}}$ is minimum at $90°$.

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

$\overrightarrow{\text{A}}+\overrightarrow{\text{B}}=\sqrt{10000{\text{cm}}^2-9600{\text{cm}}^2\sin \theta }$

Substitute $90°$ for $\theta$ in the above equation.

$\begin{array}{c}\overrightarrow{\text{A}}+\overrightarrow{\text{B}}=\sqrt{10000{\text{cm}}^2-9600{\text{cm}}^2\sin 90°}\\ =20\text{cm}\end{array}$

Conclusion:

Therefore, the value of $\theta$ for which the magnitude of $\overrightarrow{\text{A}}+\overrightarrow{\text{B}}$ is minimum is $90°$.