Chapter 3, Problem 3.56AP

The rectangle shown in Figure P3.56 has sides parallel to the x and y axes. The position vectors of two corners are  = 10.0 m at 50.0° and  = 12.0 m at 30.0°. (a) Find the perimeter of the rectangle. (b) Find the magnitude and direction of the vector from the origin to the upper-right corner of the rectangle.

To determine

The perimeter of the rectangle.

Answer to Problem 3.56AP

The perimeter of the rectangle is $11.24\text{m}$.

Explanation of Solution

Given info: The rectangle has sides parallel to the $x$ axis and $y$ axis. The position vectors of two corners is $10.0\text{m}$ for $\overrightarrow{\text{A}}$ at $50.0°$ and $12.0\text{m}$ for $\overrightarrow{\text{B}}$ at $30.0°$.

The vector component of $\overrightarrow{\text{A}}$ 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 vector component of $\overrightarrow{\text{A}}$ in $x$ direction.

$A$ is the magnitude of the 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 $10.0\text{m}$ for $A$ and $50.0°$ for ${\theta }_1$ in the above equation.

$\begin{array}{c}{\overrightarrow{\text{A}}}_{\text{x}}=10.0\text{m}\left(\text{cos}50.0°\right)\widehat{\text{i}}\\ =\left(\text{6}\text{.42}\text{m}\right)\widehat{\text{i}}\end{array}$

Thus, the vector component of $\overrightarrow{\text{A}}$ in $x$ direction is $\left(\text{6}\text{.42}\text{m}\right)\widehat{\text{i}}$.

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

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

Here,

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

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

Substitute $10.0\text{m}$ for $A$ and $50.0°$ for ${\theta }_1$ in the above equation.

$\begin{array}{c}{\overrightarrow{\text{A}}}_{\text{y}}=10.0\text{m}\left(\sin 50.0°\right)\widehat{\text{j}}\\ =\left(7.66\text{m}\right)\widehat{\text{j}}\end{array}$

Thus, the vector component of $\overrightarrow{\text{A}}$ in $y$ direction is $\left(7.66\text{m}\right)\widehat{\text{j}}$.

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

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

Here,

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

Substitute $\left(\text{6}\text{.42}\text{m}\right)\widehat{\text{i}}$ for ${\overrightarrow{\text{A}}}_{\text{x}}$, $\left(7.66\text{m}\right)\widehat{\text{j}}$ for ${\overrightarrow{\text{A}}}_{\text{y}}$ in the above equation.

$\overrightarrow{\text{A}}\text{'}=\left(\text{6}\text{.42}\text{m}\right)\widehat{\text{i}}+\left(7.66\text{m}\right)\widehat{\text{j}}$

Thus, the position vector for the vector $\overrightarrow{\text{A}}$ is $\left(\text{6}\text{.42}\text{m}\right)\widehat{\text{i}}+\left(7.66\text{m}\right)\widehat{\text{j}}$.

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

${\overrightarrow{\text{B}}}_{\text{x}}=B\left(\text{cos}{\theta }_2\right)\widehat{\text{i}}$

Here,

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

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

${\theta }_2$ is the angle of the vector $\overrightarrow{\text{B}}$.

Substitute $12.0\text{m}$ for $B$ and $30.0°$ for ${\theta }_2$ in the above equation.

$\begin{array}{c}{\overrightarrow{\text{B}}}_{\text{x}}=12.0\text{m}\left(\text{cos}30.0°\right)\widehat{\text{i}}\\ =\left(\text{10}\text{.39}\text{m}\right)\widehat{\text{i}}\\ \approx \left(\text{10}\text{.4}\text{m}\right)\widehat{\text{i}}\end{array}$

Thus, the component of vector $\overrightarrow{\text{B}}$ in $x$ direction is $\left(\text{10}\text{.4}\text{m}\right)\widehat{\text{i}}$.

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

${\overrightarrow{\text{B}}}_{\text{y}}=B\left(\sin {\theta }_2\right)\widehat{\text{j}}$

Here,

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

Substitute $12.0\text{m}$ for $B$ and $30.0°$ for ${\theta }_2$ in the above equation.

$\begin{array}{c}{\overrightarrow{\text{B}}}_{\text{y}}=12.0\text{m}\left(\sin \left(30.0°\right)\right)\widehat{\text{j}}\\ =\left(6.00\text{m}\right)\widehat{\text{j}}\end{array}$

Thus, the component of vector $\overrightarrow{\text{B}}$ in $y$ direction is $\left(6.00\text{m}\right)\widehat{\text{j}}$.

The position vector of the vector $\overrightarrow{\text{B}}$ 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 $\left(\text{10}\text{.4}\text{m}\right)\widehat{\text{i}}$ for ${\overrightarrow{\text{B}}}_{\text{x}}$, $\left(6.00\text{m}\right)\widehat{\text{j}}$ for ${\overrightarrow{\text{B}}}_{\text{y}}$ in the above equation.

$\overrightarrow{\text{B}}\text{'}=\left(\text{10}\text{.4}\text{m}\right)\widehat{\text{i}}+\left(6.00\text{m}\right)\widehat{\text{j}}$

Thus, the position vector of the vector $\overrightarrow{\text{B}}$ is $\left(\text{10}\text{.4}\text{m}\right)\widehat{\text{i}}+\left(6.00\text{m}\right)\widehat{\text{j}}$.

The length of the rectangle is,

$\overrightarrow{\text{ad}}=-R_{\text{H}}\widehat{\text{j}}$

Here,

$\overrightarrow{\text{ad}}$ is the length of the rectangle.

The width of the rectangle is,

$\overrightarrow{\text{dc}}=R_{\text{W}}\widehat{\text{i}}$

Here,

$\overrightarrow{\text{dc}}$ is the width of the rectangle.

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

$\overrightarrow{\text{B}}=\overrightarrow{\text{A}}+\overrightarrow{\text{ad}}+\overrightarrow{\text{dc}}$

Substitute $R_{\text{W}}\widehat{\text{i}}$ for $\overrightarrow{\text{dc}}$, $\left(\text{10}\text{.4}\text{m}\right)\widehat{\text{i}}+\left(6.00\text{m}\right)\widehat{\text{j}}$ for $\overrightarrow{\text{B}}$, $\left(\text{6}\text{.42}\text{m}\right)\widehat{\text{i}}+\left(7.66\text{m}\right)\widehat{\text{j}}$ for $\overrightarrow{\text{A}}$ and $-R_{\text{H}}\widehat{\text{j}}$ for $\overrightarrow{\text{ad}}$ in the above equation.

$\begin{array}{c}\left(\text{10}\text{.4}\text{m}\right)\widehat{\text{i}}+\left(6.00\text{m}\right)\widehat{\text{j}}=\left(\text{6}\text{.42}\text{m}\right)\widehat{\text{i}}+\left(7.66\text{m}\right)\widehat{\text{j}}+\left(-R_{\text{H}}\widehat{\text{j}}\right)+R_{\text{W}}\widehat{\text{i}}\\ \left[\left(\text{10}\text{.4}\text{m}\right)-\left(\text{6}\text{.42}\text{m}\right)\right]\widehat{\text{i}}+\left[\left(6.00\text{m}\right)-\left(7.66\text{m}\right)\right]\widehat{\text{j}}=\left(-R_{\text{H}}\widehat{\text{j}}\right)+R_{\text{W}}\widehat{\text{i}}\\ \left(3.96\text{m}\right)\widehat{\text{i}}+\left(-1.66\text{m}\right)\widehat{\text{j}}=R_{\text{W}}\widehat{\text{i}}+\left(-R_{\text{H}}\widehat{\text{j}}\right)\end{array}$

Formula to calculate the perimeter of the rectangle is,

$P=2\left(R_{\text{W}}+R_{\text{H}}\right)$

Here,

$P$ is the perimeter of the rectangle.

Substitute $3.96\text{m}$ for $R_{\text{W}}$ and $1.66\text{m}$ for $R_{\text{H}}$ in the above equation.

$\begin{array}{c}P=2\left(3.96\text{m}+1.66\text{m}\right)\\ =11.24\text{m}\end{array}$

Conclusion:

Therefore, the perimeter of the rectangle is $11.24\text{m}$.

To determine

The magnitude and direction of the vector from the origin to the upper right coordinate of the rectangle.

Answer to Problem 3.56AP

The magnitude and direction of the vector from the origin to the upper right coordinate of the rectangle is $12.9\text{m}$ and $36.4°$ above the positive $x$ axis.

Explanation of Solution

Given info: The rectangle has sides parallel to the $x$ axis and $y$ axis. The position vectors of two corners is $10.0\text{m}$ for $\overrightarrow{\text{A}}$ at $50.0°$ and $12.0\text{m}$ for $\overrightarrow{\text{B}}$ at $30.0°$. From the Figure (1), the point from the origin to the upper right coordinate of the rectangle is $b$.

The vector of point $b$ is $\overrightarrow{\text{C}}$.

The vector $\overrightarrow{\text{C}}$ is,

$\overrightarrow{\text{C}}=\left(B_{\text{x}}\text{'}-x_{\text{o}}\right)+\left(A_{\text{y}}\text{'}-y_{\text{o}}\right)$

Here,

$x_{\text{o}}$ is the point at origin.

$y_{\text{o}}$ is the point at origin.

Substitute $0\text{m}$ for $x_{\text{o}}$, $0\text{m}$ for $y_{\text{o}}$, $\left(\text{10}\text{.4}\text{m}\right)\widehat{\text{i}}$ for $B_{\text{x}}\text{'}$, $\left(7.66\text{m}\right)\widehat{\text{j}}$ for $A_{\text{y}}\text{'}$ in the above equation.

$\begin{array}{c}\overrightarrow{\text{C}}=\left(\left(\text{10}\text{.4}\text{m}\right)\widehat{\text{i}}-0\text{m}\right)+\left(\left(7.66\text{m}\right)\widehat{\text{j}}-0\text{m}\right)\\ =\left(\text{10}\text{.4}\text{m}\right)\widehat{\text{i}}+\left(7.66\text{m}\right)\widehat{\text{j}}\end{array}$

Formula to calculate the magnitude of vector $\overrightarrow{\text{C}}$ is,

$C=\sqrt{C_{\text{x}}{}^2+C_{\text{y}}{}^2}$

Here,

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

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

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

Substitute $\text{10}\text{.4}\text{m}$ for $C_{\text{x}}$ and $7.66\text{m}$ for $C_{\text{y}}$ in the above equation.

$\begin{array}{c}C=\sqrt{{\left(\text{10}\text{.4}\text{m}\right)}^2+{\left(7.66\text{m}\right)}^2}\\ =12.91\text{m}\\ \approx \text{12}\text{.9}\text{m}\end{array}$

Formula to calculate the angle made by the vector $\overrightarrow{\text{C}}$ is,

$\theta ={\tan }^{-1}\left(\frac{C_{\text{y}}}{C_{\text{x}}}\right)$

Here,

$\theta$ is the angle made by the vector $\overrightarrow{\text{C}}$.

Substitute $\text{10}\text{.4}\text{m}$ for $C_{\text{x}}$ and $7.66\text{m}$ for $C_{\text{y}}$ in the above equation.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{7.66\text{m}}{\text{10}\text{.4}\text{m}}\right)\\ =36.37°\\ \approx 36.4°\end{array}$

Conclusion:

Therefore, the magnitude and direction of the vector from the origin to the upper right coordinate of the rectangle is $12.9\text{m}$ and $36.4°$ above the positive $x$ axis.