Chapter 3, Problem 40P

Figure P3.28 illustrates typical proportions of male (m) and female (f) anatomies. The displacements ${\stackrel{\to }{d}}_{1\text{m}}$ and ${\stackrel{\to }{d}}_{1\text{f}}$ from the soles of the feet to the navel have magnitudes of 104 cm and 64.0 cm. respectively. The displacements ${\stackrel{\to }{d}}_{2\text{m}}$ and ${\stackrel{\to }{d}}_{2\text{f}}$ from the navel to outstretched fingertips have magnitudes of 100 cm and 86.0 cm, respectively. Find the vector sum of these displacements ${\stackrel{\to }{d}}_3={\stackrel{\to }{d}}_1+{\stackrel{\to }{d}}_2$ for both people.

Figure P3.28

To determine

The vector sum of the given displacements for both male and female.

Answer to Problem 40P

The vector sum of the displacements for male is $170.12\text{cm}$ in a direction of $57.24°$ from positive $x$ axis and the vector sum of the displacements for female is $145.72\text{cm}$ in a direction of $58.59°$ from the positive $x$ axis.

Explanation of Solution

Section 1:

To determine: The vector sum of displacements for male in unit vector notation.

Answer: The unit vector notation for vector sum of displacements for male is $\left(92.05\widehat{i}\right)\text{cm}+\left(143.07\widehat{j}\right)\text{cm}$.

Given information:

The magnitude of male displacements, ${\overrightarrow{\text{d}}}_{\text{1m}}$ and ${\overrightarrow{\text{d}}}_{2\text{m}}$ is $104\text{cm}$ and $100\text{cm}$ respectively and the magnitude of female displacements, ${\overrightarrow{\text{d}}}_{\text{1f}}$ and ${\overrightarrow{\text{d}}}_{\text{2f}}$ is $84.0\text{cm}$ and $86.0\text{cm}$ respectively.

The displacements of male in unit vector notation is given as,

${\overrightarrow{\text{d}}}_{1\text{m}}=\left(104\widehat{j}\right)\text{cm}$

${\overrightarrow{\text{d}}}_{\text{2m}}=\left(100\cos 23°\widehat{i}\right)\text{cm}+\left(100\sin 23°\widehat{j}\right)\text{cm}$

Formula to calculate the vector sum for male is given as,

${\text{d}}_{\text{sm}}={\overrightarrow{\text{d}}}_{1\text{m}}+{\overrightarrow{\text{d}}}_{\text{2m}}$ (I)

Substitute $\left(104\widehat{j}\right)\text{cm}$ for ${\overrightarrow{\text{d}}}_{1\text{m}}$ and $\left(100\cos 23°\widehat{i}\right)\text{cm}+\left(100\sin 23°\widehat{j}\right)\text{cm}$ for ${\overrightarrow{\text{d}}}_{\text{2m}}$ in equation (I).

$\begin{array}{c}{\text{d}}_{\text{sm}}=\left(104\widehat{j}\right)\text{cm}+\left(100\cos 23°\widehat{i}\right)\text{cm}+\left(100\sin 23°\widehat{j}\right)\text{cm}\\ =\left(104\widehat{j}\right)\text{cm}+\left(92.0\widehat{i}\right)\text{cm}+\left(39.07\widehat{j}\right)\text{cm}\\ =\left(92.05\widehat{i}\right)\text{cm}+\left(143.07\widehat{j}\right)\text{cm}\end{array}$

Section 2:

To determine: The magnitude of the vector sum of displacements for male.

Answer: The magnitude of the vector sum of the displacements for male is $170.12\text{cm}$.

Given information:

The magnitude of male displacements, ${\overrightarrow{\text{d}}}_{\text{1m}}$ and ${\overrightarrow{\text{d}}}_{2\text{m}}$ is $104\text{cm}$ and $100\text{cm}$ respectively and the magnitude of female displacements, ${\overrightarrow{\text{d}}}_{\text{1f}}$ and ${\overrightarrow{\text{d}}}_{\text{2f}}$ is $84.0\text{cm}$ and $86.0\text{cm}$ respectively.

Formula to calculate the magnitude of the vector sum of the displacements for male is,

$\left|{\overrightarrow{\text{d}}}_{\text{sm}}\right|=\sqrt{{\left(d_{{\text{sm}}_{\text{x}}}\right)}^2+{\left(d_{{\text{sm}}_{\text{y}}}\right)}^2}$ (II)

Substitute $92.05\text{cm}$ for $d_{{\text{sm}}_{\text{x}}}$ and $143.07\text{cm}$ for $d_{{\text{sm}}_{\text{y}}}$ in equation (II).

$\begin{array}{c}\left|{\overrightarrow{\text{d}}}_{\text{sm}}\right|=\sqrt{{\left(92.05\text{cm}\right)}^2+{\left(143.07\text{cm}\right)}^2}\\ =170.12\text{cm}\end{array}$

Conclusion:

Therefore, the magnitude of the vector sum of the displacements for male is $170.12\text{cm}$.

Section 3:

To determine: The direction for the vector sum of the displacements for male.

Answer: The direction for the vector sum of the displacements for male is $57.24°$ from the positive $x$ axis.

Given information:

The magnitude of male displacements, ${\overrightarrow{\text{d}}}_{\text{1m}}$ and ${\overrightarrow{\text{d}}}_{2\text{m}}$ is $104\text{cm}$ and $100\text{cm}$ respectively and the magnitude of female displacements, ${\overrightarrow{\text{d}}}_{\text{1f}}$ and ${\overrightarrow{\text{d}}}_{\text{2f}}$ is $84.0\text{cm}$ and $86.0\text{cm}$ respectively.

Formula to calculate the direction for the vector sum of the displacements for male is,

${\theta }_{\text{m}}={\tan }^{-1}\left(\frac{d_{{\text{sm}}_{\text{y}}}}{d_{{\text{sm}}_{\text{x}}}}\right)$

Substitute $92.05\text{cm}$ for $d_{{\text{sm}}_{\text{x}}}$ and $143.07\text{cm}$ for $d_{{\text{sm}}_{\text{y}}}$ in above equation.

$\begin{array}{c}{\theta }_{\text{m}}={\tan }^{-1}\left(\frac{143.07\text{cm}}{92.05\text{cm}}\right)\\ =57.24°\end{array}$

Conclusion:

Therefore, the direction for the vector sum of the displacements for male is $57.24°$ from the positive $x$ axis.

Section 4:

To determine: The vector sum of displacements for female in unit vector notation.

Answer: The unit vector notation for vector sum of displacements for female is $\left(75.93\widehat{i}\right)\text{cm}+\left(124.37\widehat{j}\right)\text{cm}$.

Given information:

The magnitude of male displacements, ${\overrightarrow{\text{d}}}_{\text{1m}}$ and ${\overrightarrow{\text{d}}}_{2\text{m}}$ is $104\text{cm}$ and $100\text{cm}$ respectively and the magnitude of female displacements, ${\overrightarrow{\text{d}}}_{\text{1f}}$ and ${\overrightarrow{\text{d}}}_{\text{2f}}$ is $84.0\text{cm}$ and $86.0\text{cm}$ respectively.

The displacements of female in unit vector notation is given as,

${\overrightarrow{\text{d}}}_{1\text{f}}=\left(84.0\widehat{j}\right)\text{cm}$

${\overrightarrow{\text{d}}}_{\text{2f}}=\left(86.0\cos 28°\widehat{i}\right)\text{cm}+\left(86.0\sin 28°\widehat{j}\right)\text{cm}$

Formula to calculate the vector sum for female is given as,

${\text{d}}_{\text{sf}}={\overrightarrow{\text{d}}}_{1\text{f}}+{\overrightarrow{\text{d}}}_{\text{2f}}$ (III)

Substitute $\left(84.0\widehat{j}\right)\text{cm}$ for ${\overrightarrow{\text{d}}}_{1\text{f}}$, $\left(86.0\cos 28°\widehat{i}\right)\text{cm}+\left(86.0\sin 28°\widehat{j}\right)\text{cm}$ for ${\overrightarrow{\text{d}}}_{\text{2f}}$ in equation (III).

$\begin{array}{c}{\text{d}}_{\text{sf}}=\left(84.0\widehat{j}\right)\text{cm}+\left(86.0\cos 28°\widehat{i}\right)\text{cm}+\left(86.0\sin 28°\widehat{j}\right)\text{cm}\\ =\left(84.0\widehat{j}\right)\text{cm}+\left(75.93\widehat{i}\right)\text{cm}+\left(40.7\widehat{j}\right)\text{cm}\\ \text{=}\left(75.93\widehat{i}\right)\text{cm}+\left(124.37\widehat{j}\right)\text{cm}\end{array}$

Section 5:

To determine: The magnitude of the vector sum of displacements for female.

Answer: The magnitude of the vector sum of the displacements for female is $145.72\text{cm}$.

Given information:

The magnitude of male displacements, ${\overrightarrow{\text{d}}}_{\text{1m}}$ and ${\overrightarrow{\text{d}}}_{2\text{m}}$ is $104\text{cm}$ and $100\text{cm}$ respectively and the magnitude of female displacements, ${\overrightarrow{\text{d}}}_{\text{1f}}$ and ${\overrightarrow{\text{d}}}_{\text{2f}}$ is $84.0\text{cm}$ and $86.0\text{cm}$ respectively.

Formula to calculate the magnitude of the vector sum of the displacements for female is,

$\left|{\overrightarrow{\text{d}}}_{\text{sf}}\right|=\sqrt{{\left(d_{{\text{sf}}_{\text{x}}}\right)}^2+{\left(d_{{\text{sf}}_{\text{y}}}\right)}^2}$ (IV)

Substitute $75.93\text{cm}$ for $d_{{\text{sf}}_{\text{x}}}$ and $124.37\text{cm}$ $d_{{\text{sf}}_{\text{y}}}$ in equation (IV).

$\begin{array}{c}\left|{\overrightarrow{\text{d}}}_{\text{sf}}\right|=\sqrt{{\left(75.93\text{cm}\right)}^2+{\left(124.37\text{cm}\right)}^2}\\ =145.72\text{cm}\end{array}$

Conclusion:

Therefore, the magnitude of the vector sum of the displacements for female is $145.72\text{cm}$.

Section 6:

To determine: The direction for the vector sum of the displacements for female.

Answer: The direction for the vector sum of the displacements for female is $58.59°$ from the positive $x$ axis.

Given information:

The magnitude of male displacements, ${\overrightarrow{\text{d}}}_{\text{1m}}$ and ${\overrightarrow{\text{d}}}_{2\text{m}}$ is $104\text{cm}$ and $100\text{cm}$ respectively and the magnitude of female displacements, ${\overrightarrow{\text{d}}}_{\text{1f}}$ and ${\overrightarrow{\text{d}}}_{\text{2f}}$ is $84.0\text{cm}$ and $86.0\text{cm}$ respectively.

Formula to calculate the direction for the vector sum of the displacements for male is,

${\theta }_{\text{f}}={\tan }^{-1}\left(\frac{d_{{\text{sf}}_{\text{y}}}}{d_{{\text{sf}}_{\text{x}}}}\right)$

Substitute $75.93\text{cm}$ for $d_{{\text{sf}}_{\text{x}}}$ and $124.37\text{cm}$ $d_{{\text{sf}}_{\text{y}}}$ in above equation.

$\begin{array}{c}{\theta }_{\text{m}}={\tan }^{-1}\left(\frac{124.37\text{cm}}{75.93\text{cm}}\right)\\ =58.59°\end{array}$

Conclusion:

Therefore, the direction for the vector sum of the displacements for female is $58.59°$ from the positive $x$ axis.