Chapter 8.3, Problem 41E

(a)

To determine

To Find: The diagram for the given situation.

Answer to Problem 41E

The required diagram is shown in Figure 1

Explanation of Solution

Given:

Dr. Chiaki Mukai is japans first female astronaut suppose that she is working inside the compartment that is shaped like a cube with the sides 15 feet long. She realizes that the tool she needs is diagonally in the corner of the compartment.

Calculation:

Consider the compartment shaped like the cube is with the sides 15 ft. long.

The required diagram is shown in Figure 1

Figure 1

(b)

To determine

To Find: Theminimum distance that the astronaut has to glide to secure the tool.

Answer to Problem 41E

The shortest distance between the Dr and the tool is $26\text{ft}$ .

Explanation of Solution

Consider that the position of the doctor is $\left(0,0,0\right)$ and the final coordinate is $\left(15,15,15\right)$ .

Then, the distance to which the astronaut has to glide is,

  $\begin{array}{c}\left|\overrightarrow{OT}\right|=\sqrt{{15}^2+{15}^2+{15}^2}\\ =15\sqrt{3}\\ \cong 26\text{ft}\end{array}$

Then, the shortest distance between the Dr and the tool is $26\text{ft}$ .

(c)

To determine

To Find: The angle to the floor at which the astronaut must launch herself.

Answer to Problem 41E

The required angle is $35.25°$ .

Explanation of Solution

Consider the angle to the floor at which the astronaut must launch herself is,

  $\cos \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+a_2^2+a_3^2}\cdot \sqrt{a_1^2+a_2^2+a_3^2}}$

Then,

  $\begin{array}{l}\cos \theta =\frac{15\cdot 15+15\cdot 15+15\cdot 0}{\sqrt{{\left(15\right)}^2+{\left(15\right)}^2+{\left(15\right)}^2}\cdot \sqrt{{\left(15\right)}^2+{\left(15\right)}^2+{\left(0\right)}^2}}\\ \cos \theta =0.8166\\ \theta ={\cos }^{-1}\left(0.8166\right)\\ \theta =35.25°\end{array}$