A farmer wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A, B, and C in Figure 3.60, and then correctly calculates the length and orientation of the fourth side D. What is his result?

Solution:

For the four-sided plot to be closed, the resultant displacement of the four sides should be zero. The sum of the horizontal components should be zero, and the sum of the vertical components should also be equal to zero.

We need to solve for the components of each vector. Take into consideration that rightward and upward components are positive, while the reverse is negative.

For vector A, the components are

\begin{align*}
A_x & = \left( 4.70 \ \text{km} \right) \cos 7.5^\circ \\
A_x & = 4.6598 \ \text{km}
\end{align*}

\begin{align*}
A_y & = -\left( 4.70 \ \text{km} \right) \sin 7.5^\circ \\
A_y & = -0.6135 \ \text{km}
\end{align*}

The components of vector B are

\begin{align*}
B_x & =-\left( 2.48 \ \text{km} \right) \sin 16^\circ \\
B_x & = -0.6836 \ \text{km}

\end{align*}

\begin{align*}
B_y & =\left( 2.48 \ \text{km} \right) \cos 16^\circ \\
B_y & =2.3839 \ \text{km}
\end{align*}

For vector C, the components are

\begin{align*}
C_x & = -\left( 3.02 \ \text{km} \right) \cos 19^\circ \\
C_x & = -2.8555 \ \text{km}
\end{align*}

\begin{align*}
C_y & = \left( 3.02 \ \text{km} \right) \sin 19^\circ \\
C_y & = 0.9832 \ \text{km}
\end{align*}

Now, we need to take the sum of the x-components and equate it to zero. The x-component of D is unknown.

\begin{align*}
A_x+B_x+C_x+D_x & =0 \\
4.6598 \ \text{km}-0.6836 \ \text{km}-2.8555 \ \text{km}+ D_x & =0 \\
1.1207 \ \text{km} +D_x & =0 \\
D_x & = -1.1207 \ \text{km}
\end{align*}

We also need to take the sum of the y-component and equate it to zero to solve for the y-component of D.

\begin{align*}
A_y +B_y+C_y+D_y & =0 \\
-0.6135 \ \text{km}+2.3839 \ \text{km}+0.9832 \ \text{km}+ D_y & =0 \\
2.7536 \ \text{km} +D_y & =0 \\
D_y & = -2.7536 \ \text{km}
\end{align*}

To solve for the distance of D, we shall use the Pythagorean Theorem.

\begin{align*}
D & = \sqrt{\left( D_x \right)^2+\left( D_y \right)^2} \\
D & = \sqrt{\left( -1.1207 \ \text{km} \right)^2+\left( -2.7536 \ \text{km} \right)^2} \\
D & = 2.97 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Then we can solve for θ using the tangent function. Since it is taken from the vertical axis, it can be solved by:

\begin{align*}
\theta & = \tan^{-1} \left| \frac{D_x}{D_y} \right|
 \\
\theta & = \tan^{-1} \left| \frac{-1.1207 \ \text{km}}{-2.7536 \ \text{km}} \right|
\\
\theta & = 22.1 ^ \circ  \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

You fly 32.0 km in a straight line in still air in the direction 35.0º south of west. (a) Find the distances you would have to fly straight south and then straight west to arrive at the same point. (This determination is equivalent to finding the components of the displacement along the south and west directions.) (b) Find the distances you would have to fly first in a direction 45.0º south of west and then in a direction 45.0º west of north. These are the components of the displacement along a different set of axes—one rotated 45º.

Solution:

Part A

Consider the illustration shown.

The south and west components of the 32.0 km distance are denoted by D_S and D_W, respectively. The values of these components are solved below:

\begin{align*}
D_S & = \left( 32.0\ \text{km} \right) \sin 35.0 ^\circ \\
D_S & = 18.4^\circ \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

\begin{align*}
D_W & = \left( 32.0\ \text{km} \right) \cos 35.0 ^\circ \\
D_W & = 26.2^\circ \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Part B

Consider the new set of axes (X-Y) as shown below. This new set of axes is rotated 45° from the original axes. Thus, axis X is 45° south of west, and axis Y is 45° west of north. First, we can obviously see that θ has a value of 10°.

Therefore, the components of the 32.0 km distance along X and Y axes are:

\begin{align*}
D_X & = \left( 32.0 \ \text{km} \right) \cos 10^\circ  \\
D_X & = 31.5^\circ \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

\begin{align*}
D_Y & = \left( 32.0 \ \text{km} \right) \sin 10^\circ  \\
D_Y & = 5.56^\circ \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}