College Physics by Openstax Chapter 3 Problem 22

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?

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.
Figure 3.60

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

Ax=(4.70 km)cos⁡7.5∘Ax=4.6598 km\begin{align*} A_x & = \left( 4.70 \ \text{km} \right) \cos 7.5^\circ \\ A_x & = 4.6598 \ \text{km} \end{align*}
Ay=−(4.70 km)sin⁡7.5∘Ay=−0.6135 km\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

Bx=−(2.48 km)sin⁡16∘Bx=−0.6836 km\begin{align*} B_x & =-\left( 2.48 \ \text{km} \right) \sin 16^\circ \\ B_x & = -0.6836 \ \text{km} \end{align*}
By=(2.48 km)cos⁡16∘By=2.3839 km\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

Cx=−(3.02 km)cos⁡19∘Cx=−2.8555 km\begin{align*} C_x & = -\left( 3.02 \ \text{km} \right) \cos 19^\circ \\ C_x & = -2.8555 \ \text{km} \end{align*}
Cy=(3.02 km)sin⁡19∘Cy=0.9832 km\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.

Ax+Bx+Cx+Dx=04.6598 km−0.6836 km−2.8555 km+Dx=01.1207 km+Dx=0Dx=−1.1207 km\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.

Ay+By+Cy+Dy=0−0.6135 km+2.3839 km+0.9832 km+Dy=02.7536 km+Dy=0Dy=−2.7536 km\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.

D=(Dx)2+(Dy)2D=(−1.1207 km)2+(−2.7536 km)2D=2.97 km  (Answer)\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:

θ=tan⁡−1∣DxDy∣θ=tan⁡−1∣−1.1207 km−2.7536 km∣θ=22.1∘  (Answer)\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*}
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