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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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College Physics by Openstax Chapter 3 Problem 21

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 DS and DW, respectively. The values of these components are solved below:

DS=(32.0 km)sin⁡35.0∘DS=18.4∘  (Answer)\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*}
DW=(32.0 km)cos⁡35.0∘DW=26.2∘  (Answer)\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:

DX=(32.0 km)cos⁡10∘DX=31.5∘  (Answer)\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*}
DY=(32.0 km)sin⁡10∘DY=5.56∘  (Answer)\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*}
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College Physics by Openstax Chapter 3 Problem 20

A new landowner has a triangular piece of flat land she wishes to fence. Starting at the west corner, she measures the first side to be 80.0 m long and the next to be 105 m. These sides are represented as displacement vectors A from B in Figure 3.59. She then correctly calculates the length and orientation of the third side C. What is her result?

Figure 3.59

Solution:

Consider the illustration shown.

We need to solve for an interior angle of the triangle. So, we need to solve for the value of α first. This can be done by simply subtracting the sum of 21 and 11 degrees from 90 degrees.

α=90∘−(21∘+11∘)α=58∘\begin{align*} \alpha & = 90 ^ \circ -\left( 21^\circ +11^\circ \right) \\ \alpha & = 58^\circ \end{align*}

Then, using the cosine law, we can now solve for the magnitude of vector C. That is

C2=A2+B2−2ABcos⁡αC2=(80 m)2+(105 m)2−2(80 m)(105 m)cos⁡58∘C2=8522.3564C=8522.3564C=92.3 m  (Answer)\begin{align*} C^2 & = A^2 + B^2 - 2AB \cos \alpha \\ C^2 & = \left( 80\ \text{m} \right)^2+\left( 105\ \text{m} \right)^2-2\left( 80\ \text{m} \right)\left( 105\ \text{m} \right) \cos 58^\circ \\ C^2 & = 8522.3564 \\ C & = \sqrt{8522.3564} \\ C & = 92.3 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

Before we can solve for the value of θ, we need to know the value of β first. This can be done by using the sine law.

sin⁡β80 m=sin⁡58∘92.3 msin⁡β=(80 m)sin⁡58∘92.3 mβ=arcsin⁡[(80 m)sin⁡58∘92.3 m]β=47.3∘\begin{align*} \frac{\sin \beta}{80\ \text{m}} & = \frac{\sin 58^\circ }{92.3 \ \text{m}} \\ \sin \beta & = \frac{\left( 80 \ \text{m} \right)\sin 58^\circ }{92.3 \ \text{m}} \\ \beta & = \arcsin \left[ \frac{\left( 80 \ \text{m} \right)\sin 58^\circ }{92.3 \ \text{m}} \right] \\ \beta & = 47.3^\circ \end{align*}

Finally, we can solve for θ.

θ=(90∘+11∘)−βθ=(90∘+11∘)−47.3∘θ=53.7∘  (Answer)\begin{align*} \theta & = \left( 90 ^\circ +11^\circ \right) - \beta \\ \theta & = \left( 90 ^\circ +11^\circ \right) - 47.3^\circ \\ \theta & = 53.7^\circ \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}
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College Physics by Openstax Chapter 3 Problem 18

You drive 7.50 km in a straight line in a direction 15º east of north. (a) Find the distances you would have to drive straight east and then straight north to arrive at the same point. (This determination is equivalent to find the components of the displacement along the east and north directions.) (b) Show that you still arrive at the same point if the east and north legs are reversed in order.

Solution:

Part A

Consider the illustration shown.

Let DE be the east component of the distance, and DN be the north component of the distance.

DE=7.50 sin⁡15∘DE=1.9411 kmDE=1.94 km  (Answer)\begin{align*} D_E & = 7.50 \ \sin 15^\circ \\ D_E & = 1.9411 \ \text{km} \\ D_E & = 1.94 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}
DN=7.50 cos⁡15∘DN=7.2444 kmDN=7.24 km  (Answer)\begin{align*} D_N & = 7.50 \ \cos 15^\circ \\ D_N & = 7.2444\ \text{km} \\ D_N & = 7.24 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

Part B

It can be obviously seen from the figure below that you still arrive at the same point if the east and north legs are reversed in order.

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College Physics by Openstax Chapter 3 Problem 17

Repeat Exercise 3.16 using analytical techniques, but reverse the order of the two legs of the walk and show that you get the same final result. (This problem shows that adding them in reverse order gives the same result—that is, B + A = A + B .) Discuss how taking another path to reach the same point might help to overcome an obstacle blocking your other path.

Figure 3.58 The two displacements A and B add to give a total displacement R having magnitude R and direction θ.

Solution:

Considering the right triangle formed by the vectors A, B, and R. We can solve for the magnitude of R using the Pythagorean Theorem. That is

R=A2+B2=(18.0m)2+(25.0m)2=30.806m≈30.8m(Answer)\begin{align*} R & = \sqrt{A^2+B^2} \\ & = \sqrt{\left( 18.0 \text{m} \right)^2+\left( 25.0 \text{m} \right)^2} \\ & =30.806 \text{m} \\ & \approx 30.8 \text{m} \qquad {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

Then we solve for the compass direction by solving the value θ using the same right triangle.

θ=arctan⁡(BA)=arctan⁡(25.0m18.0m)=54.246∘≈54.2∘\begin{align*} \theta & = \arctan \left( \frac{B}{A} \right) \\ & = \arctan \left( \frac{25.0 \text{m}}{18.0 \text{m}} \right) \\ & = 54.246 ^\circ \\ & \approx 54.2 ^ \circ \\ \end{align*}

Therefore, the compass direction of the resultant is 54.2° North of West.

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College Physics by Openstax Chapter 3 Problem 13

Find the following for path C in Figure 3.56: (a) the total distance traveled and (b) the magnitude and direction of the displacement from start to finish. In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition.

Figure 3.56 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side.

Solution:

Part A

Looking at path C, it moves 1 block upward, 5 blocks to the right, 2 blocks downward, 1 block to the left, 1 block upward, and 3 blocks to the left. So, the total distance is

distance = (1×120 m)+(5×120 m)+(2×120 m)+(1×120 m)+(1×120 m)+(3×120 m)= 120 m+600 m+240 m+120 m+120 m+360 m= 1560 m  (Answer)\begin{align*} \text{distance} \ = \ &\left( 1\times 120 \ \text{m} \right)+\left( 5\times 120\ \text{m} \right)+\left( 2\times 120\ \text{m} \right)+\left( 1\times 120\ \text{m} \right) \\ & +\left( 1\times 120\ \text{m} \right)+\left( 3\times 120\ \text{m} \right) \\ = \ & 120\ \text{m}+600 \ \text{m}+240 \ \text{m} + 120 \ \text{m}+ 120 \ \text{m}+360\ \text{m} \\ = \ & 1560 \ \text{m} \ \qquad \ {\color{DarkOrange}\left( \text{Answer} \right) } \end{align*}

Part B

It can be seen from the figure that the end of path C is just one block to the right from the starting point. Therefore, the magnitude of the displacement is

displacement=120 m  (Answer)\begin{align*} \text{displacement} = 120\ \text{m} \ \qquad \ {\color{DarkOrange}\left( \text{Answer} \right) } \end{align*}

The direction is to the right or is equivalent to 0° measured from the positive x-axis.

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College Physics by Openstax Chapter 3 Problem 24

Suppose a pilot flies 40.0 km in a direction 60º north of east and then flies 30.0 km in a direction 15º north of east as shown in Figure 3.61. Find her total distance R from the starting point and the direction θ of the straight-line path to the final position. Discuss qualitatively how this flight would be altered by a wind from the north and how the effect of the wind would depend on both wind speed and the speed of the plane relative to the air mass.

Figure 3.61

Solution:

The pilot’s displacement is characterized by 2 vectors, A and B, as depicted in Figure 3.61. To determine her total displacement R from the starting point, we need to add the two given vectors. To do this, we individually get the x and y components of each vector. This is presented in the table that follows:

Vectorx-componenty-component
A40 cos⁡60∘=20 km40\:\cos 60^{\circ} =20\:\text{km} 40 sin⁡60∘=34.6410 km40\:\sin 60^{\circ} =34.6410\:\text{km}
B 30 cos⁡15∘=28.9778 km30\:\cos 15^{\circ} =28.9778\:\text{km} 30 sin⁡15∘=7.7646 km30\:\sin 15^{\circ} =7.7646\:\text{km}
Sum 48.9778 km48.9778\: \text{km} 42.4056 km42.4056 \:\text{km}

The table above indicates east and north as positive components, while west and south indicate negative components. The last row is the sum of the components. These are also the x and y components of the resultant vector.

To calculate the magnitude of the resultant, we simply use the Pythagorean Theorem as follows:

R=(48.9778 km)2+(42.4056 km)2R=64.8 km  (Answer)\begin{align*} \text{R} & = \sqrt{\left(48.9778\:\text{km}\right)^2+\left(42.4056\:\text{km}\right)^2} \\ \text{R} & = 64.8\:\text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \\ \end{align*}

The direction of the resultant is calculated as follows:

θ=tan⁡−1(42.405648.9778)θ=40.9∘  (Answer)\begin{align*} \theta & =\tan ^{-1}\left(\frac{42.4056}{48.9778}\right) \\ \theta & =40.9^{\circ} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

Therefore, the pilot’s resultant displacement is about 64.8 km directed 40.9° North of East from the starting island.

Discussion:

If the wind speed is less than the speed of the plane, it is possible to travel to the northeast, but she will travel more to the east than without the wind. If the wind speed is greater than the speed of the plane, then it is no longer possible for the plane to travel to the northeast, it will end up traveling southeast.

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College Physics by Openstax Chapter 3 Problem 23

In an attempt to escape his island, Gilligan builds a raft and sets to sea. The wind shifts a great deal during the day, and he is blown along the following straight lines: 2.50 km 45.0º north of west; then 4.70 km 60.0º south of east; then 1.30 km 25.0º south of west; then 5.10 km straight east; then 1.70 km 5.00º east of north; then 7.20 km 55.0º south of west; and finally 2.80 km 10.0º north of east. What is his final position relative to the island?

Solution:

Gilligan’s displacement is characterized by 7 vectors. To determine his final position relative to the starting point, we simply need to add the vectors. To do this, we individually get the x and y components of each vector. This is presented in the table that follows:

VectorX-ComponentY-Component
(1) −2.5 cos⁡45∘=−1.7678 km -2.5\:\cos 45^{\circ} =-1.7678\:\text{km} +2.5 sin⁡45∘=+1.7678 km +2.5\:\sin 45^{\circ} =+1.7678\:\text{km}
(2) +4.70 cos⁡60∘=+2.3500 km +4.70\:\cos 60^{\circ} =+2.3500\:\text{km} −4.70 sin⁡60∘=−4.0703 km -4.70\:\sin 60^{\circ} =-4.0703\:\text{km}
(3) −1.30 cos⁡25∘=−1.1782 km -1.30\:\cos 25^{\circ} =-1.1782\:\text{km} −1.30 sin⁡25∘=−0.5494 km -1.30\:\sin 25^{\circ} =-0.5494\:\text{km}
(4) +5.1000 km +5.1000\:\text{km} 0 0
(5) +1.70 sin⁡5∘=+0.1482 km +1.70\:\sin 5^{\circ} =+0.1482\:\text{km} +1.70 cos⁡5∘=+1.6935 km +1.70\:\cos 5^{\circ} =+1.6935\:\text{km}
(6) −7.20 cos⁡55∘=−4.1298 km -7.20\:\cos 55^{\circ} =-4.1298\:\text{km} −7.20 sin⁡55∘=−5.8979 km -7.20\:\sin 55^{\circ} =-5.8979\:\text{km}
(7) +2.80 cos⁡10∘=+2.7575 km +2.80\:\cos 10^{\circ} =+2.7575\:\text{km} +2.80 sin⁡10∘=+0.4862 km +2.80\:\sin 10^{\circ} =+0.4862\:\text{km}
Sum 3.2799 km 3.2799\:\text{km} −6.5701 km -6.5701\:\text{km}

The table above indicates east and north as positive components, while west and south indicate negative components. The last row is the sum of the components. This is also the x and y components of the resultant vector.

To calculate the magnitude of the resultant, we simply use the Pythagorean Theorem as follows:

R=(3.2799 km)2+(−6.5701 km)2R=7.34 km  (Answer)\begin{align*} \text{R} & =\sqrt{\left(3.2799\:\text{km}\right)^2+\left(-6.5701\:\text{km}\right)^2} \\ \text{R} & =7.34\:\text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \\ \end{align*}

The direction of the resultant is calculated as follows:

θ=tan⁡−1(6.57013.2799)θ=63.47∘  (Answer)\begin{align*} \theta & =\tan ^{-1}\left(\frac{6.5701}{3.2799}\right) \\ \theta & =63.47^{\circ} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \\ \end{align*}

Therefore, Gilligan is about 7.34 km directed 63.47° South of East from the starting island.

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