Tag Archives: graphical method of vector sum

College Physics by Openstax Chapter 3 Problem 2

Find the following for path B in Figure 3.52:
(a) The total distance traveled, and
(b) The magnitude and direction of the displacement from start to finish.

Figure 3.54 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side
Figure 3.52 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side

Solution:

Part A

The total distance traveled is 

d=(4×120m)+(3×120 m)+(3×120 m)d=1200 m  (Answer)\begin{align*} \text{d} & = \left(4 \times 120 \text{m} \right) + \left(3 \times 120\ \text{m} \right) + \left(3 \times 120\ \text{m} \right) \\ \text{d} & = 1 200\ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}\\ \end{align*}

Part B

The magnitude of the displacement is 

s=(sx)2+(sy)2s=(1×120 m)2+(3×120 m)2s=(120 m)2+(360 m)2s=379 m  (Answer)\begin{align*} \text{s} & = \sqrt{\left( s_x \right)^2+\left( s_y \right)^2} \\ \text{s} & = \sqrt{\left( 1 \times 120\ \text{m} \right)^2+ \left( 3 \times 120 \ \text{m} \right)^2} \\ \text{s} & = \sqrt{\left( 120\ \text{m} \right)^2+ \left( 360 \ \text{m} \right)^2} \\ \text{s} & = 379 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

The direction is

θ=arctan⁡(sysx)θ=arctan⁡(360 m120 m)θ=71.6∘, N of E (Answer)\begin{align*} \theta & = \arctan \left( \frac{s_y}{s_x} \right) \\ \theta & = \arctan \left( \frac{360\ \text{m}}{120\ \text{m}} \right) \\ \theta & = 71.6^\circ , \ \text{N of E} \ \qquad {\color{DarkOrange} \left( \text{Answer} \right)}\\ \end{align*}
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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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College Physics by Openstax Chapter 3 Problem 9

Show that the sum of the vectors discussed in Example 3.2 gives the result shown in Figure 3.24.

Figure 3.24

Solution:

So, we are given the two vectors shown below.

Vectors A and B

If we use the graphical method of adding vectors, we can join the two vectors using head-tail addition and come up with the following:

Figure 3.9B: Vectors A and B added graphically

The resultant is drawn from the tail of the first vectors (the origin) to the head of the last vector. The resultant is shown in red in the figure below.

Solve for the value of the angle 𝛼 by geometry.

α=66∘+(180∘−112∘)=134∘\alpha = 66^\circ +\left( 180^\circ-112^\circ \right) = 134^\circ

Solve for the magnitude of the resultant using cosine law.

R2=A2+B2−2ABcos⁡αR=A2+B2−2ABcos⁡αR=(27.5 m)2+(30.0 m)2−2(27.5 m)(30.0 m)cos⁡134∘R=52.9380 mR=52.9 m  (Answer)\begin{align*} R^2 & = A^2+B^2-2AB\cos \alpha \\ R & = \sqrt{A^2+B^2-2AB\cos \alpha} \\ R & = \sqrt{\left( 27.5 \ \text{m} \right)^2+\left( 30.0 \ \text{m} \right)^2-2\left( 27.5\ \text{m} \right)\left( 30.0\ \text{m} \right) \cos 134^\circ} \\ R & =52.9380 \ \text{m} \\ R & = 52.9 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

Solve for 𝛽 using sine law.

sin⁡βB=sin⁡αRβ=sin⁡−1(Bsin⁡αR)β=sin⁡−1(30.0 msin⁡134∘52.9380 m)β=24.0573∘\begin{align*} \frac{\sin \beta}{B} & = \frac{\sin \alpha}{R} \\ \beta & = \sin ^{-1} \left( \frac{B \sin \alpha }{R} \right) \\ \beta & = \sin ^{-1} \left( \frac{30.0\ \text{m} \sin 134^\circ}{52.9380 \ \text{m}} \right) \\ \beta & = 24.0573^\circ \end{align*}

Finally, solve for 𝜃.

θ=66∘+24.0573∘=90.1∘  (Answer)\theta = 66^\circ+24.0573^\circ = 90.1^\circ \ \qquad \ {\color{Orange} \left( \text{Answer} \right)}

The result is in conformity with that in figure 3.24 shown on the question shown above.

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College Physics by Openstax Chapter 3 Problem 1

Find the following for path A in Figure 3.52:
(a) The total distance traveled, and
(b) The magnitude and direction of the displacement from start to finish.

Figure 3.54 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side
Figure 3.52 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side

Solution:

Part A

The total distance traveled is 

d=(3×120 m)+(1×120 m)d=480 m  (Answer)\begin{align*} \text{d} & =\left(3\times 120\ \text{m}\right)+\left(1\times 120\:\text{m}\right) \\ \text{d} & =480\:\text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}\\ \end{align*}

Part B

The magnitude of the displacement is 

s =(sx)2 +(sy)2s =(1×120 m)2+(3×120 m)2s =379 m  (Answer)\begin{align*} \text{s }& =\sqrt{\left( s_x \right)^{2\:}+\left( s_y \right)^2} \\ \text{s }& = \sqrt{\left(1\times 120\:\text{m}\right)^2+\left(3\times 120\:\text{m}\right)^2} \\ \text{s }& = 379\ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

The direction is

θ=tan⁡−1(sxsy)θ=tan⁡−1(1×120 m3×120 m)θ=71.6∘, E of N  (Answer)\begin{align*} \theta & = \tan^{-1}\left(\frac{s_x}{s_y}\right) \\ \theta & = \tan^{-1}\left(\frac{1\times 120\:\text{m}}{3\times 120 \ \text{m}}\right) \\ \theta & =71.6^{\circ} ,\:\text{E of N} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}
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