Tag Archives: vector resultant

Hibbeler Statics 14E P2.1 — Solving for the Magnitude and Direction of the Resultant of Two Coplanar-Concurrent Forces

If θ=60°\theta = 60 \degree and F=450 N \textbf{F} = 450 \ \text{N}, determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.

Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 2-1

Solution:

The parallelogram law and the triangulation rule are shown in the figures below.

(a) Parallelogram Law
(b) Triangulation Rule

Considering figure (b), we can solve for the magnitude of FR \textbf{F}_R using the cosine law.

FR=7002+4502−2(700)(450)cos⁡45∘=497.01 N=497 N\begin{align*} \textbf{F}_R & = \sqrt{700^2+450^2-2\left( 700 \right)\left( 450 \right)\cos45^{\circ}}\\ & = 497.01 \ \text{N}\\ & = 497 \ \text{N} \end{align*}

Then we use the sine law to solve for the interior angle θ\theta.

sin⁡θ700=sin⁡45∘497.01sin⁡θ=700 sin⁡45∘497.01θ=sin⁡−1(700 sin⁡45∘497.01)This is an ambiguous case θ=84.81∘ or θ=95.19∘\begin{align*} \frac{\sin \theta}{700} & = \frac{\sin 45^{\circ}}{497.01}\\ \sin \theta & =\frac{700\ \sin 45^{\circ }}{497.01}\\ \theta & = \sin^{-1} \left( \frac{700\ \sin 45^{\circ }}{497.01} \right)\\ & \text{This is an ambiguous case }\\ \theta & = 84.81^\circ \ or \ \theta =95.19^\circ \\ \end{align*}

In here, the correct angle measurement is θ=95.19∘ \theta = 95.19^{\circ}.

Thus, the direction angle ϕ \phi of FR \textbf{F}_R measured counterclockwise from the positive x-axis, is

ϕ=θ+60∘=95.19∘+60∘=155∘\begin{align*} \phi & = \theta +60^\circ \\ & = 95.19^\circ +60^\circ \\ & = 155^\circ \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 19

Do Exercise 3.16 again using analytical techniques and change the second leg of the walk to 25.0 m straight south. (This is equivalent to subtracting B from A — that is, finding R’=A – B ) (b) Repeat again, but now you first walk 25.0 m north and then 18.0 m east. (This is equivalent to subtract A from B —that is, to find A=B+C . Is that consistent with your result?)

Solution:

Part A

From the given statement, you first walk 18.0 m straight west and then 25.0 straight south. These vectors are represented by the graph shown below.

To solve for the resultant, we simply need to use the Pythagorean theorem to solve for the hypotenuse of the right triangle formed. That is,

R=(18.0 m)2+(25.0 m)2R=30.8058 mR=30.8 m  (Answer)\begin{align*} R & = \sqrt{\left( 18.0 \ \text{m} \right)^2+\left( 25.0\ \text{m} \right)^2} \\ R & = 30.8058 \ \text{m} \\ R & = 30.8 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

To solve for the angle, θ, we shall use the tangent function.

θ=arctan⁡(25.0 m18.0 m)θ=54.2461∘θ=54.2∘\begin{align*} \theta & = \arctan \left( \frac{25.0 \ \text{m}}{18.0 \ \text{m}} \right) \\ \theta & = 54.2461^\circ \\ \theta & = 54.2^\circ \end{align*}

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

Part B

From the statement, you walk 25.0 m north first and then 18.0 m east. This is represented by the figure shown below.

R=(18.0 m)2+(25.0 m)2R=30.8058 mR=30.8 m  (Answer)\begin{align*} R & = \sqrt{\left( 18.0 \ \text{m} \right)^2+\left( 25.0\ \text{m} \right)^2} \\ R & = 30.8058 \ \text{m} \\ R & = 30.8 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

So, we have a right triangle with legs 25.0 m and 18.0 m. We are tasked to solve for the value of R, and the angle θ for the compass direction. The value of R can be solved using the Pythagorean Theorem as in Part A.

To solve for the angle, θ, we shall use the tangent function.

θ=arctan⁡(18.0 m25.0 m)θ=35.7539∘θ=35.8∘\begin{align*} \theta & = \arctan \left( \frac{18.0 \ \text{m}}{25.0 \ \text{m}} \right) \\ \theta & = 35.7539^\circ \\ \theta & = 35.8^\circ \end{align*}

Therefore, the compass direction of the resultant is 35.8° East of North.

This result is consistent with the previous results.

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Hibbeler Statics 14E P2.2 — Resultant of a System of Two Forces

Determine the magnitude of the resultant force FR=F1+F2\textbf{F}_{\text{R}} = \textbf{F}_1 + \textbf{F}_2 and its direction, measured counterclockwise from the positive x axis. 

Engineering Mechanics: Statics figure for Problem 2-3

Engineering Mechanics: Statics 13th Edition by RC Hibbeler, Problem 2-1
Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 2-3

SOLUTION:

The parallelogram law of the force system is shown.

Consider the triangle AOB.

Using cosine law to solve for the resultant force FR\textbf{F}_{\text{R}}

FR=(250)2+(375)2−2(250)(375)cos⁡ 75∘=393.2 lb=393 lb\begin{align*} \textbf{F}_\text{R} & =\sqrt{\left(250\right)^2+\left(375\right)^2-2\left(250\right)\left(375\right) \cos\:75^{\circ} }\\ & =393.2 \ \text{lb}\\ & =393\:\text{lb}\\ \end{align*}

The value of angle θ can be solved using sine law. 

393.2sin⁡ (75∘)=250sin⁡ θsin⁡θ=250 sin⁡75°393.2θ=sin⁡−1(250 sin⁡75°393.2)θ=37.89∘\begin{align*} \frac{393.2}{\sin\:\left(75^{\circ} \right)} & = \frac{250}{\sin\:\theta } \\ \sin \theta & = \frac{250 \ \sin75 \degree}{393.2}\\ \theta & =\sin^{-1} \left(\frac{250 \ \sin75 \degree}{393.2}\right)\\ \theta & = 37.89^{\circ}\\ \end{align*}

Solve for the unknown angle ϕ \phi .

ϕ=360∘−45∘+37.89∘=353∘\phi =360^{\circ} -45^{\circ} +37.89^{\circ} =353^{\circ}

The resultant force has a magnitude of 393 lb and is located 353º measured counterclockwise from the positive x-axis.

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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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