Tag Archives: Vector Addition

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,

\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.

\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.

\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.

\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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College Physics by Openstax Chapter 3 Problem 12

Find the components of v_tot along a set of perpendicular axes rotated 30º counterclockwise relative to those in Figure 3.55.

The figure shows v_A directed 22.5° from the positive x-axis, and v_B started from the head of v_A and is directed 23.0° from the resultant. The resultant is given to be 6.72 m/s and is directed 26.5° from v_A. In total, the resultant is measured 49° from the positive x-axis. — ***Figure 3.55***

Solution:

By isolating the v_tot from the rest of the other vectors, we come up with the following figure.

The resultant velocity has a magnitude of 6.72 m/s and is directed 49° from the positive x-axis. Now, we shall create another set of axes rotated at 30° counterclockwise. We call the axes x’ and y’ axes. The figure is shown below.

***The resultant velocity with the rotated axes.***

From the figure, we can see that the resultant velocity is 19° from the x’ axis. Therefore, the x’ and y’ components are:

\text{x'-component}=\left(6.72\:\text{m/s}\right)\cos 19^{\circ} =6.35\:\text{m/s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

\text{y'-component}=\left(6.72\:\text{m/s}\right)\sin 19^{\circ} =2.19\:\text{m/s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

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College Physics by Openstax Chapter 3 Problem 11

Find the components of v_tot along the x- and y-axes in Figure 3.55.

Solution:

By isolating the v_tot from the rest of the other vectors, we come up with the following figure. Also, the x and y-components are shown.

***The resultant velocity and its x and y components***

The resultant velocity has a magnitude of 6.72 m/s and is directed 49° from the positive x-axis. To solve for the x and y components, we just need to solve the legs of the right triangle formed by the three vectors. That is,

\text{x-component}=\left(6.72\:\text{m/s}\right)\cos 49^{\circ} =4.41\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

\text{y-component}=\left(6.72\:\text{m/s}\right)\sin 49^{\circ} =5.07\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

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College Physics by Openstax Chapter 3 Problem 10

Find the magnitudes of velocities v_A and v_B in Figure 3.55

Solution:

Basically, we are given an oblique triangle. First, we shall determine the value of the interior angle at the intersection of v_A and v_B. We can solve this knowing that the sum of the interior angles of a triangle is 180°.

To solve for v_A and v_B, we will use the sine law.

\begin{align*} \frac{\text{v}_{\text{A}}}{\sin 23^{\circ} } & =\frac{6.72\:\text{m/s}}{\sin 130.5^{\circ} } \\ \text{v}_{\text{A}} & =\frac{6.72\:\text{m/s}\:\sin \:23^{\circ }\:}{\sin \:130.5^{\circ }\:} \\ \text{v}_{\text{A}} & =3.45\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Using the same law to solve for the value of v_B, we have

\begin{align*} \frac{\text{v}_{\text{B}}}{\sin 26.5^{\circ} } & =\frac{6.72\:\text{m/s}}{\sin 130.5^{\circ} } \\ \text{v}_{\text{B}} & =\frac{6.72\:\text{m/s}\:\sin \:26.5^{\circ }\:\:}{\sin \:130.5^{\circ }\:} \\ \text{v}_{\text{B}} & =3.94\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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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.

Solution:

So, we are given the two vectors shown below.

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.

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

Solve for the magnitude of the resultant using cosine law.

\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.

\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.

\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 8

Show that the order of addition of three vectors does not affect their sum. Show this property by choosing any three vectors A, B, and C, all having different lengths and directions. Find the sum A + B + C then find their sum when added in a different order and show the result is the same. (There are five other orders in which A, B, and C can be added; choose only one.)

Solution:

Consider the three vectors shown in the figures below:

Vector A

Vector B

Vector C

First, we shall add them A+B+C. Using the head-tail or graphical method of vector addition, we have the figure shown below.

***Figure 3.8B: The resultant force of A+B+C***

Now, let us try to find the sum of the three vectors by reordering vectors A, B, and C. Let us try to find the sum of C+B+A in that order. The result is shown below.

***Figure 3.8C: The resultant of 3 vectors added in different order.***

We can see that the resultant is the same directed from the origin upward. This proves that the resultant must be the same even if the vectors are added in different order.

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Arizona State University| PHY 121: Univ Physics I: Mechanics|Homework 1-1| Problem 2

Learning Goal:

To practice Tactics Box 1.2 Vector Subtraction.

Vector subtraction has some similarities to the subtraction of two scalar quantities. With numbers, subtraction is the same as the addition of a negative number. For example, 5−3 is the same as 5+(−3). Similarly, A⃗ −B⃗ =A⃗ +(−B⃗ ). We can use the rules for vector addition and the fact that −B⃗ is a vector opposite in direction to B⃗ to form rules for vector subtraction, as explained in this tactics box.

The figure shows two vectors. Vector A is on the left and is directed upwards to the right. Vector B is on the right and is directed downwards to the right.

To subtract B⃗ from A⃗ (Figure 1), perform these steps:

Draw A⃗ .
Draw −B⃗ and place its tail at the tip of A⃗ .
Draw an arrow from the tail of A⃗ to the tip of −B⃗ . This is vector A⃗ −B⃗ .

Part A

Draw vector C⃗ =A⃗ −B⃗ by following the steps in the tactics box above. When drawing −B⃗ , keep in mind that it has the same magnitude as B⃗ but opposite direction.

Part B

Draw vector F⃗ =D⃗ −E⃗ by following the steps in the tactics box above. When drawing −E⃗ , keep in mind that it has the same magnitude as E⃗ but opposite direction.

ANSWERS:

Part A

Part B

Arizona State University| PHY 121: Univ Physics I: Mechanics|Homework 1-1| Problem 1

Learning Goal:
To practice Tactics Box 1.1 Vector Addition.

Vector addition obeys rules that are different from those for the addition of two scalar quantities. When you add two vectors, their directions, as well as their magnitudes, must be taken into account. This Tactics Box explains how to add vectors graphically.

To add B⃗ to A⃗ (Figure 1), perform these steps:

Draw A⃗ .
Place the tail of B⃗ at the tip of A⃗ .
Draw an arrow from the tail of A⃗ to the tip of B⃗ . This is vector A⃗ +B⃗ .

Part A

Create the vector R⃗ =A⃗ +B⃗ by following the steps in the Tactics Box above. When moving vector B⃗ , keep in mind that its direction should remain unchanged.

Part B

Create the vector R⃗ =C⃗ +D⃗ by following the steps in the Tactics Box above. When moving vector D⃗ , keep in mind that its direction should remain unchanged. The location, orientation, and length of your vectors will be graded.

ANSWERS:

Part A

Part B

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

\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

\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

\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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Tag Archives: Vector Addition

College Physics by Openstax Chapter 3 Problem 19

College Physics by Openstax Chapter 3 Problem 12

Find the components of v_tot along a set of perpendicular axes rotated 30º counterclockwise relative to those in Figure 3.55.

College Physics by Openstax Chapter 3 Problem 11

Find the components of v_tot along the x- and y-axes in Figure 3.55.

College Physics by Openstax Chapter 3 Problem 10

Find the magnitudes of velocities v_A and v_B in Figure 3.55

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.

College Physics by Openstax Chapter 3 Problem 8

Arizona State University| PHY 121: Univ Physics I: Mechanics|Homework 1-1| Problem 2

Arizona State University| PHY 121: Univ Physics I: Mechanics|Homework 1-1| Problem 1

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.

Find the components of vtot along a set of perpendicular axes rotated 30º counterclockwise relative to those in Figure 3.55.

Find the components of vtot along the x- and y-axes in Figure 3.55.

Find the magnitudes of velocities vA and vB in Figure 3.55

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

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.

Find the components of v_tot along a set of perpendicular axes rotated 30º counterclockwise relative to those in Figure 3.55.

Find the components of v_tot along the x- and y-axes in Figure 3.55.

Find the magnitudes of velocities v_A and v_B in Figure 3.55

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.