Physics

5.2Vector Addition and Subtraction: Analytical Methods

Physics5.2 Vector Addition and Subtraction: Analytical Methods

Section Learning Objectives

By the end of this section, you will be able to do the following:

• Define components of vectors
• Describe the analytical method of vector addition and subtraction
• Use the analytical method of vector addition and subtraction to solve problems

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• (3) Scientific processes. The student uses critical thinking, scientific reasoning, and problem solving to make informed decisions within and outside the classroom. The student is expected to:
• (F) express and interpret relationships symbolically in accordance with accepted theories to make predictions and solve problems mathematically, including problems requiring proportional reasoning and graphical vector addition
• (4) Science concepts. The student knows and applies the laws governing motion in two dimensions for a variety of situations. The student is expected to:
• (E) develop and interpret free-body force diagrams;
• (F) identify and describe motion relative to different frames of reference.

In addition, the High School Physics Laboratory Manual addresses content in this section in the lab titled: Motion in Two Dimensions, as well as the following standards:

• (3) Scientific processes. The student uses critical thinking, scientific reasoning, and problem solving to make informed decisions within and outside the classroom. The student is expected to:
• (F) express and interpret relationships symbolically in accordance with accepted theories to make predictions and solve problems mathematically, including problems requiring proportional reasoning and graphical vector addition.

Section Key Terms

 analytical method component (of a two-dimensional vector)

Components of Vectors

For the analytical method of vector addition and subtraction, we use some simple geometry and trigonometry, instead of using a ruler and protractor as we did for graphical methods. However, the graphical method will still come in handy to visualize the problem by drawing vectors using the head-to-tail method. The analytical method is more accurate than the graphical method, which is limited by the precision of the drawing. For a refresher on the definitions of the sine, cosine, and tangent of an angle, see Figure 5.17.

Figure 5.17 For a right triangle, the sine, cosine, and tangent of θ are defined in terms of the adjacent side, the opposite side, or the hypotenuse. In this figure, x is the adjacent side, y is the opposite side, and h is the hypotenuse.

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[BL][OL] Review trigonometric concepts of sine, cosine, tangent and the Pythagorean theorem.

Since, by definition, $cosθ=x/hcosθ=x/h$, we can find the length x if we know h and $θθ$ by using $x=hcosθx=hcosθ$ . Similarly, we can find the length of y by using $y=hsinθy=hsinθ$ . These trigonometric relationships are useful for adding vectors.

When a vector acts in more than one dimension, it is useful to break it down into its x and y components. For a two-dimensional vector, a component is a piece of a vector that points in either the x- or y-direction. Every 2-d vector can be expressed as a sum of its x and y components.

For example, given a vector like $A A$ in Figure 5.18, we may want to find what two perpendicular vectors, $A x A x$ and $A y A y$, add to produce it. In this example, $A x A x$ and $A y A y$ form a right triangle, meaning that the angle between them is 90 degrees. This is a common situation in physics and happens to be the least complicated situation trigonometrically.

Figure 5.18 The vector $A A$, with its tail at the origin of an x- y-coordinate system, is shown together with its x- and y-components, $A x A x$ and $A y . A y .$ These vectors form a right triangle.

$A x A x$ and $A y A y$ are defined to be the components of $A A$ along the x- and y-axes. The three vectors, $A A$, $A x A x$, and $A y A y$, form a right triangle.

If the vector $A A$ is known, then its magnitude $A A$ (its length) and its angle $θ θ$ (its direction) are known. To find $A x A x$ and $A y A y$, its x- and y-components, we use the following relationships for a right triangle:

$A x =Acosθ A x =Acosθ$

and

$A y =Asinθ, A y =Asinθ,$

where $A x A x$ is the magnitude of A in the x-direction, $A y A y$ is the magnitude of A in the y-direction, and $θ θ$ is the angle of the resultant with respect to the x-axis, as shown in Figure 5.19.

Figure 5.19 The magnitudes of the vector components $A x A x$ and $A y A y$ can be related to the resultant vector $A A$ and the angle $θ θ$ with trigonometric identities. Here we see that $A x =Acosθ A x =Acosθ$ and $A y =Asinθ. A y =Asinθ.$

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[BL][OL][AL] Derive the formula for getting the magnitude and direction of a vector.

Students might be confused between the relationship , which shows the addition of vectors and $A= A x 2 + A y 2 A= A x 2 + A y 2$ which shows the addition of magnitudes of vectors.

Suppose, for example, that $A A$ is the vector representing the total displacement of the person walking in a city, as illustrated in Figure 5.20.

Figure 5.20 We can use the relationships $A x =Acosθ A x =Acosθ$ and $A y =Asinθ A y =Asinθ$ to determine the magnitude of the horizontal and vertical component vectors in this example.

Then A = 10.3 blocks and $θ= 29.1 ∘ θ= 29.1 ∘$, so that

5.6

This magnitude indicates that the walker has traveled 9 blocks to the east—in other words, a 9-block eastward displacement. Similarly,

5.7

indicating that the walker has traveled 5 blocks to the north—a 5-block northward displacement.

Analytical Method of Vector Addition and Subtraction

Calculating a resultant vector (or vector addition) is the reverse of breaking the resultant down into its components. If the perpendicular components $A x A x$ and $A y A y$ of a vector $A A$ are known, then we can find $A A$ analytically. How do we do this? Since, by definition,

we solve for $θ θ$ to find the direction of the resultant.

$θ= tan −1 ( A y / A x ) θ= tan −1 ( A y / A x )$

Since this is a right triangle, the Pythagorean theorem (x2 + y2 = h2) for finding the hypotenuse applies. In this case, it becomes

$A 2 = A x 2 + A y 2 . A 2 = A x 2 + A y 2 .$

Solving for A gives

$A= A x 2 + A y 2 . A= A x 2 + A y 2 .$

In summary, to find the magnitude $A A$ and direction $θ θ$ of a vector from its perpendicular components $A x A x$ and $A y A y$, as illustrated in Figure 5.21, we use the following relationships:

$A = A x 2 + A y 2 θ= tan −1 ( A y / A x ) A = A x 2 + A y 2 θ= tan −1 ( A y / A x )$
Figure 5.21 The magnitude and direction of the resultant vector $A A$ can be determined once the horizontal components $A x A x$ and $A y A y$ have been determined.

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[BL][OL][AL] Demonstrate a problem involving displacement by physically walking along the specified direction. Show how this can be represented on a graph. Explain that even when solving problems analytically; representing it on a graph would make it easier to visualize the problem.

Sometimes, the vectors added are not perfectly perpendicular to one another. An example of this is the case below, where the vectors $A A$ and $B B$ are added to produce the resultant $R , R ,$ as illustrated in Figure 5.22.

Figure 5.22 Vectors $A A$ and $B B$ are two legs of a walk, and $R R$ is the resultant or total displacement. You can use analytical methods to determine the magnitude and direction of $R R$ .

If $A A$ and $B B$ represent two legs of a walk (two displacements), then $R R$ is the total displacement. The person taking the walk ends up at the tip of $R R$ . There are many ways to arrive at the same point. The person could have walked straight ahead first in the x-direction and then in the y-direction. Those paths are the x- and y-components of the resultant, $R x R x$ and $R y . R y .$ If we know $R x R x$ and $R y R y$, we can find $R R$ and $θ θ$ using the equations $R= R x 2 + R y 2 R= R x 2 + R y 2$ and $θ=ta n –1 ( R y / R x ) θ=ta n –1 ( R y / R x )$ .

1. Draw in the x and y components of each vector (including the resultant) with a dashed line. Use the equations $A x =Acosθ A x =Acosθ$ and $A y =Asinθ A y =Asinθ$ to find the components. In Figure 5.23, these components are $A x A x$, $A y A y$, $B x B x$, and $B y . B y .$ Vector $A A$ makes an angle of $θ A θ A$ with the x-axis, and vector $B B$ makes and angle of $θ B θ B$ with its own x-axis (which is slightly above the x-axis used by vector A).
Figure 5.23 To add vectors $A A$ and $B, B,$ first determine the horizontal and vertical components of each vector. These are the dotted vectors $A x , A x ,$ $A y A y$ $B y B y$ shown in the image.
2. Find the x component of the resultant by adding the x component of the vectors
$R x = A x + B x R x = A x + B x$

and find the y component of the resultant (as illustrated in Figure 5.24) by adding the y component of the vectors.

$R y = A y + B y . R y = A y + B y .$
Figure 5.24 The vectors $A x A x$ and $B x B x$ add to give the magnitude of the resultant vector in the horizontal direction, $R x . R x .$ Similarly, the vectors $A y A y$ and $B y B y$ add to give the magnitude of the resultant vector in the vertical direction, $R y . R y .$

Now that we know the components of $R, R,$ we can find its magnitude and direction.

3. To get the magnitude of the resultant R, use the Pythagorean theorem.
$R= R x 2 + R y 2 R= R x 2 + R y 2$
4. To get the direction of the resultant
$θ= tan −1 ( R y / R x ). θ= tan −1 ( R y / R x ).$

Watch Physics

Classifying Vectors and Quantities Example

This video contrasts and compares three vectors in terms of their magnitudes, positions, and directions.

Grasp Check

Three vectors, $ModifyingAbove u With right-arrow$, $ModifyingAbove v With right-arrow$, and $ModifyingAbove w With right-arrow$, have the same magnitude of $5 units$. Vector $ModifyingAbove v With right-arrow$ points to the northeast. Vector $ModifyingAbove w With right-arrow$ points to the southwest exactly opposite to vector $ModifyingAbove u With right-arrow$. Vector $ModifyingAbove u With right-arrow$ points in the northwest. If the vectors $ModifyingAbove u With right-arrow$, $ModifyingAbove v With right-arrow$, and $ModifyingAbove w With right-arrow$ were added together, what would be the magnitude of the resultant vector? Why?
1. $0 units$. All of them will cancel each other out.
2. $5 units$. Two of them will cancel each other out.
3. $10 units$. Two of them will add together to give the resultant.
4. $15$ units. All of them will add together to give the resultant.

Tips For Success

In the video, the vectors were represented with an arrow above them rather than in bold. This is a common notation in math classes.

Using the Analytical Method of Vector Addition and Subtraction to Solve Problems

Figure 5.25 uses the analytical method to add vectors.

Worked Example

An Accelerating Subway Train

Add the vector $A A$ to the vector $B B$ shown in Figure 5.25, using the steps above. The x-axis is along the east–west direction, and the y-axis is along the north–south directions. A person first walks in a direction $20.0° 20.0°$ north of east, represented by vector $A. A.$ The person then walks in a direction $63.0° 63.0°$ north of east, represented by vector $B. B.$

Figure 5.25 You can use analytical models to add vectors.

Strategy

The components of $A A$ and $B B$ along the x- and y-axes represent walking due east and due north to get to the same ending point. We will solve for these components and then add them in the x-direction and y-direction to find the resultant.

Discussion

This example shows vector addition using the analytical method. Vector subtraction using the analytical method is very similar. It is just the addition of a negative vector. That is, $A−B≡A+(−B) A−B≡A+(−B)$ . The components of – $B B$ are the negatives of the components of $B B$ . Therefore, the x- and y-components of the resultant $A−B=R A−B=R$ are

$R x = A x +- B x R x = A x +- B x$

and

$R y = A y +- B y R y = A y +- B y$

and the rest of the method outlined above is identical to that for addition.

Practice Problems

5.

What is the magnitude of a vector whose x-component is 4 cm and whose y-component is 3 cm?

1. 1 cm
2. 5 cm
3. 7 cm
4. 25 cm
6.

What is the magnitude of a vector that makes an angle of 30° to the horizontal and whose x-component is 3 units?

1. 2.61 units
2. 3.00 units
3. 3.46 units
4. 6.00 units

7.

Between the analytical and graphical methods of vector additions, which is more accurate? Why?

1. The analytical method is less accurate than the graphical method, because the former involves geometry and trigonometry.
2. The analytical method is more accurate than the graphical method, because the latter involves some extensive calculations.
3. The analytical method is less accurate than the graphical method, because the former includes drawing all figures to the right scale.
4. The analytical method is more accurate than the graphical method, because the latter is limited by the precision of the drawing.
8.

What is a component of a two dimensional vector?

1. A component is a piece of a vector that points in either the x or y direction.
2. A component is a piece of a vector that has half of the magnitude of the original vector.
3. A component is a piece of a vector that points in the direction opposite to the original vector.
4. A component is a piece of a vector that points in the same direction as original vector but with double of its magnitude.
9.
How can we determine the global angle $theta$ (measured counter-clockwise from positive $x$) if we know $upper A Subscript x$ and $upper A Subscript y$?
1. $theta equals cosine Superscript negative 1 Baseline StartFraction upper A Subscript y Baseline Over upper A Subscript x Baseline EndFraction$
2. $theta equals cotangent Superscript negative 1 Baseline StartFraction upper A Subscript y Baseline Over upper A Subscript x Baseline EndFraction$
3. $theta equals sine Superscript negative 1 Baseline StartFraction upper A Subscript y Baseline Over upper A Subscript x Baseline EndFraction$
4. $theta equals tangent Superscript negative 1 Baseline StartFraction upper A Subscript y Baseline Over upper A Subscript x Baseline EndFraction$
10.

How can we determine the magnitude of a vector if we know the magnitudes of its components?

1. $|A→|=Ax+Ay|A→|=Ax+Ay$
2. $|A→|=Ax2+Ay2|A→|=Ax2+Ay2$
3. $|A→|=Ax2+Ay2|A→|=Ax2+Ay2$
4. $|A→|=(Ax2+Ay2)2|A→|=(Ax2+Ay2)2$

Teacher Support

Teacher Support

Use the Check Your Understanding questions to assess whether students achieve the learning objectives for this section. If students are struggling with a specific objective, the Check Your Understanding will help identify which objective is causing the problem and direct students to the relevant content.

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