Vector Addition and Subtraction: Analytical Methods

LEARNING OBJECTIVES

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

• Understand the rules of vector addition and subtraction using analytical methods.
• Apply analytical methods to determine vertical and horizontal component vectors.
• Apply analytical methods to determine the magnitude and direction of a resultant vector.

Analytical methods of vector addition and subtraction employ geometry and simple trigonometry rather than the ruler and protractor of graphical methods. Part of the graphical technique is retained, because vectors are still represented by arrows for easy visualization. However, analytical methods are more concise, accurate, and precise than graphical methods, which are limited by the accuracy with which a drawing can be made. Analytical methods are limited only by the accuracy and precision with which physical quantities are known.

Resolving a Vector into Perpendicular Components

Analytical techniques and right triangles go hand-in-hand in physics because (among other things) motions along perpendicular directions are independent. We very often need to separate a vector into perpendicular components. For example, given a vector like A𝐴 in Figure 3.24, we may wish to find which two perpendicular vectors, Ax𝐴𝑥 and Ay𝐴𝑦, add to produce it.

Figure 3.24 The vector A𝐴, with its tail at the origin of an x, y-coordinate system, is shown together with its x– and y-components, Ax𝐴𝑥 and Ay𝐴𝑦. These vectors form a right triangle. The analytical relationships among these vectors are summarized below.

Ax𝐴𝑥 and Ay𝐴𝑦 are defined to be the components of A𝐴 along the x– and y-axes. The three vectors A𝐴, Ax𝐴𝑥, and Ay𝐴𝑦 form a right triangle:

Ax + Ay = A.𝐴𝑥 + A𝑦 = A.

3.3

Note that this relationship between vector components and the resultant vector holds only for vector quantities (which include both magnitude and direction). The relationship does not apply for the magnitudes alone. For example, if Ax=3 mA𝑥=3 m east, Ay=4 mA𝑦=4 m north, and A=5 mA=5 m north-east, then it is true that the vectors Ax + Ay = A𝐴𝑥 + A𝑦 = A. However, it is not true that the sum of the magnitudes of the vectors is also equal. That is,

3 m+4 m ≠ 5 m3 m+4 m ≠ 5 m

3.4

Thus,

Ax+Ay≠A𝐴𝑥+𝐴𝑦≠𝐴

3.5

If the vector A𝐴 is known, then its magnitude A𝐴 (its length) and its angle θ𝜃 (its direction) are known. To find Ax𝐴𝑥 and Ay𝐴𝑦, its x– and y-components, we use the following relationships for a right triangle.

Ax=Acosθ𝐴𝑥=𝐴cos𝜃

3.6

and

Ay=Asinθ.𝐴𝑦=𝐴sin𝜃.

3.7

Figure 3.25 The magnitudes of the vector components Ax𝐴𝑥 and Ay𝐴𝑦 can be related to the resultant vector A𝐴 and the angle θ𝜃 with trigonometric identities. Here we see that Ax=Acosθ𝐴𝑥=𝐴cos𝜃 and Ay=Asinθ𝐴𝑦=𝐴sin𝜃.

Suppose, for example, that A𝐴 is the vector representing the total displacement of the person walking in a city considered in Kinematics in Two Dimensions: An Introduction and Vector Addition and Subtraction: Graphical Methods.

Figure 3.26 We can use the relationships Ax=Acosθ𝐴𝑥=𝐴cos𝜃 and Ay=Asinθ𝐴𝑦=𝐴sin𝜃 to determine the magnitude of the horizontal and vertical component vectors in this example.

Then A=10.3𝐴=10.3 blocks and θ=29.1º𝜃=29.1º , so that

Ax=Acosθ=(10.3 blocks)(cos29.1º)=9.0 blocks𝐴𝑥=𝐴cos𝜃=(10.3 blocks)(cos29.1º)=9.0 blocks

3.8

Ay=Asinθ=(10.3 blocks)(sin29.1º)=5.0 blocks.𝐴𝑦=𝐴sin𝜃=(10.3 blocks)(sin29.1º)=5.0 blocks.

3.9

Calculating a Resultant Vector

If the perpendicular components Ax𝐴𝑥 and Ay𝐴𝑦 of a vector A𝐴 are known, then A𝐴 can also be found analytically. To find the magnitude A𝐴 and direction θ𝜃 of a vector from its perpendicular components Ax𝐴𝑥 and Ay𝐴𝑦, relative to the x-axis, we use the following relationships:

A=Ax2+Ay2−−−−−−−−√𝐴=𝐴𝑥2+𝐴𝑦2

3.10

θ=tan−1(Ay/Ax).𝜃=tan−1(𝐴𝑦/𝐴𝑥).

3.11

Figure 3.27 The magnitude and direction of the resultant vector can be determined once the horizontal and vertical components Ax𝐴𝑥 and Ay𝐴𝑦 have been determined.

Note that the equation A=A2x+A2y−−−−−−−√𝐴=𝐴𝑥2+𝐴𝑦2 is just the Pythagorean theorem relating the legs of a right triangle to the length of the hypotenuse. For example, if Ax𝐴𝑥 and Ay𝐴𝑦 are 9 and 5 blocks, respectively, then A=92+52−−−−−√=10.3𝐴=92+52=10.3 blocks, again consistent with the example of the person walking in a city. Finally, the direction is θ=tan–1(5/9)=29.1º𝜃=tan–1(5/9)=29.1º , as before.

DETERMINING VECTORS AND VECTOR COMPONENTS WITH ANALYTICAL METHODS

Equations Ax=Acosθ𝐴𝑥=𝐴cos𝜃 and Ay=Asinθ𝐴𝑦=𝐴sin𝜃 are used to find the perpendicular components of a vector—that is, to go from A𝐴 and θ𝜃 to Ax𝐴𝑥 and Ay𝐴𝑦. Equations A=A2x+A2y−−−−−−−√𝐴=𝐴𝑥2+𝐴𝑦2 and θ=tan–1(Ay/Ax)𝜃=tan–1(𝐴𝑦/𝐴𝑥) are used to find a vector from its perpendicular components—that is, to go from Ax𝐴𝑥 and Ay𝐴𝑦 to A𝐴 and θ𝜃. Both processes are crucial to analytical methods of vector addition and subtraction.

Adding Vectors Using Analytical Methods

To see how to add vectors using perpendicular components, consider Figure 3.28, in which the vectors A𝐴 and B𝐵 are added to produce the resultant R𝑅.

Figure 3.28 Vectors A𝐴 and B𝐵 are two legs of a walk, and R𝑅 is the resultant or total displacement. You can use analytical methods to determine the magnitude and direction of R𝑅.

If A𝐴 and B𝐵 represent two legs of a walk (two displacements), then R𝑅 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. In particular, the person could have walked first in the x-direction and then in the y-direction. Those paths are the x– and y-components of the resultant, Rx𝑅𝑥 and Ry𝑅𝑦. If we know RxR𝑥 and Ry𝑅𝑦, we can find R𝑅 and θ𝜃 using the equations A=Ax2+Ay2−−−−−−−−√𝐴=𝐴𝑥2+𝐴𝑦2 and θ=tan–1(Ay/Ax)𝜃=tan–1(𝐴𝑦/𝐴𝑥). When you use the analytical method of vector addition, you can determine the components or the magnitude and direction of a vector.

Step 1. Identify the x- and y-axes that will be used in the problem. Then, find the components of each vector to be added along the chosen perpendicular axes. Use the equations Ax=Acosθ𝐴𝑥=𝐴cos𝜃 and Ay=Asinθ𝐴𝑦=𝐴sin𝜃 to find the components. In Figure 3.29, these components are Ax𝐴𝑥, Ay𝐴𝑦, Bx𝐵𝑥, and By𝐵𝑦. The angles that vectors A𝐴 and B𝐵 make with the x-axis are θA𝜃A and θB𝜃B, respectively.

Figure 3.29 To add vectors A𝐴 and B𝐵, first determine the horizontal and vertical components of each vector. These are the dotted vectors Ax𝐴𝑥, Ay𝐴𝑦, Bx𝐵𝑥 and ByB𝑦 shown in the image.

Step 2. Find the components of the resultant along each axis by adding the components of the individual vectors along that axis. That is, as shown in Figure 3.30,

Rx=Ax+Bx𝑅𝑥=𝐴𝑥+𝐵𝑥

3.12

and

Ry=Ay+By.𝑅𝑦=𝐴𝑦+𝐵𝑦.

3.13

Figure 3.30 The magnitude of the vectors Ax𝐴𝑥 and Bx𝐵𝑥 add to give the magnitude Rx𝑅𝑥 of the resultant vector in the horizontal direction. Similarly, the magnitudes of the vectors Ay𝐴𝑦 and By𝐵𝑦 add to give the magnitude Ry𝑅𝑦 of the resultant vector in the vertical direction.

Components along the same axis, say the x-axis, are vectors along the same line and, thus, can be added to one another like ordinary numbers. The same is true for components along the y-axis. (For example, a 9-block eastward walk could be taken in two legs, the first 3 blocks east and the second 6 blocks east, for a total of 9, because they are along the same direction.) So resolving vectors into components along common axes makes it easier to add them. Now that the components of R𝑅 are known, its magnitude and direction can be found.

Step 3. To get the magnitude R𝑅 of the resultant, use the Pythagorean theorem:

R=R2x+R2y−−−−−−−√.𝑅=𝑅𝑥2+𝑅𝑦2.

3.14

Step 4. To get the direction of the resultant relative to the x-axis:

θ=tan−1(Ry/Rx).𝜃=tan−1(𝑅𝑦/𝑅𝑥).

3.15

The following example illustrates this technique for adding vectors using perpendicular components.

EXAMPLE 3.3

Adding Vectors Using Analytical Methods

Add the vector A𝐴 to the vector B𝐵 shown in Figure 3.31, using perpendicular components along the x– and y-axes. The x– and y-axes are along the east–west and north–south directions, respectively. Vector A𝐴 represents the first leg of a walk in which a person walks 53.0 m53.0 m in a direction 20.0º20.0º north of east. Vector B𝐵 represents the second leg, a displacement of 34.0 m34.0 m in a direction 63.0º63.0º north of east.

Figure 3.31 Vector A𝐴 has magnitude 53.0 m53.0 m and direction 20.0º20.0º north of the x-axis. Vector B𝐵 has magnitude 34.0 m34.0 m and direction 63.0º63.0º north of the x-axis. You can use analytical methods to determine the magnitude and direction of R𝑅.

Strategy

The components of A𝐴 and B𝐵 along the x– and y-axes represent walking due east and due north to get to the same ending point. Once found, they are combined to produce the resultant.

Solution

Following the method outlined above, we first find the components of A𝐴 and B𝐵 along the x– and y-axes. Note that A=53.0 m𝐴=53.0 m, θA=20.0º𝜃A=20.0º, B=34.0 m𝐵=34.0 m, and θB=63.0º𝜃B=63.0º. We find the x-components by using Ax=Acosθ𝐴𝑥=𝐴cos𝜃, which gives

Ax==AcosθA=(53.0 m)(cos 20.0º)(53.0 m)(0.940)=49.8 m𝐴𝑥=𝐴cos𝜃A=(53.0 m)(cos 20.0º)=(53.0 m)(0.940)=49.8 m

3.16

and

Bx==BcosθB=(34.0 m)(cos 63.0º)(34.0 m)(0.454)=15.4 m.𝐵𝑥=𝐵cos𝜃B=(34.0 m)(cos 63.0º)=(34.0 m)(0.454)=15.4 m.

3.17

Similarly, the y-components are found using Ay=AsinθA𝐴𝑦=𝐴sin𝜃A:

Ay==AsinθA=(53.0 m)(sin 20.0º)(53.0 m)(0.342)=18.1 m𝐴𝑦=𝐴sin𝜃A=(53.0 m)(sin 20.0º)=(53.0 m)(0.342)=18.1 m

3.18

and

By==BsinθB=(34.0 m)(sin 63.0º)(34.0 m)(0.891)=30.3 m.𝐵𝑦=𝐵sin𝜃B=(34.0 m)(sin 63.0º)=(34.0 m)(0.891)=30.3 m.

3.19

The x– and y-components of the resultant are thus

Rx=Ax+Bx=49.8 m+15.4 m=65.2 m𝑅𝑥=𝐴𝑥+𝐵𝑥=49.8 m+15.4 m=65.2 m

3.20

and

Ry=Ay+By=18.1 m+30.3 m=48.4 m.𝑅𝑦=𝐴𝑦+𝐵𝑦=18.1 m+30.3 m=48.4 m.

3.21

Now we can find the magnitude of the resultant by using the Pythagorean theorem:

R=R2x+R2y−−−−−−−√=(65.2)2+(48.4)2m−−−−−−−−−−−−−−√𝑅=𝑅𝑥2+𝑅𝑦2=(65.2)2+(48.4)2m

3.22

so that

R=81.2 m.𝑅=81.2 m.

3.23

Finally, we find the direction of the resultant:

θ=tan−1(Ry/Rx)=+tan−1(48.4/65.2).𝜃=tan−1(𝑅𝑦/𝑅𝑥)=+tan−1(48.4/65.2).

3.24

Thus,

θ=tan−1(0.742)=36.6º.𝜃=tan−1(0.742)=36.6º.

3.25

Figure 3.32 Using analytical methods, we see that the magnitude of R𝑅 is 81.2 m81.2 m and its direction is 36.6º36.6º north of east.

Discussion

This example illustrates the addition of vectors using perpendicular components. Vector subtraction using perpendicular components is very similar—it is just the addition of a negative vector.

Subtraction of vectors is accomplished by the addition of a negative vector. That is, A−B≡A+(–B)𝐴−𝐵≡𝐴+(–B). Thus, the method for the subtraction of vectors using perpendicular components is identical to that for addition. The components of –B–B are the negatives of the components of B𝐵. The x– and y-components of the resultant A−B = R𝐴−B = R are thus

Rx=Ax+(–Bx)𝑅𝑥=𝐴𝑥+(–𝐵𝑥)

3.26

and

Ry=Ay+(–By)𝑅𝑦=𝐴𝑦+(–𝐵𝑦)

3.27

and the rest of the method outlined above is identical to that for addition. (See Figure 3.33.)

Analyzing vectors using perpendicular components is very useful in many areas of physics, because perpendicular quantities are often independent of one another. The next module, Projectile Motion, is one of many in which using perpendicular components helps make the picture clear and simplifies the physics.

Figure 3.33 The subtraction of the two vectors shown in Figure 3.28. The components of –B–B are the negatives of the components of B𝐵. The method of subtraction is the same as that for addition.