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College Physics 2e

3.3 Vector Addition and Subtraction: Analytical Methods

College Physics 2e3.3 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.

In the given figure a dotted vector A sub x is drawn from the origin along the x axis. From the head of the vector A sub x another vector A sub y is drawn in the upward direction. Their resultant vector A is drawn from the tail of the vector A sub x to the head of the vector A sub y at an angle theta from the x axis. On the graph a vector A, inclined at an angle theta with x axis is shown. Therefore vector A is the sum of the vectors A sub x and A sub y.
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 m 3 m+4 m   5 m
3.4

Thus,

A x + A y A A x + A y A
3.5

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

A x = A cos θ A x = A cos θ
3.6

and

Ay=Asinθ.Ay=Asinθ.
3.7
]A dotted vector A sub x whose magnitude is equal to A cosine theta is drawn from the origin along the x axis. From the head of the vector A sub x another vector A sub y whose magnitude is equal to A sine theta is drawn in the upward direction. Their resultant vector A is drawn from the tail of the vector A sub x to the head of the vector A-y at an angle theta from the x axis. Therefore vector A is the sum of the vectors A sub x and A sub y.
Figure 3.25 The magnitudes of the vector components AxAx and AyAy can be related to the resultant vector AA and the angle θ θ 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.

In the given figure a vector A of magnitude ten point three blocks is inclined at an angle twenty nine point one degrees to the positive x axis. The horizontal component A sub x of vector A is equal to A cosine theta which is equal to ten point three blocks multiplied to cosine twenty nine point one degrees which is equal to nine blocks east. Also the vertical component A sub y of vector A is equal to A sin theta is equal to ten point three blocks multiplied to sine twenty nine point one degrees,  which is equal to five point zero blocks north.
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 = 29.1º θ = 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 θ θ of a vector from its perpendicular components AxAx and AyAy, relative to the x-axis, we use the following relationships:

A= A x 2 + A y 2 A= A x 2 + A y 2
3.10
θ = tan 1 ( A y / A x ) . θ = tan 1 ( A y / A x ) .
3.11
Vector A is shown with its horizontal and vertical components A sub x and A sub y respectively. The magnitude of vector A is equal to the square root of A sub x squared plus A sub y squared. The angle theta of the vector A with the x axis is equal to inverse tangent of A sub y over A sub x
Figure 3.27 The magnitude and direction of the resultant vector can be determined once the horizontal and vertical components A x A x and A y A y have been determined.

Note that the equation A = A x 2 + A y 2 A = A x 2 + A y 2 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 ) =29.1º θ = 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 θ θ to AxAx and AyAy. Equations A=Ax2+Ay2A=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 θ θ. 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.

Two vectors A and B are shown. The tail of vector B is at the head of vector A and the tail of the vector A is at origin. Both the vectors are in the first quadrant. The resultant R of these two vectors extending from the tail of vector A to the head of vector B is also shown.
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 θ θ using the equations A = A x 2 + Ay 2 A = A x 2 + Ay 2 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.

Two vectors A and B are shown. The tail of the vector B is at the head of vector A and the tail of the vector A is at origin. Both the vectors are in the first quadrant. The resultant R of these two vectors extending from the tail of vector A to the head of vector B is also shown. The horizontal and vertical components of the vectors A and B are shown with the help of dotted lines. The vectors labeled as A sub x and A sub y are the components of vector A, and B sub x and B sub y as the components of vector B..
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,

R x = A x + B x R x = A x + B x
3.12

and

R y = A y + B y . R y = A y + B y .
3.13
Two vectors A and B are shown. The tail of vector B is at the head of vector A and the tail of the vector A is at origin. Both the vectors are in the first quadrant. The resultant R of these two vectors extending from the tail of vector A to the head of vector B is also shown. The vectors A and B are resolved into the horizontal and vertical components shown as dotted lines parallel to x axis and y axis respectively. The horizontal components of vector A and vector B are labeled as A sub x and B sub x and the horizontal component of the resultant R is labeled at R sub x and is equal to A sub x plus B sub x. The vertical components of vector A and vector B are labeled as A sub y and B sub y and the vertical components of the resultant R is labeled as R sub y is equal to A sub y plus B sub y.
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=Rx2+Ry2.R=Rx2+Ry2.
3.14

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

θ=tan1(Ry/Rx).θ=tan1(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.

Two vectors A and B are shown. The tail of the vector A is at origin. Both the vectors are in the first quadrant. Vector A is of magnitude fifty three units and is inclined at an angle of twenty degrees to the horizontal. From the head of the vector A another vector B of magnitude 34 units is drawn and is inclined at angle sixty three degrees with the horizontal. The resultant of two vectors is drawn from the tail of the vector A to the head of the vector B.
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

A x = A cos θ A = ( 53. 0 m ) ( cos 20.0º ) = ( 53. 0 m ) ( 0 .940 ) = 49. 8 m A x = A cos θ A = ( 53. 0 m ) ( cos 20.0º ) = ( 53. 0 m ) ( 0 .940 ) = 49. 8 m
3.16

and

B x = B cos θ B = ( 34 . 0 m ) ( cos 63.0º ) = ( 34 . 0 m ) ( 0 . 454 ) = 15 . 4 m . B x = B cos θ B = ( 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:

A y = A sin θ A = ( 53 . 0 m ) ( sin 20.0º ) = ( 53 . 0 m ) ( 0 . 342 ) = 18 . 1 m A y = A sin θ A = ( 53 . 0 m ) ( sin 20.0º ) = ( 53 . 0 m ) ( 0 . 342 ) = 18 . 1 m
3.18

and

B y = B sin θ B = ( 34 . 0 m ) ( sin 63 . 0 º ) = ( 34 . 0 m ) ( 0 . 891 ) = 30 . 3 m . B y = B sin θ B = ( 34 . 0 m ) ( sin 63 . 0 º ) = ( 34 . 0 m ) ( 0 . 891 ) = 30 . 3 m .
3.19

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

R x = A x + B x = 49 . 8 m + 15 . 4 m = 65 . 2 m R x = A x + B x = 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 = R x 2 + R y 2 = ( 65 . 2 ) 2 + ( 48 . 4 ) 2 m R = R x 2 + R y 2 = ( 65 . 2 ) 2 + ( 48 . 4 ) 2 m
3.22

so that

R = 81.2 m. R = 81.2 m.
3.23

Finally, we find the direction of the resultant:

θ=tan1(Ry/Rx)=+tan1(48.4/65.2).θ=tan1(Ry/Rx)=+tan1(48.4/65.2).
3.24

Thus,

θ=tan1(0.742)=36.6º.θ=tan1(0.742)=36.6º.
3.25
The addition of two vectors A and B is shown. Vector A is of magnitude fifty three units and is inclined at an angle of twenty degrees to the horizontal. Vector B is of magnitude thirty four units and is inclined at angle sixty three degrees to the horizontal. The components of vector A are shown as dotted vectors A X is equal to forty nine point eight meter along x axis and A Y is equal to eighteen point one meter along Y axis. The components of vector B are also shown as dotted vectors B X is equal to fifteen point four meter and B Y is equal to thirty point three meter. The horizontal component of the resultant R X is equal to A X plus B X is equal to sixty five point two meter. The vertical component of the resultant R Y is equal to A Y plus B Y is equal to forty eight point four meter. The magnitude of the resultant of two vectors is eighty one point two meters. The direction of the resultant R is in thirty six point six degree from the vector A in anticlockwise direction.
Figure 3.32 Using analytical methods, we see that the magnitude of RR is 81.2 m81.2 m and its direction is 36.36. 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, ABA+(–B)ABA+(–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 AB = RAB = R are thus

R x = A x + ( B x ) R x = A x + ( B x )
3.26

and

R y = A y + ( B y ) R y = A y + ( B y )
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.

In this figure, the subtraction of two vectors A and B is shown. A red colored vector A is inclined at an angle theta A to the positive of x axis. From the head of vector A a blue vector negative B is drawn. Vector B is in west of south direction. The resultant of the vector A and vector negative B is shown as a black vector R from the tail of vector A to the head of vector negative B. The resultant R is inclined to x axis at an angle theta below the x axis. The components of the vectors are also shown along the coordinate axes as dotted lines of their respective colors.
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.

PhET Explorations

Vector Addition

Learn how to add vectors. Drag vectors onto a graph, change their length and angle, and sum them together. The magnitude, angle, and components of each vector can be displayed in several formats.

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