Contents

Vector Addition and Subtraction: Graphical Methods

Some Hawaiian Islands like Kauai Oahu, Molokai, Lanai, Maui, Kahoolawe, and Hawaii are shown. On the scale map of Hawaiian Islands the path of a journey is shown moving from Hawaii to Molokai. The path of the journey is turning at different angles and finally reaching its destination. The displacement of the journey is shown with the help of a straight line connecting its starting point and the destination.

Figure 3.8 Displacement can be determined graphically using a scale map, such as this one of the Hawaiian Islands. A journey from Hawai’i to Moloka’i has a number of legs, or journey segments. These segments can be added graphically with a ruler to determine the total two-dimensional displacement of the journey. (credit: US Geological Survey)

Vectors in Two Dimensions

A vector is a quantity that has magnitude and direction. Displacement, velocity, acceleration, and force, for example, are all vectors. In one-dimensional, or straight-line, motion, the direction of a vector can be given simply by a plus or minus sign. In two dimensions (2-d), however, we specify the direction of a vector relative to some reference frame (i.e., coordinate system), using an arrow having length proportional to the vector’s magnitude and pointing in the direction of the vector.

Figure 3.9 shows such a graphical representation of a vector, using as an example the total displacement for the person walking in a city considered in Kinematics in Two Dimensions: An Introduction. We shall use the notation that a boldface symbol, such as $D D size 12{D} {}$ , stands for a vector. Its magnitude is represented by the symbol in italics, $D D size 12{D} {}$ , and its direction by $θ θ size 12{θ} {}$ .

Vectors in this Text

In this text, we will represent a vector with a boldface variable. For example, we will represent the quantity force with the vector $F F size 12{F} {}$ , which has both magnitude and direction. The magnitude of the vector will be represented by a variable in italics, such as $F F size 12{F} {}$ , and the direction of the variable will be given by an angle $θ θ size 12{θ} {}$ .

A graph is shown. On the axes the scale is set to one block is equal to one unit. A helicopter starts moving from the origin at an angle of twenty nine point one degrees above the x axis. The current position of the helicopter is ten point three blocks along its line of motion. The destination of the helicopter is the point which is nine blocks in the positive x direction and five blocks in the positive y direction. The positive direction of the x axis is east and the positive direction of the y axis is north.

Figure 3.9 A person walks 9 blocks east and 5 blocks north. The displacement is 10.3 blocks at an angle

29 .1º 29 .1º size 12{"29" "." "1º"} {}

north of east.

On a graph a vector is shown. It is inclined at an angle theta equal to twenty nine point one degrees above the positive x axis. A protractor is shown to the right of the x axis to measure the angle. A ruler is also shown parallel to the vector to measure its length. The ruler shows that the length of the vector is ten point three units.

Figure 3.10 To describe the resultant vector for the person walking in a city considered in Figure 3.9 graphically, draw an arrow to represent the total displacement vector

D D size 12{D} {}

. Using a protractor, draw a line at an angle

θ θ size 12{θ} {}

relative to the east-west axis. The length

D D size 12{D} {}

of the arrow is proportional to the vector’s magnitude and is measured along the line with a ruler. In this example, the magnitude

D D size 12{D} {}

of the vector is 10.3 units, and the direction

θ θ size 12{θ} {}

29.1º 29.1º size 12{"29" "." 1 rSup { size 12{º} } } {}

north of east.

Vector Addition: Head-to-Tail Method

The head-to-tail method is a graphical way to add vectors, described in Figure 3.11 below and in the steps following. The tail of the vector is the starting point of the vector, and the head (or tip) of a vector is the final, pointed end of the arrow.

In part a, a vector of magnitude of nine units and making an angle of theta is equal to zero degrees is drawn from the origin and along the positive direction of x axis. In part b a vector of magnitude of nine units and making an angle of theta is equal to zero degree is drawn from the origin and along the positive direction of x axis. Then a vertical arrow from the head of the horizontal arrow is drawn. In part c a vector D of magnitude ten point three is drawn from the tail of the horizontal vector at an angle theta is equal to twenty nine point one degrees from the positive direction of x axis. The head of the vector D meets the head of the vertical vector. A scale is shown parallel to the vector D to measure its length. Also a protractor is shown to measure the inclination of the vectorD.

Figure 3.11 Head-to-Tail Method: The head-to-tail method of graphically adding vectors is illustrated for the two displacements of the person walking in a city considered in Figure 3.9. (a) Draw a vector representing the displacement to the east. (b) Draw a vector representing the displacement to the north. The tail of this vector should originate from the head of the first, east-pointing vector. (c) Draw a line from the tail of the east-pointing vector to the head of the north-pointing vector to form the sum or resultant vector

D D size 12{A} {}

. The length of the arrow

D D size 12{A} {}

is proportional to the vector’s magnitude and is measured to be 10.3 units . Its direction, described as the angle with respect to the east (or horizontal axis)

θ θ size 12{θ} {}

is measured with a protractor to be

29 . 1º 29 . 1º size 12{"29" "." 1°} {}

Step 1. Draw an arrow to represent the first vector (9 blocks to the east) using a ruler and protractor.

In part a, a vector of magnitude of nine units and making an angle theta is equal to zero degree is drawn from the origin and along the positive direction of x axis.

Figure 3.12

Step 2. Now draw an arrow to represent the second vector (5 blocks to the north). Place the tail of the second vector at the head of the first vector.

In part b, a vector of magnitude of nine units and making an angle theta is equal to zero degree is drawn from the origin and along the positive direction of x axis. Then a vertical vector from the head of the horizontal vector is drawn.

Figure 3.13

Step 3. If there are more than two vectors, continue this process for each vector to be added. Note that in our example, we have only two vectors, so we have finished placing arrows tip to tail.

Step 4. Draw an arrow from the tail of the first vector to the head of the last vector. This is the resultant, or the sum, of the other vectors.

In part c, a vector D of magnitude ten point three is drawn from the tail of the horizontal vector at an angle theta is equal to twenty nine point one degrees from the positive direction of the x axis. The head of the vector D meets the head of the vertical vector. A scale is shown parallel to the vector D to measure its length. Also a protractor is shown to measure the inclination of the vector D.

Figure 3.14

Step 5. To get the magnitude of the resultant, measure its length with a ruler. (Note that in most calculations, we will use the Pythagorean theorem to determine this length.)

Step 6. To get the direction of the resultant, measure the angle it makes with the reference frame using a protractor. (Note that in most calculations, we will use trigonometric relationships to determine this angle.)

The graphical addition of vectors is limited in accuracy only by the precision with which the drawings can be made and the precision of the measuring tools. It is valid for any number of vectors.

Example 3.1

Adding Vectors Graphically Using the Head-to-Tail Method: A Woman Takes a Walk

Use the graphical technique for adding vectors to find the total displacement of a person who walks the following three paths (displacements) on a flat field. First, she walks 25.0 m in a direction $49.0º 49.0º size 12{"49" "." "0º"} {}$ north of east. Then, she walks 23.0 m heading $15.0º 15.0º size 12{"15" "." "º°"} {}$ north of east. Finally, she turns and walks 32.0 m in a direction 68.0° south of east.

Strategy

Represent each displacement vector graphically with an arrow, labeling the first $A A size 12{A} {}$ , the second $B B size 12{B} {}$ , and the third $C C size 12{C} {}$ , making the lengths proportional to the distance and the directions as specified relative to an east-west line. The head-to-tail method outlined above will give a way to determine the magnitude and direction of the resultant displacement, denoted $R R size 12{R} {}$ .

Solution

(1) Draw the three displacement vectors.

On the graph a vector of magnitude twenty three meters and inclined above the x axis at an angle theta-b equal to fifteen degrees is shown. This vector is labeled as B.

Figure 3.15

(2) Place the vectors head to tail retaining both their initial magnitude and direction.

In this figure a vector A with a positive slope is drawn from the origin. Then from the head of the vector A another vector B with positive slope is drawn and then another vector C with negative slope from the head of the vector B is drawn which cuts the x axis.

Figure 3.16

(3) Draw the resultant vector, $R R size 12{R} {}$ .

Figure 3.17

(4) Use a ruler to measure the magnitude of $R R size 12{R} {}$ , and a protractor to measure the direction of $R R size 12{R} {}$ . While the direction of the vector can be specified in many ways, the easiest way is to measure the angle between the vector and the nearest horizontal or vertical axis. Since the resultant vector is south of the eastward pointing axis, we flip the protractor upside down and measure the angle between the eastward axis and the vector.

Figure 3.18

In this case, the total displacement $R R size 12{R} {}$ is seen to have a magnitude of 50.0 m and to lie in a direction $7.0º 7.0º size 12{7 "." 0°} {}$ south of east. By using its magnitude and direction, this vector can be expressed as $R = 50.0 m R = 50.0 m size 12{R" = 50" "." "0 m"} {}$ and $θ = 7 . 0º θ = 7 . 0º size 12{θ=7 "." "0°"} {}$ south of east.

Discussion

The head-to-tail graphical method of vector addition works for any number of vectors. It is also important to note that the resultant is independent of the order in which the vectors are added. Therefore, we could add the vectors in any order as illustrated in Figure 3.19 and we will still get the same solution.

In this figure a vector C with a negative slope is drawn from the origin. Then from the head of the vector C another vector A with positive slope is drawn and then another vector B with negative slope from the head of the vector A is drawn. From the tail of the vector C a vector R of magnitude of fifty point eight meters and with negative slope of five point four seven degrees is drawn. The head of this vector R meets the head of the vector B. The vector R is known as the resultant vector.

Figure 3.19

Here, we see that when the same vectors are added in a different order, the result is the same. This characteristic is true in every case and is an important characteristic of vectors. Vector addition is commutative. Vectors can be added in any order.

A + B = B + A . A + B = B + A . size 12{"A+B=B+A"} {}

3.1

(This is true for the addition of ordinary numbers as well—you get the same result whether you add $2 + 3 2 + 3 size 12{"2+3"} {}$ or $3 + 2 3 + 2 size 12{"3+2"} {}$ , for example).

Vector Subtraction

Vector subtraction is a straightforward extension of vector addition. To define subtraction (say we want to subtract $B B size 12{B} {}$ from $A A size 12{A} {}$ , written $A - B A - B size 12{ "A" "-B"} {}$ , we must first define what we mean by subtraction. The negative of a vector $B B$ is defined to be $–B –B$ ; that is, graphically the negative of any vector has the same magnitude but the opposite direction, as shown in Figure 3.20. In other words, $B B size 12{B} {}$ has the same length as $–B –B size 12{"-" "B"} {}$ , but points in the opposite direction. Essentially, we just flip the vector so it points in the opposite direction.

Two vectors are shown. One of the vectors is labeled as vector in north east direction. The other vector is of the same magnitude and is in the opposite direction to that of vector B. This vector is denoted as negative B.

Figure 3.20 The negative of a vector is just another vector of the same magnitude but pointing in the opposite direction. So

B B size 12{B} {}

is the negative of

–B –B size 12{ ital "-B"} {}

; it has the same length but opposite direction.

The subtraction of vector $B B$ from vector $A A$ is then simply defined to be the addition of $–B –B$ to $A A$ . Note that vector subtraction is the addition of a negative vector. The order of subtraction does not affect the results.

A – B = A + (–B) . A – B = A + (–B) . size 12{ bold "A – B = A + " \(  bold "–B" \) } {}

3.2

This is analogous to the subtraction of scalars (where, for example, $5 – 2 = 5 + (–2) 5 – 2 = 5 + (–2) size 12{"5 – 2 = 5 + " $ "–2" $ } {}$ ). Again, the result is independent of the order in which the subtraction is made. When vectors are subtracted graphically, the techniques outlined above are used, as the following example illustrates.

Example 3.2

Subtracting Vectors Graphically: A Woman Sailing a Boat

A woman sailing a boat at night is following directions to a dock. The instructions read to first sail 27.5 m in a direction $66.0º 66.0º size 12{"66" "." 0º} {}$ north of east from her current location, and then travel 30.0 m in a direction $112º 112º size 12{"112"º} {}$ north of east (or $22.0º 22.0º size 12{"22" "." 0º} {}$ west of north). If the woman makes a mistake and travels in the opposite direction for the second leg of the trip, where will she end up? Compare this location with the location of the dock.

A vector of magnitude twenty seven point five meters is shown. It is inclined to the horizontal at an angle of sixty six degrees. Another vector of magnitude thirty point zero meters is shown. It is inclined to the horizontal at an angle of one hundred and twelve degrees.

Figure 3.21

Strategy

We can represent the first leg of the trip with a vector $A A$ , and the second leg of the trip with a vector $B B size 12{B} {}$ . The dock is located at a location $A + B A + B$ . If the woman mistakenly travels in the opposite direction for the second leg of the journey, she will travel a distance $B B$ (30.0 m) in the direction $180º - 112º = 68º 180º - 112º = 68º$ south of east. We represent this as $–B –B$ , as shown below. The vector $–B –B$ has the same magnitude as $B B$ but is in the opposite direction. Thus, she will end up at a location $A + (–B) A + (–B)$ , or $A - B A - B$ .

A vector labeled negative B is inclined at an angle of sixty-eight degrees below a horizontal line. A dotted line in the reverse direction inclined at one hundred and twelve degrees above the horizontal line is also shown.

Figure 3.22

We will perform vector addition to compare the location of the dock, $A + B A + B size 12{ ital "A ""+ "B} {}$ , with the location at which the woman mistakenly arrives, $A + (–B) A + (–B) size 12{ bold "A + " $ bold "–B" $ } {}$ .

Solution

(1) To determine the location at which the woman arrives by accident, draw vectors $A A size 12{A} {}$ and $–B –B$ .

(2) Place the vectors head to tail.

(3) Draw the resultant vector $R R size 12{R} {}$ .

(4) Use a ruler and protractor to measure the magnitude and direction of $R R size 12{R} {}$ .

Vectors A and negative B are connected in head to tail method. Vector A is inclined with horizontal with positive slope and vector negative B with a negative slope. The resultant of these two vectors is shown as a vector R from tail of A to the head of negative B. The length of the resultant is twenty three point zero meters and has a negative slope of seven point five degrees.

Figure 3.23

In this case, $R = 23 . 0 m R = 23 . 0 m size 12{R"=23" "." "0 m"} {}$ and $θ = 7 . 5º θ = 7 . 5º size 12{θ=7 "." "5° south of east"} {}$ south of east.

(5) To determine the location of the dock, we repeat this method to add vectors $A A size 12{A} {}$ and $B B size 12{B} {}$ . We obtain the resultant vector $R' R' size 12{R'} {}$ :

A vector A inclined at sixty six degrees with horizontal is shown. From the head of this vector another vector B is started. Vector B is inclined at one hundred and twelve degrees with the horizontal. Another vector labeled as R prime from the tail of vector A to the head of vector B is drawn. The length of this vector is fifty two point nine meters and its inclination with the horizontal is shown as ninety point one degrees. Vector R prime is equal to the sum of vectors A and B.

Figure 3.24

In this case $R = 52.9 m R = 52.9 m size 12{R" = 52" "." "9 m"} {}$ and $θ = 90.1º θ = 90.1º size 12{θ="90" "." "1° north of east "} {}$ north of east.

We can see that the woman will end up a significant distance from the dock if she travels in the opposite direction for the second leg of the trip.

Discussion

Because subtraction of a vector is the same as addition of a vector with the opposite direction, the graphical method of subtracting vectors works the same as for addition.

Multiplication of Vectors and Scalars

If we decided to walk three times as far on the first leg of the trip considered in the preceding example, then we would walk $3 \times 27 . 5 m 3 \times 27 . 5 m size 12{"3 " times " 27" "." "5 m"} {}$ , or 82.5 m, in a direction $66.0 º 66.0 º size 12{"66" "." 0 { size 12{º} } } {}$ north of east. This is an example of multiplying a vector by a positive scalar. Notice that the magnitude changes, but the direction stays the same.

If the scalar is negative, then multiplying a vector by it changes the vector’s magnitude and gives the new vector the opposite direction. For example, if you multiply by –2, the magnitude doubles but the direction changes. We can summarize these rules in the following way: When vector $A A size 12{A} {}$ is multiplied by a scalar $c c size 12{c} {}$ ,

the magnitude of the vector becomes the absolute value of $c c size 12{c} {}$ $A A size 12{A} {}$ ,
if $c c size 12{A} {}$ is positive, the direction of the vector does not change,
if $c c size 12{A} {}$ is negative, the direction is reversed.

In our case, $c = 3 c = 3 size 12{c=3}$ and $A = 27.5 m A = 27.5 m size 12{"A= 27.5 m"}$ . Vectors are multiplied by scalars in many situations. Note that division is the inverse of multiplication. For example, dividing by 2 is the same as multiplying by the value (1/2). The rules for multiplication of vectors by scalars are the same for division; simply treat the divisor as a scalar between 0 and 1.

Resolving a Vector into Components

In the examples above, we have been adding vectors to determine the resultant vector. In many cases, however, we will need to do the opposite. We will need to take a single vector and find what other vectors added together produce it. In most cases, this involves determining the perpendicular components of a single vector, for example the x- and y-components, or the north-south and east-west components.

For example, we may know that the total displacement of a person walking in a city is 10.3 blocks in a direction $29 .0º 29 .0º size 12{"29" "." 0º} } {}$ north of east and want to find out how many blocks east and north had to be walked. This method is called finding the components (or parts) of the displacement in the east and north directions, and it is the inverse of the process followed to find the total displacement. It is one example of finding the components of a vector. There are many applications in physics where this is a useful thing to do. We will see this soon in Projectile Motion, and much more when we cover forces in Dynamics: Newton’s Laws of Motion. Most of these involve finding components along perpendicular axes (such as north and east), so that right triangles are involved. The analytical techniques presented in Vector Addition and Subtraction: Analytical Methods are ideal for finding vector components.

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Figure 3.25