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Chapter 3 motion in two d

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Chapter 3 Kinematics in Two Dimensions; Vectors
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Page 1: Chapter 3 motion in two d

Chapter 3

Kinematics in Two Dimensions; Vectors

Page 2: Chapter 3 motion in two d

Vectors

An Introduction

Page 3: Chapter 3 motion in two d

There are two kinds of quantities…

• Scalars are quantities that have magnitude only, such as– position

– speed

– time

– mass

• Vectors are quantities that have both magnitude and direction, such as– displacement

– velocity

– acceleration

Page 4: Chapter 3 motion in two d

Notating vectors

• This is how you notate a vector…

• This is how you draw a vector…

R R

R headtail

Page 5: Chapter 3 motion in two d

Direction of Vectors

• Vector direction is the direction of the arrow, given by an angle.

• This vector has an angle that is between 0o and 90o.

Ax

Page 6: Chapter 3 motion in two d

Vector angle ranges

x

y

Quadrant I0 < < 90o

Quadrant II90o < < 180o

Quadrant III180o < < 270o

Quadrant IV

270o < < 360o

Page 7: Chapter 3 motion in two d

Direction of Vectors

• What angle range would this vector have?• What would be the exact angle, and how

would you determine it?

Bx

Between 180o and 270o

or between- 90o and -180o

Page 8: Chapter 3 motion in two d

Magnitude of Vectors

• The best way to determine the magnitude (or size) of a vector is to measure its length.

• The length of the vector is proportional to the magnitude (or size) of the quantity it represents.

Page 9: Chapter 3 motion in two d

Sample Problem

• If vector A represents a displacement of three miles to the north, then what does vector B represent? Vector C?

A

B

C

Page 10: Chapter 3 motion in two d

Equal Vectors

• Equal vectors have the same length and direction, and represent the same quantity (such as force or velocity).

• Draw several equal vectors.

Page 11: Chapter 3 motion in two d

Inverse Vectors

• Inverse vectors have the same length, but opposite direction.

• Draw a set of inverse vectors.

A

-A

Page 12: Chapter 3 motion in two d

The Right Triangle

θ

op

po

site

adjacent

hypotenuse

Page 13: Chapter 3 motion in two d

Pythagorean Theorem

• hypotenuse2 = opposite2 + adjacent2

• c2 = a2 + b2

θ

op

po

site

adjacent

hypotenuse

Page 14: Chapter 3 motion in two d

Basic Trigonometry functions

• sin θ = opposite/hypotenuse• cos θ = adjacent/hypotenuse• tan θ = opposite/adjacent

θ

op

po

site

adjacent

hypotenuseSOHCAHTOA

Page 15: Chapter 3 motion in two d

Inverse functions

• θ = sin-1(opposite/hypotenuse)• θ = cos-1(adjacent/hypotenuse)• θ = tan-1(opposite/adjacent)

θ

op

po

site

adjacent

hypotenuseSOHCAHTOA

Page 16: Chapter 3 motion in two d

Sample problem

• A surveyor stands on a riverbank directly across the river from a tree on the opposite bank. She then walks 100 m downstream, and determines that the angle from her new position to the tree on the opposite bank is 50o. How wide is the river, and how far is she from the tree in her new location?

Page 17: Chapter 3 motion in two d

Sample problem

• You are standing at the very top of a tower and notice that in order to see a manhole cover on the ground 50 meters from the base of the tower, you must look down at an angle 75o below the horizontal. If you are 1.80 m tall, how high is the tower?

Page 18: Chapter 3 motion in two d

Vectors: x-component

• The x-component of a vector is the “shadow” it casts on the x-axis.

• cos θ = adjacent ∕ hypotenuse

• cos θ = Ax ∕ A

• Ax = A cos

A

x

Ax

Page 19: Chapter 3 motion in two d

Vectors: y-component

• The y-component of a vector is the “shadow” it casts on the y-axis.

• sin θ = opposite ∕ hypotenuse

• sin θ = Ay ∕ A

• Ay = A sin

A

x

y

Ay Ay

Page 20: Chapter 3 motion in two d

Vectors: angle

• The angle a vector makes with the x-axis can be determined by the components.

• It is calculated by the inverse tangent function

• = tan-1 (Ay/Ax)

x

y

Rx

Ry

Page 21: Chapter 3 motion in two d

Vectors: magnitude

• The magnitude of a vector can be determined by the components.

• It is calculated using the Pythagorean Theorem.

• R2 = Rx2 + Ry

2x

y

Rx

Ry

R

Page 22: Chapter 3 motion in two d

Practice Problem

• You are driving up a long inclined road. After 1.5 miles you notice that signs along the roadside indicate that your elevation has increased by 520 feet.

a) What is the angle of the road above the horizontal?

Page 23: Chapter 3 motion in two d

Practice Problem

• You are driving up a long inclined road. After 1.5 miles you notice that signs along the roadside indicate that your elevation has increased by 520 feet.

b) How far do you have to drive to gain an additional 150 feet of elevation?

Page 24: Chapter 3 motion in two d

Practice Problem

• Find the x- and y-components of the following vectors

a) R = 175 meters @ 95o

Page 25: Chapter 3 motion in two d

Practice Problem

• Find the x- and y-components of the following vectors

b) v = 25 m/s @ -78o

Page 26: Chapter 3 motion in two d

Practice Problem

• Find the x- and y-components of the following vectors

c) a = 2.23 m/s2 @ 150o

Page 27: Chapter 3 motion in two d

Graphical Addition of Vectors

Day 2

Page 28: Chapter 3 motion in two d

Graphical Addition of Vectors

1) Add vectors A and B graphically by drawing them together in a head to tail arrangement.

2) Draw vector A first, and then draw vector B such that its tail is on the head of vector A.

3) Then draw the sum, or resultant vector, by drawing a vector from the tail of A to the head of B.

4) Measure the magnitude and direction of the resultant vector.

Page 29: Chapter 3 motion in two d

A

B

RA + B = R

Practice Graphical Addition

R is called the resultant vector!

B

Page 30: Chapter 3 motion in two d

The Resultant and the Equilibrant

• The sum of two or more vectors is called the resultant vector.

• The resultant vector can replace the vectors from which it is derived.

• The resultant is completely canceled out by adding it to its inverse, which is called the equilibrant.

Page 31: Chapter 3 motion in two d

A

B

R A + B = R

The Equilibrant Vector

The vector -R is called the equilibrant.If you add R and -R you get a null (or zero) vector.

-R

Page 32: Chapter 3 motion in two d

Graphical Subtraction of Vectors

1) Subtract vectors A and B graphically by adding vector A with the inverse of vector B (-B).

2) First draw vector A, then draw -B such that its tail is on the head of vector A.

3) The difference is the vector drawn from the tail of vector A to the head of -B.

Page 33: Chapter 3 motion in two d

A

B

A - B = C

Practice Graphical Subtraction

-B

C

Page 34: Chapter 3 motion in two d

Practice Problem

• Vector A points in the +x direction and has a magnitude of 75 m. Vector B has a magnitude of 30 m and has a direction of 30o relative to the x axis. Vector C has a magnitude of 50 m and points in a direction of -60o relative to the x axis.

a) Find A + Bb) Find A + B + Cc) Find A – B.

Page 35: Chapter 3 motion in two d

a)

Page 36: Chapter 3 motion in two d

b)

Page 37: Chapter 3 motion in two d

c)

Page 38: Chapter 3 motion in two d

Vector Addition Laboratory

Page 39: Chapter 3 motion in two d

Vector Addition Lab

1. Attach spring scales to force board such that they all have different readings.

2. Slip graph paper between scales and board and carefully trace your set up.

3. Record readings of all three spring scales.4. Detach scales from board and remove graph paper.5. On top of your tracing, draw a force diagram by constructing vectors

proportional in length to the scale readings. Point the vectors in the direction of the forces they represent. Connect the tails of the vectors to each other in the center of the drawing.

6. On a separate sheet of graph paper, add the three vectors together graphically. Identify your resultant, if any.

7. Did you get a resultant? Did you expect one?8. You must have a separate set of drawings for each member of

your lab group, so work efficiently

In C

lass

Ho

mew

ork

Page 40: Chapter 3 motion in two d

Vector Addition by Component

Page 41: Chapter 3 motion in two d

Component Addition of Vectors

1) Resolve each vector into its x- and y-components.

Ax = Acos Ay = Asin

Bx = Bcos By = Bsin

Cx = Ccos Cy = Csin etc.

2) Add the x-components (Ax, Bx, etc.) together to get Rx and the y-components (Ay, By, etc.) to get Ry.

Page 42: Chapter 3 motion in two d

Component Addition of Vectors

3) Calculate the magnitude of the resultant with the Pythagorean Theorem (R = Rx

2 + Ry2).

4) Determine the angle with the equation = tan-1 Ry/Rx.

Page 43: Chapter 3 motion in two d

Practice Problem• In a daily prowl through the neighborhood, a cat makes a

displacement of 120 m due north, followed by a displacement of 72 m due west. Find the magnitude and displacement required if the cat is to return home.

Page 44: Chapter 3 motion in two d

Practice Problem• If the cat in the previous problem takes 45 minutes to complete

the first displacement and 17 minutes to complete the second displacement, what is the magnitude and direction of its average velocity during this 62-minute period of time?

Page 45: Chapter 3 motion in two d

Relative Motion

Day 3

Page 46: Chapter 3 motion in two d

Relative Motion

• Relative motion problems are difficult to do unless one applies vector addition concepts.

• Define a vector for a swimmer’s velocity relative to the water, and another vector for the velocity of the water relative to the ground. Adding those two vectors will give you the velocity of the swimmer relative to the ground.

Page 47: Chapter 3 motion in two d

Relative Motion

Vs

Vw

Vt = Vs + Vw

Vw

Page 48: Chapter 3 motion in two d

Relative Motion

Vs

Vw

Vt = Vs + Vw

Vw

Page 49: Chapter 3 motion in two d

Relative Motion

Vs

Vw

Vt = Vs + Vw Vw

Page 50: Chapter 3 motion in two d

Practice Problem

• You are paddling a canoe in a river that is flowing at 4.0 mph east. You are capable of paddling at 5.0 mph.

a) If you paddle east, what is your velocity relative to the shore?

b) If you paddle west, what is your velocity relative to the shore?

c) You want to paddle straight across the river, from the south to the north.At what angle to you aim your boat relative to the shore? Assume east is 0o.

Page 51: Chapter 3 motion in two d

Practice Problem

• You are flying a plane with an airspeed of 400 mph. If you are flying in a region with a 80 mph west wind, what must your heading be to fly due north?

Page 52: Chapter 3 motion in two d

Solving 2-D Problems

• Resolve all vectors into components– x-component – Y-component

• Work the problem as two one-dimensional problems.– Each dimension can obey different equations of

motion.

• Re-combine the results for the two components at the end of the problem.

Page 53: Chapter 3 motion in two d

Sample Problem• You run in a straight line at a speed of 5.0 m/s in a direction

that is 40o south of west.a) How far west have you traveled in 2.5 minutes?

b) How far south have you traveled in 2.5 minutes?

Page 54: Chapter 3 motion in two d

Sample Problem• A roller coaster rolls down a 20o incline with an

acceleration of 5.0 m/s2.a) How far horizontally has the coaster traveled in 10 seconds?

b) How far vertically has the coaster traveled in 10 seconds?

Page 55: Chapter 3 motion in two d

Sample ProblemA particle passes through the origin with a speed of 6.2 m/s traveling along the y axis. If the particle accelerates in the negative x direction at 4.4 m/s2

.

a) What are the x and y positions at 5.0 seconds?

Page 56: Chapter 3 motion in two d

Sample ProblemA particle passes through the origin with a speed of 6.2 m/s traveling along the y axis. If the particle accelerates in the negative x direction at 4.4 m/s2

.

b) What are the x and y components of velocity at this time?

Page 57: Chapter 3 motion in two d

Projectile Motion

• Day 4

Page 58: Chapter 3 motion in two d

3-5 Projectile Motion

A projectile is an object moving in two dimensions under the influence of Earth's gravity; its path is a parabola.

Page 59: Chapter 3 motion in two d

Projectile Motion

• Something is fired, thrown, shot, or hurled near the earth’s surface.

• Horizontal velocity is constant.

• Vertical velocity is accelerated.

• Air resistance is ignored.

Page 60: Chapter 3 motion in two d

1-Dimensional Projectile

• Definition: A projectile that moves in a vertical direction only, subject to acceleration by gravity.

• Examples:– Drop something off a cliff.– Throw something straight up and catch it.

• You calculate vertical motion only.• The motion has no horizontal component.

Page 61: Chapter 3 motion in two d

2-Dimensional Projectile

• Definition: A projectile that moves both horizontally and vertically, subject to acceleration by gravity in vertical direction.

• Examples:– Throw a softball to someone else.– Fire a cannon horizontally off a cliff.– Shoot a monkey with a blowgun.

• You calculate vertical and horizontal motion.

Page 62: Chapter 3 motion in two d

Horizontal Component of Velocity

• Is constant

• Not accelerated

• Not influence by gravity

• Follows equation:

• x = Vo,xt

Page 63: Chapter 3 motion in two d

Horizontal Component of Velocity

Page 64: Chapter 3 motion in two d

Vertical Component of Velocity

• Undergoes accelerated motion

• Accelerated by gravity (9.8 m/s2 down)

• Vy = Vo,y - gt

• y = yo + Vo,yt - 1/2gt2

• Vy2 = Vo,y

2 - 2g(y – yo)

Page 65: Chapter 3 motion in two d

Horizontal and Vertical

Page 66: Chapter 3 motion in two d

Horizontal and Vertical

Page 67: Chapter 3 motion in two d

Zero Launch Angle Projectiles

Page 68: Chapter 3 motion in two d

Launch angle

• Definition: The angle at which a projectile is launched.

• The launch angle determines what the trajectory of the projectile will be.

• Launch angles can range from -90o (throwing something straight down) to +90o (throwing something straight up) and everything in between.

Page 69: Chapter 3 motion in two d

Zero Launch angle

• A zero launch angle implies a perfectly horizontal launch.

vo

Page 70: Chapter 3 motion in two d

Sample Problem• The Zambezi River flows over Victoria Falls in Africa. The falls are

approximately 108 m high. If the river is flowing horizontally at 3.6 m/s just before going over the falls, what is the speed of the water when it hits the bottom? Assume the water is in freefall as it drops.

Page 71: Chapter 3 motion in two d

Sample Problem• An astronaut on the planet Zircon tosses a rock horizontally with a

speed of 6.75 m/s. The rock falls a distance of 1.20 m and lands a horizontal distance of 8.95 m from the astronaut. What is the acceleration due to gravity on Zircon?

Page 72: Chapter 3 motion in two d

Sample Problem• Playing shortstop, you throw a ball horizontally to the second baseman

with a speed of 22 m/s. The ball is caught by the second baseman 0.45 s later.

a) How far were you from the second baseman?b) What is the distance of the vertical drop?

Page 73: Chapter 3 motion in two d

General Launch Angle Projectiles

Day 5

Page 74: Chapter 3 motion in two d

General launch angle

vo

• Projectile motion is more complicated when the launch angle is not straight up or down (90o or –90o), or perfectly horizontal (0o).

Page 75: Chapter 3 motion in two d

General launch angle

vo

• You must begin problems like this by resolving the velocity vector into its components.

Page 76: Chapter 3 motion in two d

Resolving the velocity

• Use speed and the launch angle to find horizontal and vertical velocity components

VoVo,y = Vo sin

Vo,x = Vo cos

Page 77: Chapter 3 motion in two d

Resolving the velocity

• Then proceed to work problems just like you did with the zero launch angle problems.

VoVo,y = Vo sin

Vo,x = Vo cos

Page 78: Chapter 3 motion in two d

Sample problem• A soccer ball is kicked with a speed of 9.50 m/s at an angle of

25o above the horizontal. If the ball lands at the same level from which is was kicked, how long was it in the air?

Page 79: Chapter 3 motion in two d

Sample problem• Snowballs are thrown with a speed of 13 m/s from a roof 7.0 m

above the ground. Snowball A is thrown straight downward; snowball B is thrown in a direction 25o above the horizontal. When the snowballs land, is the speed of A greater than, less than, or the same speed of B? Verify your answer by calculation of the landing speed of both snowballs.

Page 80: Chapter 3 motion in two d

Projectiles launched over level ground

• These projectiles have highly symmetric characteristics of motion.

• It is handy to know these characteristics, since a knowledge of the symmetry can help in working problems and predicting the motion.

• Lets take a look at projectiles launched over level ground.

Page 81: Chapter 3 motion in two d

Trajectory of a 2-D Projectile

x

y

• Definition: The trajectory is the path traveled by any projectile. It is plotted on an x-y graph.

Page 82: Chapter 3 motion in two d

Trajectory of a 2-D Projectile

x

y

• Mathematically, the path is defined by a parabola.

Page 83: Chapter 3 motion in two d

Trajectory of a 2-D Projectile

x

y

• For a projectile launched over level ground, the symmetry is apparent.

Page 84: Chapter 3 motion in two d

Range of a 2-D Projectile

x

y

Range

• Definition: The RANGE of the projectile is how far it travels horizontally.

Page 85: Chapter 3 motion in two d

Maximum height of a projectile

x

y

Range

MaximumHeight

• The MAXIMUM HEIGHT of the projectile occurs when it stops moving upward.

Page 86: Chapter 3 motion in two d

Maximum height of a projectile

x

y

Range

MaximumHeight

• The vertical velocity component is zero at maximum height.

Page 87: Chapter 3 motion in two d

Maximum height of a projectile

x

y

Range

MaximumHeight

• For a projectile launched over level ground, the maximum height occurs halfway through the flight of the projectile.

Page 88: Chapter 3 motion in two d

Acceleration of a projectile

g

g

g

g

g

x

y

• Acceleration points down at 9.8 m/s2 for the entire trajectory of all projectiles.

Page 89: Chapter 3 motion in two d

Velocity of a projectile

vo

vf

v

v

v

x

y

• Velocity is tangent to the path for the entire trajectory.

Page 90: Chapter 3 motion in two d

Velocity of a projectile

vy

vx

vx

vy

vx

vy

vx

x

y

vx

vy

• The velocity can be resolved into components all along its path.

Page 91: Chapter 3 motion in two d

Velocity of a projectile

vy

vx

vx

vy

vx

vy

vx

x

y

vx

vy

• Notice how the vertical velocity changes while the horizontal velocity remains constant.

Page 92: Chapter 3 motion in two d

Velocity of a projectile

vy

vx

vx

vy

vx

vy

vx

x

y

vx

vy

• Maximum speed is attained at the beginning, and again at the end, of the trajectory if the projectile is launched over level ground.

Page 93: Chapter 3 motion in two d

vo -

vo

Velocity of a projectile

• Launch angle is symmetric with landing angle for a projectile launched over level ground.

Page 94: Chapter 3 motion in two d

to = 0

t

Time of flight for a projectile

• The projectile spends half its time traveling upward…

Page 95: Chapter 3 motion in two d

Time of flight for a projectile

to = 0

t

2t

• … and the other half traveling down.

Page 96: Chapter 3 motion in two d

Position graphs for 2-D projectiles

x

y

t

y

t

x

Page 97: Chapter 3 motion in two d

Velocity graphs for 2-D projectiles

t

Vy

t

Vx

Page 98: Chapter 3 motion in two d

Acceleration graphs for 2-D projectiles

t

ay

t

ax

Page 99: Chapter 3 motion in two d

Projectile Lab

Page 100: Chapter 3 motion in two d

Projectile LabThe purpose is to collect data to plot a trajectory for a projectile launched horizontally, and to calculate the launch velocity of the projectile. Equipment is provided, you figure out how to use it.• What you turn in:

1. a table of data

2. a graph of the trajectory

3. a calculation of the launch velocity of the ball obtained from the data

• Hints and tips:

1. The thin paper strip is pressure sensitive. Striking the paper produces a mark.

2. You might like to hang a sheet of your own graph paper on the brown board.

Page 101: Chapter 3 motion in two d

More on Projectile Motion

Page 102: Chapter 3 motion in two d

The Range Equation

• Derivation is an important part of physics.• Your book has many more equations than

your formula sheet.• The Range Equation is in your textbook,

but not on your formula sheet. You can use it if you can memorize it or derive it!

Page 103: Chapter 3 motion in two d

The Range Equation

• R = vo2sin(2)/g.

– R: range of projectile fired over level ground

– vo: initial velocity

– g: acceleration due to gravity : launch angle

Page 104: Chapter 3 motion in two d

Deriving the Range Equation

Page 105: Chapter 3 motion in two d

Review Day

Page 106: Chapter 3 motion in two d

Sample problem• A golfer tees off on level ground, giving the ball an initial

speed of 42.0 m/s and an initial direction of 35o above the horizontal.

a) How far from the golfer does the ball land?

Page 107: Chapter 3 motion in two d

Sample problem• A golfer tees off on level ground, giving the ball an initial

speed of 42.0 m/s and an initial direction of 35o above the horizontal.

b) The next golfer hits a ball with the same initial speed, but at a greater angle than 45o. The ball travels the same horizontal distance. What was the initial direction of motion?

Page 108: Chapter 3 motion in two d

It can be understood by analyzing the horizontal and vertical motions separately.

3-5 Projectile Motion

4 monkey problem

Page 109: Chapter 3 motion in two d

3-5 Projectile Motion

The speed in the x-direction is constant; in the y-direction the object moves with constant acceleration g.

This photograph shows two balls that start to fall at the same time. The one on the right has an initial speed in the x-direction. It can be seen that vertical positions of the two balls are identical at identical times, while the horizontal position of the yellow ball increases linearly.

Page 110: Chapter 3 motion in two d

3-5 Projectile Motion

If an object is launched at an initial angle of θ0 with the horizontal, the analysis is similar except that the initial velocity has a vertical component.

Page 111: Chapter 3 motion in two d

3-6 Solving Problems Involving Projectile Motion

Projectile motion is motion with constant acceleration in two dimensions, where the acceleration is g and is down.

Page 112: Chapter 3 motion in two d

3-6 Solving Problems Involving Projectile Motion

1. Read the problem carefully, and choose the object(s) you are going to analyze.

2. Draw a diagram.

3. Choose an origin and a coordinate system.

4. Decide on the time interval; this is the same in both directions, and includes only the time the object is moving with constant acceleration g.

5. Examine the x and y motions separately.

Page 113: Chapter 3 motion in two d

3-6 Solving Problems Involving Projectile Motion

6. List known and unknown quantities. Remember that vx never changes, and that vy = 0 at the highest point.

7. Plan how you will proceed. Use the appropriate equations; you may have to combine some of them.

Page 114: Chapter 3 motion in two d

3-7 Projectile Motion Is ParabolicIn order to demonstrate that projectile motion is parabolic, we need to write y as a function of x. When we do, we find that it has the form:

This is indeed the equation for a parabola.

Page 115: Chapter 3 motion in two d

3-8 Relative Velocity

We already considered relative speed in one dimension; it is similar in two dimensions except that we must add and subtract velocities as vectors.

Each velocity is labeled first with the object, and second with the reference frame in which it has this velocity. Therefore, vWS is the velocity of the water in the shore frame, vBS is the velocity of the boat in the shore frame, and vBW is the velocity of the boat in the water frame.

Page 116: Chapter 3 motion in two d

3-8 Relative Velocity

In this case, the relationship between the three velocities is:

(3-6)

Page 117: Chapter 3 motion in two d

Summary of Chapter 3

• A quantity with magnitude and direction is a vector.

• A quantity with magnitude but no direction is a scalar.

• Vector addition can be done either graphically or using components.

• The sum is called the resultant vector.

• Projectile motion is the motion of an object near the Earth’s surface under the influence of gravity.

Page 118: Chapter 3 motion in two d

3-1 Vectors and Scalars

A vector has magnitude as well as direction.

Some vector quantities: displacement, velocity, force, momentum

A scalar has only a magnitude.

Some scalar quantities: mass, time, temperature

1

Page 119: Chapter 3 motion in two d

3-2 Addition of Vectors – Graphical Methods

For vectors in one dimension, simple addition and subtraction are all that is needed.

You do need to be careful about the signs, as the figure indicates.

Page 120: Chapter 3 motion in two d

3-2 Addition of Vectors – Graphical MethodsIf the motion is in two dimensions, the situation is somewhat more complicated.

Here, the actual travel paths are at right angles to one another; we can find the displacement by

using the Pythagorean Theorem.

Page 121: Chapter 3 motion in two d

3-2 Addition of Vectors – Graphical Methods

Adding the vectors in the opposite order gives the same result:

Page 122: Chapter 3 motion in two d

3-2 Addition of Vectors – Graphical Methods

Even if the vectors are not at right angles, they can be added graphically by using the “tail-to-tip” method.

Page 123: Chapter 3 motion in two d

3-2 Addition of Vectors – Graphical Methods

The parallelogram method may also be used; here again the vectors must be “tail-to-tip.”

Page 124: Chapter 3 motion in two d

3-3 Subtraction of Vectors, and Multiplication of a Vector by a Scalar

In order to subtract vectors, we define the negative of a vector, which has the same magnitude but points in the opposite direction.

Then we add the negative vector:

Page 125: Chapter 3 motion in two d

3-3 Subtraction of Vectors, and Multiplication of a Vector by a Scalar

A vector V can be multiplied by a scalar c; the result is a vector cV that has the same direction but a magnitude cV. If c is negative, the resultant vector points in the opposite direction.

Page 126: Chapter 3 motion in two d

3-4 Adding Vectors by Components

Any vector can be expressed as the sum of two other vectors, which are called its components. Usually the other vectors are chosen so that they are perpendicular to each other.

Page 127: Chapter 3 motion in two d

3-4 Adding Vectors by Components

If the components are perpendicular, they can be found using trigonometric functions.

Page 128: Chapter 3 motion in two d

3-4 Adding Vectors by Components

The components are effectively one-dimensional, so they can be added arithmetically:

Page 129: Chapter 3 motion in two d

3-4 Adding Vectors by Components

Adding vectors:

1. Draw a diagram; add the vectors graphically.

2. Choose x and y axes.

3. Resolve each vector into x and y components.

4. Calculate each component using sines and cosines.

5. Add the components in each direction.

6. To find the length and direction of the vector, use:


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