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Chapter 3
Kinematics in Two Dimensions; Vectors
Vectors
An Introduction
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
Notating vectors
• This is how you notate a vector…
• This is how you draw a vector…
R R
R headtail
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
Vector angle ranges
x
y
Quadrant I0 < < 90o
Quadrant II90o < < 180o
Quadrant III180o < < 270o
Quadrant IV
270o < < 360o
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
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.
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
Equal Vectors
• Equal vectors have the same length and direction, and represent the same quantity (such as force or velocity).
• Draw several equal vectors.
Inverse Vectors
• Inverse vectors have the same length, but opposite direction.
• Draw a set of inverse vectors.
A
-A
The Right Triangle
θ
op
po
site
adjacent
hypotenuse
Pythagorean Theorem
• hypotenuse2 = opposite2 + adjacent2
• c2 = a2 + b2
θ
op
po
site
adjacent
hypotenuse
Basic Trigonometry functions
• sin θ = opposite/hypotenuse• cos θ = adjacent/hypotenuse• tan θ = opposite/adjacent
θ
op
po
site
adjacent
hypotenuseSOHCAHTOA
Inverse functions
• θ = sin-1(opposite/hypotenuse)• θ = cos-1(adjacent/hypotenuse)• θ = tan-1(opposite/adjacent)
θ
op
po
site
adjacent
hypotenuseSOHCAHTOA
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?
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?
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
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
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
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
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?
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?
Practice Problem
• Find the x- and y-components of the following vectors
a) R = 175 meters @ 95o
Practice Problem
• Find the x- and y-components of the following vectors
b) v = 25 m/s @ -78o
Practice Problem
• Find the x- and y-components of the following vectors
c) a = 2.23 m/s2 @ 150o
Graphical Addition of Vectors
Day 2
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.
A
B
RA + B = R
Practice Graphical Addition
R is called the resultant vector!
B
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.
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
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.
A
B
A - B = C
Practice Graphical Subtraction
-B
C
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.
a)
b)
c)
Vector Addition Laboratory
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
Vector Addition by Component
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.
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.
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.
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?
Relative Motion
Day 3
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.
Relative Motion
Vs
Vw
Vt = Vs + Vw
Vw
Relative Motion
Vs
Vw
Vt = Vs + Vw
Vw
Relative Motion
Vs
Vw
Vt = Vs + Vw Vw
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.
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?
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.
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?
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?
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?
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?
Projectile Motion
• Day 4
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.
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.
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.
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.
Horizontal Component of Velocity
• Is constant
• Not accelerated
• Not influence by gravity
• Follows equation:
• x = Vo,xt
Horizontal Component of Velocity
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)
Horizontal and Vertical
Horizontal and Vertical
Zero Launch Angle Projectiles
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.
Zero Launch angle
• A zero launch angle implies a perfectly horizontal launch.
vo
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.
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?
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?
General Launch Angle Projectiles
Day 5
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).
General launch angle
vo
• You must begin problems like this by resolving the velocity vector into its components.
Resolving the velocity
• Use speed and the launch angle to find horizontal and vertical velocity components
VoVo,y = Vo sin
Vo,x = Vo cos
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
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?
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.
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.
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.
Trajectory of a 2-D Projectile
x
y
• Mathematically, the path is defined by a parabola.
Trajectory of a 2-D Projectile
x
y
• For a projectile launched over level ground, the symmetry is apparent.
Range of a 2-D Projectile
x
y
Range
• Definition: The RANGE of the projectile is how far it travels horizontally.
Maximum height of a projectile
x
y
Range
MaximumHeight
• The MAXIMUM HEIGHT of the projectile occurs when it stops moving upward.
Maximum height of a projectile
x
y
Range
MaximumHeight
• The vertical velocity component is zero at maximum height.
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.
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.
Velocity of a projectile
vo
vf
v
v
v
x
y
• Velocity is tangent to the path for the entire trajectory.
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.
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.
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.
vo -
vo
Velocity of a projectile
• Launch angle is symmetric with landing angle for a projectile launched over level ground.
to = 0
t
Time of flight for a projectile
• The projectile spends half its time traveling upward…
Time of flight for a projectile
to = 0
t
2t
• … and the other half traveling down.
Position graphs for 2-D projectiles
x
y
t
y
t
x
Velocity graphs for 2-D projectiles
t
Vy
t
Vx
Acceleration graphs for 2-D projectiles
t
ay
t
ax
Projectile Lab
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.
More on Projectile Motion
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!
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
Deriving the Range Equation
Review Day
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?
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?
It can be understood by analyzing the horizontal and vertical motions separately.
3-5 Projectile Motion
4 monkey problem
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.
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.
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.
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.
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.
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.
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.
3-8 Relative Velocity
In this case, the relationship between the three velocities is:
(3-6)
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.
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
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.
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.
3-2 Addition of Vectors – Graphical Methods
Adding the vectors in the opposite order gives the same result:
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.
3-2 Addition of Vectors – Graphical Methods
The parallelogram method may also be used; here again the vectors must be “tail-to-tip.”
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:
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.
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.
3-4 Adding Vectors by Components
If the components are perpendicular, they can be found using trigonometric functions.
3-4 Adding Vectors by Components
The components are effectively one-dimensional, so they can be added arithmetically:
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: