Chapter 7: Using Vectors: Motion
and Force
 7.1 Vectors and Direction
 7.2 Projectile Motion and the Velocity
Vector
 7.3 Forces in Two Dimensions
Chapter 7 Objectives

Add and subtract displacement vectors to describe changes in
position.

Calculate the x and y components of a displacement, velocity, and
force vector.

Write a velocity vector in polar and x-y coordinates.

Calculate the range of a projectile given the initial velocity vector.

Use force vectors to solve two-dimensional equilibrium problems with
up to three forces.

Calculate the acceleration on an inclined plane when given the angle
of incline.
Chapter 7 Vocabulary
 Cartesian
coordinates
 polar
coordinates
 component
 projectile
 cosine
 Pythagorean
theorem
 displacement
 inclined plane
 magnitude
 parabola
 range
 resolution
 resultant
 right triangle
 scalar
 scale
 sine
 tangent
 trajectory
 velocity vector
 x-component
 y-component
Inv 7.1 Vectors and Direction
Investigation Key Question:
How do you give directions in physics?
7.1 Vectors and Direction
 A scalar is a quantity that
can be completely
described by one value:
the magnitude.
 You can think of
magnitude as size or
amount, including units.
7.1 Vectors and Direction
 A vector is a quantity that
includes both magnitude
and direction.
 Vectors require more than
one number.
 The information “1 kilometer,
40 degrees east of north” is
an example of a vector.
7.1 Vectors and Direction
 In drawing a vector as
an arrow you must
choose a scale.
 If you walk five meters
east, your displacement
can be represented by a
5 cm arrow pointing to
the east.
7.1 Vectors and Direction
 Suppose you walk 5 meters
east, turn, go 8 meters north,
then turn and go 3 meters
west.
 Your position is now 8 meters
north and 2 meters east of
where you started.
 The diagonal vector that
connects the starting position
with the final position is
called the resultant.
7.1 Vectors and Direction
 The resultant is the sum of
two or more vectors added
together.
 You could have walked a
shorter distance by going 2 m
east and 8 m north, and still
ended up in the same place.
 The resultant shows the most
direct line between the
starting position and the final
position.
7.1 Representing vectors with
components
 Every displacement vector in
two dimensions can be
represented by its two
perpendicular component
vectors.
 The process of describing a
vector in terms of two
perpendicular directions is
called resolution.
7.1 Representing vectors with
components
 Cartesian coordinates are also known as x-y
coordinates.
 The vector in the east-west direction is called the x-component.
 The vector in the north-south direction is called the ycomponent.
 The degrees on a compass are an example of a polar
coordinates system.
 Vectors in polar coordinates are usually converted first
to Cartesian coordinates.
7.1 Adding Vectors
 Writing vectors in components make it easy to add
them.
7.1 Subtracting Vectors
 To subtract one vector from another vector, you subtract
the components.
Calculating the resultant vector
by adding components
An ant walks 2 meters West, 3 meters
North, and 6 meters East. What is the
displacement of the ant?
1.
2.
3.
You are asked for the resultant vector.
You are given 3 displacement vectors.
Sketch, then add the displacement vectors by
components.
Add the x and y coordinates for each vector:
4.


X1 = (-2, 0) m + X2 = (0, 3) m + X3 = (6, 0) m
= (-2 + 0 + 6, 0 + 3 + 0) m = (4, 3) m
 The final displacement is 4 meters east and 3 meters
north from where the ant started.
7.1 Calculating Vector Components
 Finding components
graphically makes use of
a protractor.
 Draw a displacement
vector as an arrow of
appropriate length at the
specified angle.
 Mark the angle and use a
ruler to draw the arrow.
7.1 Finding components mathematically
 Finding components using trigonometry is
quicker and more accurate than the graphical
method.
 The triangle is a right triangle since the sides are
parallel to the x- and y-axes.
 The ratios of the sides of a right triangle are
determined by the angle and are called sine and
cosine.
7.1 Finding the Magnitude of a Vector
 When you know the x- and y- components of a vector,
and the vectors form a right triangle, you can find the
magnitude using the Pythagorean theorem.
Finding two vectors…
Robots are programmed to move with vectors. A robot
must be told exactly how far to go and in which
direction for every step of a trip. A trip of many steps is
communicated to the robot as series of vectors. A maildelivery robot needs to get from where it is to the mail
bin on the map. Find a sequence of two displacement
vectors that will allow the robot to avoid hitting the desk
in the middle?
1.
You are asked to find two displacement vectors.
2.
You are given the starting (1, 1) and final positions (5,5)
3.
Add components (5, 5) m – (1, 1) m = (4, 4) m.
4.
Use right triangle to find vector coordinates x1 = (0, 4) m, x2 = (4, 0) m
 Check the resultant: (4, 0) m + (0, 4) m = (4, 4) m
Chapter 7: Using Vectors: Motion
and Force
 7.1 Vectors and Direction
 7.2 Projectile Motion and the Velocity
Vector
 7.3 Forces in Two Dimensions
Inv 7.2 Projectile Motion
Investigation Key Question:
How can you predict the range of a launched
marble?
7.2 Projectile Motion and the Velocity Vector
 Any object that is
moving through the air
affected only by gravity
is called a projectile.
 The path a projectile
follows is called its
trajectory.
7.2 Projectile Motion and the Velocity Vector
 The trajectory of a
thrown basketball
follows a special type
of arch-shaped curve
called a parabola.
 The distance a
projectile travels
horizontally is called its
range.
7.2 The velocity vector
 The velocity vector (v) is a
way to precisely describe
the speed and direction of
motion.
 There are two ways to
represent velocity.
 Both tell how fast and in
what direction the ball
travels.
Drawing a velocity vector
to calculate speed
Draw the velocity vector v = (5, 5) m/sec
and calculate the magnitude of the velocity
(the speed), using the Pythagorean
theorem.
1.
2.
3.
You are asked to sketch a velocity vector and calculate its speed.
You are given the x-y component form of the velocity.
Set a scale of 1 cm = 1 m/s. Draw the sketch. Measure the resulting
line segment or use the Pythagorean theorem: a2 + b2 = c2
4.
Solve: v2 = (5 m/s)2 + (5 m/s)2 = 50 m2 /s2
 v = 0 m2 /s2 = 7.07 m/s
7.2 The components of the velocity vector
 Suppose a car is driving
20 meters per second.
 The direction of the
vector is 127 degrees.
 The polar representation
of the velocity is v = (20
m/sec, 127°).
Calculating the components
of a velocity vector
A soccer ball is kicked at a speed of 10 m/s
and an angle of 30 degrees. Find the
horizontal and vertical components of the
ball’s initial velocity.
1.
2.
3.
4.
You are asked to calculate the components of the velocity vector.
You are given the initial speed and angle.
Draw a diagram to scale or use vx = v cos θ and vy = v sin θ.
Solve:
 vx = (10 m/s)(cos 30o) = (10 m/s)(0.87) = 8.7 m/s
 vy = (10 m/s)(sin 30o) = (10 m/s)(0.5) = 5 m/s
7.2 Adding velocity vectors
 Sometimes the total velocity of an object is a
combination of velocities.
 One example is the motion of a boat on a river.
 The boat moves with a certain velocity relative to the
water.
 The water is also moving with another velocity relative to
the land.
7.2 Adding Velocity Components
 Velocity vectors are added by components, just
like displacement vectors.
 To calculate a resultant velocity, add the x
components and the y components separately.
Calculating the components
of a velocity vector
An airplane is moving at a velocity of 100 m/s in a
direction 30 degrees northeast relative to the air. The
wind is blowing 40 m/s in a direction 45 degrees
southeast relative to the ground. Find the resultant
velocity of the airplane relative to the ground.
1.
2.
3.
4.
You are asked to calculate the resultant velocity vector.
You are given the plane’s velocity and the wind velocity
Draw diagrams, use Pythagorean theorem.
Solve and add the components to get the resultant velocity :
 Plane: vx = 100 cos 30o = 86.6 m/s, vy = 100 sin 30o = 50 m/s
 Wind: vx = 40 cos 45o = 28.3 m/s, vy = - 40 sin 45o = -28.3 m/s
 v = (86.6 + 28.3, 50 – 28.3) = (114.9, 21.7) m/s or (115, 22) m/s
7.2 Projectile motion
 When we drop a ball
from a height we
know that its speed
increases as it falls.
 The increase in
speed is due to the
acceleration gravity,
g = 9.8 m/sec2.
Vx
Vy
y
x
7.2 Horizontal motion
 The ball’s horizontal
velocity remains constant
while it falls because
gravity does not exert any
horizontal force.
 Since there is no force, the
horizontal acceleration is
zero (ax = 0).
 The ball will keep moving
to the right at 5 m/sec.
7.2 Horizontal motion
 The horizontal distance a projectile moves can
be calculated according to the formula:
7.2 Vertical motion
 The vertical speed (vy) of the
ball will increase by 9.8
m/sec after each second.
 After one second has
passed, vy of the ball will be
9.8 m/sec.
 After 2 seconds have
passed, vy will be 19.6 m/sec
and so on.
Analyzing a projectile
A stunt driver steers a car off a cliff at a speed
of 20 meters per second. He lands in the lake
below two seconds later. Find the height of
the cliff and the horizontal distance the car
travels.
1.
2.
3.
4.
You are asked for the vertical and horizontal distances.
You know the initial speed and the time.
Use relationships: y = voyt – ½ gt2 and x = vox t
The car goes off the cliff horizontally, so assume voy = 0. Solve:
 y = – (1/2)(9.8 m/s2)(2 s)2 y = –19.6 m. (negative means the car is
below its starting point)
 Use x = voxt, to find the horizontal distance: x = (20 m/s)(2 s) x = 40 m.
7.2 Projectiles launched at an angle
 A soccer ball
kicked off the
ground is also a
projectile, but it
starts with an
initial velocity
that has both
vertical and
*The launch angle determines how the initial
horizontal
divides between vertical (y) and
components. velocity
horizontal (x) directions.
7.2 Steep Angle
 A ball launched at a
steep angle will have
a large vertical
velocity component
and a small
horizontal velocity.
7.2 Shallow Angle
 A ball launched at
a low angle will
have a large
horizontal velocity
component and a
small vertical one.
7.2 Projectiles Launched at an Angle
 The initial velocity components of an object launched at
a velocity vo and angle θ are found by breaking the
velocity into x and y components.
7.2 Range of a Projectile
 The range, or horizontal distance, traveled by a
projectile depends on the launch speed and the
launch angle.
7.2 Range of a Projectile
 The range of a projectile is calculated from the
horizontal velocity and the time of flight.
7.2 Range of a Projectile
 A projectile travels farthest when launched at
45 degrees.
7.2 Range of a Projectile
 The vertical velocity is responsible for giving
the projectile its "hang" time.
7.2 "Hang Time"
 You can easily calculate your own hang time.
 Run toward a doorway and jump as high as you can,
touching the wall or door frame.
 Have someone watch to see exactly how high you
reach.
 Measure this distance with a meter stick.
 The vertical distance formula can be rearranged to
solve for time:
Chapter 7: Using Vectors: Motion
and Force
 7.1 Vectors and Direction
 7.2 Projectile Motion and the Velocity
Vector
 7.3 Forces in Two Dimensions
Inv 7.3 Forces in Two Dimensions
Investigation Key Question:
How do forces balance in two dimensions?
7.3 Forces in Two Dimensions
 Force is also represented by x-y components.
7.3 Force Vectors
 If an object is in
equilibrium, all of the
forces acting on it are
balanced and the net
force is zero.
 If the forces act in two
dimensions, then all of
the forces in the xdirection and y-direction
balance separately.
7.3 Equilibrium and Forces
 It is much more difficult for
a gymnast to hold his arms
out at a 45-degree angle.
 To see why, consider that
each arm must still support
350 newtons vertically to
balance the force of gravity.
7.3 Forces in Two Dimensions
 Use the y-component to find the total force in
the gymnast’s left arm.
7.3 Forces in Two Dimensions
 The force in the right arm must also be 495
newtons because it also has a vertical
component of 350 N.
7.3 Forces in Two Dimensions
 When the gymnast’s arms
are at an angle, only part
of the force from each
arm is vertical.
 The total force must be
larger because the
vertical component of
force in each arm must
still equal half his weight.
7.3 The inclined plane
 An inclined plane is a straight surface,
usually with a slope.
 Consider a block sliding
down a ramp.
 There are three forces
that act on the block:
 gravity (weight).
 friction
 the reaction force acting on
the block.
7.3 Forces on an inclined plane
 When discussing forces, the word “normal”
means “perpendicular to.”
 The normal force
acting on the block is
the reaction force from
the weight of the block
pressing against the
ramp.
7.3 Forces on an inclined plane
 The normal force
on the block is
equal and
opposite to the
component of the
block’s weight
perpendicular to
the ramp (Fy).
7.3 Forces on an inclined plane
 The force parallel
to the surface (Fx)
is given by
Fx = mg sinθ.
7.3 Forces on an inclined plane
 The magnitude of
the friction force
between two sliding
surfaces is roughly
proportional to the
force holding the
surfaces together:
Ff = -mg cosθ.
7.3 Motion on an inclined plane
 Newton’s second law can be used to calculate
the acceleration once you know the
components of all the forces on an incline.
 According to the second law:
Acceleration
(m/sec2)
a=F
m
Force (kg . m/sec2)
Mass (kg)
7.3 Motion on an inclined plane
 Since the block can only accelerate along the ramp, the
force that matters is the net force in the x direction,
parallel to the ramp.
 If we ignore friction, and substitute Newtons' 2nd Law,
the net force is:
Fx = m g sin θ
a =F
m
7.3 Motion on an inclined plane
 To account for friction, the horizontal component of
acceleration is reduced by combining equations:
Fx = mg sin θ -  mg cos θ
7.3 Motion on an inclined plane
 For a smooth surface, the coefficient of friction
(μ) is usually in the range 0.1 - 0.3.
 The resulting equation for acceleration is:
Calculating acceleration
A skier with a mass of 50 kg is on a hill
making an angle of 20 degrees. The
friction force is 30 N. What is the skier’s
acceleration?
1.
2.
3.
4.
You are asked to find the acceleration.
You know the mass, friction force, and angle.
Use relationships: a = F ÷ m and Fx = mg sinθ.
Calculate the x component of the skier’s weight:
 Fx = (50 kg)(9.8 m/s2) × (sin 20o) = 167.6 N
 Calculate the force: F = 167.6 N – 30 N = 137.6 N
 Calculate the acceleration: a = 137.6 N ÷ 50 kg = 2.75 m/s2
7.3 The vector form of Newton’s 2nd law
 An object moving in three dimensions can be
accelerated in the x, y, and z directions.
 The acceleration vector can be written in a similar
way to the velocity vector: a = (ax, ay, az) m/s2.
7.3 The vector form of Newton’s 2nd law
 If you know the forces acting on an object, you
can predict its motion in three dimensions.
 The process of calculating three-dimensional
motion from forces and accelerations is called
dynamics.
 Computers that control space missions
determine when and for how long to run the
rocket engines by finding the magnitude and
direction of the required acceleration.
Calculating acceleration
A 100-kg satellite has many small rocket engines
pointed in different directions that allow it to
maneuver in three dimensions. If the engines make
the following forces, what is the acceleration of the
satellite?
F1 = (0, 0, 50) N F2 = (25, 0, –50) N F3 = (25, 0, 0) N
1.
2.
You are asked to find the acceleration of the satellite.
You know the mass, forces, and assume no friction in space.
3.
4.
Use relationships: F = net force and a = F ÷ m
Calculate the net force by adding components. F = (50, 0, 0) N
5.
Calculate acceleration: ay = az = 0 ax = 50 N ÷ 100 kg = 0.5 m/s2
 a = (0.5, 0, 0) m/s2
Robot Navigation
 A Global Positioning System
(GPS) receiver determines
position to within a few meters
anywhere on Earth’s surface.
 The receiver works by comparing
signals from three different GPS
satellites.
 About twenty-four satellites orbit
Earth and transmit radio signals
as part of this positioning or
navigation system.