Vectors phy2 chp7

Chapter7: Using Vectors: Motion and
Force
 7.1 Vectors and Direction
 7.2 Projectile Motion and the Velocity
Vector
 7.3 Forces in Two Dimensions

Chapter7 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.

Chapter7 Vocabulary
 Cartesian
coordinates
 component
 cosine
 displacement
 inclined plane
 magnitude
 parabola
 polar
coordinates
 projectile
 Pythagorean
theorem
 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:
Ho w do yo u g ive dire ctio ns in physics?

7.1 Vectors and Direction
 A scalaris a quantity that
can be completely
described by one value: the
magnitude.
 You can thinkof magnitude
as size oramount, including
units.

 A vectoris 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.

 In drawing a vectoras an
arrow you must choose a
scale.
 If you walkfive meters
east, yourdisplacement
can be represented by a 5
cmarrow pointing to the
east.

 Suppose you walk5 meters
east, turn, go 8 meters north,
then turn and go 3 meters west.
 Yourposition is now 8 meters
north and 2 meters east of
where you started.
 The diagonal vectorthat
connects the starting position
with the final position is called
the resultant.

 The resultant is the sum of two
ormore vectors added together.
 You could have walked a
shorterdistance 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 vectorin two
dimensions can be represented by
its two perpendicularcomponent
vectors.
 The process of describing a vector
in terms of two perpendicular
directions is called resolution.

7.1 Representing vectors with
components
 Cartesiancoordinates are also known as x-ycoordinates.
 The vectorin the east-west direction is called the x-component.
 The vectorin the north-south direction is called the y-component.
 The degrees on a compass are an example of a polar
coordinates system.
 Vectors in polarcoordinates 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 vectorfrom anothervector, you subtract
the components.

1. You are asked for the resultant vector.
2. You are given 3 displacement vectors.
3. Sketch, then add the displacement vectors by
components.
4. Add the x and y coordinates for each vector:
 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.
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?

7.1 Calculating VectorComponents
 Finding components
graphically makes use of a
protractor.
 Draw a displacement vector
as an arrow of appropriate
length at the specified
angle.
 Markthe angleand use a
rulerto 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 trianglesince 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 sineand
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 Pythagoreantheorem.

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
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
mail-delivery 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?

Inv 7.2 Projectile Motion
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 velocityvector(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.

Draw the velocity vector v = (5, 5) m/sec
and calculate the magnitude of the velocity
(the speed), using the Pythagorean
theorem.
1. You are asked to sketch a velocity vector and calculate its speed.
2. You are given the x-y component form of the velocity.
3. 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
Drawing a velocity vector
to calculate speed

7.2 The components of the velocity vector
 Suppose a caris driving 20
meters persecond.
 The direction of the vector
is 127 degrees.
 The polarrepresentationof
the velocity is v = (20
m/sec, 127°).

1. You are asked to calculate the components of the velocity vector.
2. You are given the initial speed and angle.
3. Draw a diagram to scale or use vx = v cos θ and vy = v sin θ.
4. 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
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.

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 wateris also moving with anothervelocity 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 ycomponents separately.

1. You are asked to calculate the resultant velocity vector.
2. You are given the plane’s velocity and the wind velocity
3. Draw diagrams, use Pythagorean theorem.
4. 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
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.

7.2 Projectile motion
Vx
Vy
x
y
 When we drop a ball
froma 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
.

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 increaseby 9.8 m/sec
aftereach second.
 Afterone second has passed,
vy of the ball will be 9.8 m/sec.
 After2 seconds have passed, vy
will be 19.6 m/sec and so on.

1. You are asked for the vertical and horizontal distances.
2. You know the initial speed and the time.
3. Use relationships: y = voyt – ½ g t2
and x = vox t
4. 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.
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.

7.2 Projectiles launched at an angle
 A soccerball
kicked off the
ground is also a
projectile, but it
starts with an
initial velocity
that has both
vertical and
horizontal
components.
*The launch angle determines how the initial
velocity divides between vertical (y) and
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 are found by breaking the velocityθ
into xand ycomponents.

7.2 Range of a Projectile
 The range, orhorizontal distance, traveled by a
projectile depends on the launch speed and the
launch angle.

 The range of a projectile is calculated fromthe
horizontal velocity and the time of flight.

 A projectile travels farthest when launched at 45
degrees.

 The vertical velocity is responsible forgiving the
projectile its "hang"time.

7.2 "Hang Time"
 You can easily calculate yourown hang time.
 Run toward a doorway and jump as high as you can,
touching the wall ordoorframe.
 Have someone watch to see exactly how high you reach.
 Measure this distance with a meterstick.
 The vertical distance formula can be rearranged to solve
fortime:

Inv 7.3 Forces in Two Dimensions
Ho w do fo rce s balance in two dim e nsio ns?

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 x-direction
and y-direction balance
separately.

7.3 Equilibrium and Forces
 It is much more difficult fora
gymnast to hold his arms out
at a 45-degree angle.
 To see why, considerthat
each armmust still support
350 newtons vertically to
balance the force of gravity.

 Use the y-component to find the total force in the
gymnast’s left arm.

 The force in the right armmust also be 495
newtons because it also has a vertical component
of 350 N.

 When the gymnast’s arms
are at an angle, only part of
the force fromeach armis
vertical.
 The total force must be
largerbecause the vertical
component of force in each
armmust still equal half his
weight.

7.3 The inclined plane
 An inclinedplaneis a straight surface, usually
with a slope.
 Considera blocksliding
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 “perpendicularto.”
 The normal force acting
on the blockis the
reaction force fromthe
weight of the block
pressing against the
ramp.

 The normal force
on the blockis
equal and opposite
to the component
of the block’s
weight
perpendicularto
the ramp (Fy).

 The force parallel to
the surface (Fx) is
given by
Fx = mgsinθ.

 The magnitude of the
friction force
between two sliding
surfaces is roughly
proportional to the
force holding the
surfaces together:
Ff =-µmgcos .θ

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:
a = F
m
Force (kg .
m/sec2
)
Mass (kg)
Acceleration
(m/sec2
)

 Since the blockcan only accelerate along the ramp, the
force that matters is the net force in the xdirection,
parallel to the ramp.
 If we ignore friction, and substitute Newtons' 2nd Law, the
net force is:
Fx =
a
m sin θg
F
m
=

 To account forfriction, the horizontal component of
acceleration is reduced by combining equations:
Fx=mgsin -θ µ mg cos θ

 Fora smooth surface, the coefficient of friction
( ) is usually in the range 0.1 - 0.3.μ
 The resulting equation foracceleration is:

1. You are asked to find the acceleration.
2. You know the mass, friction force, and angle.
3. Use relationships: a = F ÷ m and Fx = m g sinθ.
4. 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
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?

7.3 The vectorformof 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 vectorform 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 fromforces and accelerations is called
dynamics.
 Computers that control space missions determine
when and forhow long to run the rocket engines by
finding the magnitude and direction of the required
acceleration.

1. You are asked to find the acceleration of the satellite.
2. You know the mass, forces, and assume no friction in space.
3. Use relationships: F = net force and a = F ÷ m
4. 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
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

 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.
Robot Navigation

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