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.

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.

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.

Calculate a resultant vector
• An ant walks 2 meters West, 3 meters North, and 6 meters East.
• What is the displacement of the ant?

7.1 Finding Vector Components
• Draw a displacement vector as an arrow of appropriate length at the specified angle.
Graphically
• Mark the angle and use a ruler to draw the arrow.

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 .

7.1 Adding Vectors
• Writing vectors in components make it easy to add them.

7.1 Subtracting Vectors

7.1 Calculate vector magnitude
• 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.

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 archshaped curve called a parabola .
• The distance a projectile travels horizontally is called its range .

7.2 Projectile Motion and 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.

7.2 Calculate magnitude
Draw the velocity vector v =
(5, 5) m/sec and calculate the magnitude of the velocity (the speed), using the Pythagorean theorem.

7.2 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°).

7.2 Calculate velocity
• 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 Components
• 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

7.2 Calculate velocity components
• An airplane is moving at a velocity of 100 m/s in a direction 30 degrees NE relative to the air.
• The wind is blowing 40 m/s in a direction 45 degrees SE relative to the ground.
• Find the resultant velocity of the airplane relative to the ground.

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.
y
V y
V x x

7.2 Horizontal Speed
• 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 Speed
• The horizontal distance a projectile moves can be calculated according to the formula:

7.2 Vertical Speed
• The vertical speed (v y
) of the ball will increase by 9.8 m/sec after each second.
• After one second has passed, v y of the ball will be 9.8 m/sec.
• After the 2nd second has passed, v y and so on.
will be 19.6 m/sec

7.2 Calculate using projectile motion
• 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 soccer ball 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 v o 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:

7.2 Projectile Motion and the Velocity
Vector
Key Question:
Can you predict the landing spot of a projectile?
*Students read Section 7.2 BEFORE Investigation 7.2

y
Marble’s Path
V x
V y t = ?
x = ?

In order to solve “x” we must know “t”
2y = g t 2
Y = v o t – ½ g t 2 v o t = 0 (zero)
Y = ½ g t 2 t 2 = 2y g t = 2y g

7.3 Forces in Two Dimensions
• Force is also represented in 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 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 Forces and Inclined Planes
• 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 and Inclined Planes
• 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 and Inclined Planes
• The normal force on the block is equal and opposite to the component of the block’s weight perpendicular to the ramp ( F y
).

7.3 Forces and Inclined Planes
• The force parallel to the surface ( F x
) is given by
F x
= mg sinθ.

7.3 Acceleration on a Ramp
• 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:
Force (kg .
m/sec 2 )
Acceleration
(m/sec 2 ) a = F m
Mass (kg)

7.3 Acceleration on a Ramp
• 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:
F x
= m g sin θ a = F m

7.3 Acceleration on a Ramp
• To account for friction, the horizontal component of acceleration is reduced by combining equations:
F x
= mg sin θ m mg cos θ

7.3 Acceleration on a Ramp
• For a smooth surface, the coefficient of friction (μ) is usually in the range 0.1 - 0.3.
• The resulting equation for acceleration is:

7.3 Calculate acceleration on a ramp
• 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 Vectors and Direction
Key Question:
How do forces balance in two dimensions?
*Students read Section 7.3 BEFORE Investigation 7.3

Application: Robot Navigation