1. 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

2. 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.

3. Chapter 7 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

4. Inv 7.1 Vectors and Direction Investigation Key Question: How do you give directions in physics?

5. 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.

6. 7.1 Vectors and Direction  A vector is a of magnitude and direction.  Vectors require more than thing.  The information “1 kilometer, 40 degrees east of north” is an example of a vector.

7. 7.1 Vectors and Direction  To draw a vector arrow you must choose a scale.  You walk five meters east,

8. 7.1 Vectors and Direction  You walk 5 meters east, turn, go 8 meters north, then turn and go 3 meters west.  Your are now 8 meters north and 2 meters east of your start.  The diagonal vector that connects the starting position with the final position is called the resultant.

9. 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.

11. 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.

12. 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 y- component.  The degrees on a compass are an example of a polar coordinates system.  Vectors in polar coordinates are usually converted first to Cartesian coordinates.

13. 7.1 Adding Vectors  Writing vectors in components make it easy to add them.

14. 7.1 Subtracting Vectors  To subtract one vector from another vector, you subtract the components.

15.  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:  X 1 = (-2, 0) m + X 2 = (0, 3) m + X 3 = (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?

16. 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.

17. 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.

19. 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.

21.  You are asked to find two displacement vectors.  You are given the starting (1, 1) and final positions (5,5)  Add components (5, 5) m – (1, 1) m = (4, 4) m.  Use right triangle to find vector coordinates x 1 = (0, 4) m, x 2 = (4, 0) m  Check the resultant: (4, 0) m + (0, 4) m = (4, 4) m Finding two 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?

22. 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

23. Inv 7.2 Projectile Motion Investigation Key Question: How can you predict the range of a launched marble?

24. 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.

25. 7.2 Projectile Motion and the Velocity Vector  The trajectory of a thrown object follow a parabola.  The distance a projectile travels horizontally is called its range. Range

26. 7.2 The velocity vector  The velocity vector (v) is a way to precisely describe speed and direction.  There are two ways to represent velocity.

27. Draw the velocity vector v = (5, 5) m/sec and calculate the magnitude of the velocity (the speed), using the Pythagorean theorem. Drawing a velocity vector to calculate speed

28. Draw the velocity vector v = (5, 5) m/sec and calculate the magnitude of the velocity (the speed), using the Pythagorean theorem.  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: a 2 + b 2 = c 2  Solve: v 2 = (5 m/s) 2 + (5 m/s) 2 = 50 m 2 /s 2  v =  0 m 2 /s 2 = 7 m/s 2 sf Drawing a velocity vector to calculate speed

29. 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°).

30. 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 (Vo).

31.  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 v x = v cos θ and v y = v sin θ.  Solve:  v x = (10 m/s)(cos 30 o ) = (10 m/s)(0.87) = 8.7 m/s  v y = (10 m/s)(sin 30 o ) = (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 (Vo).

32. 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.

33. 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.

34.  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: v x = 100 cos 30 o = 86.6 m/s, v y = 100 sin 30 o = 50 m/s  Wind: v x = 40 cos 45 o = 28.3 m/s, v y = - 40 sin 45 o = -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 north of east relative to the air. The wind is blowing 40. m/s in a direction 45 degrees south of east relative to the ground. Find the resultant velocity of the airplane relative to the ground.

35. 7.2 Projectile motion VxVx VyVy x y  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/sec 2. down

36. 7.2 Projectile motion  When a ball is thrown up, we know it starts to slow down too.  You can use ΔV y =a y t to find out how long it takes to get to the top

37. 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 (a x = 0).  The ball will keep moving to the right at 5 m/sec.

38. 7.2 Horizontal motion  The horizontal distance a projectile moves can be calculated according to the formula:

39. 7.2 Vertical motion  The vertical speed (v y ) of the ball will change by 9.8 m/sec down after each second.  After one second v y = -9.8 m/sec.  After 2 seconds v y = -19.6 m/sec and so on.

41.  You are asked for the vertical and horizontal distances.  You know the initial speed and the time.  Use relationships: y = v oy t – ½ gt 2 and x = v ox t  The car goes off the cliff horizontally, so assume v oy = 0. Solve:  y = – (1/2)(9.8 m/s 2 )(2 s) 2 y = –19.6 m. (negative means the car is below its starting point)  Use x = v ox t, 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.

42. 7.2 Projectiles launched at an angle  A kicked soccer ball kicked is also a projectile,  but it starts with an initial velocity with vertical and horizontal components. *The launch angle determines how the initial velocity divides between vertical (y) and horizontal (x) directions.

43. 7.2 Steep Angle  steep angle will have a large vertical velocity component and a small horizontal one.

44. 7.2 Shallow Angle  A ball launched at a low angle will have a large horizontal velocity component and a small one.

45. 7.2 Projectiles Launched at an Angle  The initial velocity components of projectile with velocity v o and angle θ are found by breaking the velocity into x and y components.

46. 7.2 Range of a Projectile  The range, or horizontal distance, traveled by a projectile depends on the launch speed and the launch angle.

47. 7.2 Range of a Projectile  The range of a projectile is calculated from the horizontal velocity and the time of flight.

48. 7.2 Range of a Projectile  A projectile travels farthest when launched at 45 degrees without air resistance. Range: If there is no change in height

49. 7.2 Range of a Projectile  The vertical velocity is responsible for giving the projectile its "hang" time.

50. 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:

51. 7.2 Problem Solving  Find the knowns and unknown and write them down  Draw a triangle and find the x and y velocities  Find the time of flight  ΔV y =at (same start and end height)  Y=Y o + V y t + 1/2gt^2 (different start and end height  Find the range x=V x t + 1/2a x t^2

52. Olympic gold medalist long jumper Greg Rutherford Analyzing a projectile A long jumper jumps at 12 m/s 30° above vertical. Determine time of flight, peak height, and the range of the jumper.

53. 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

54. Inv 7.3 Forces in Two Dimensions Investigation Key Question: How do forces balance in two dimensions?

55. 7.3 Forces in Two Dimensions  Force is also represented by x-y components.

56. 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.

57. 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.

58. 7.3 Forces in Two Dimensions  Use the y-component to find the total force in the gymnast’s left arm.

59. 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.

60. 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.

61. 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.

63. 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.

64. 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 (F y ).

65. 7.3 Forces on an inclined plane  The force parallel to the surface (F x ) is given by F x = mg sinθ.

66. 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: F f = -  mg cosθ.

67. 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/sec 2 ) Mass (kg) Acceleration (m/sec 2 )

68. 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: F x = a msin θg F m =

69. 7.3 Motion on an inclined plane  To account for friction, the horizontal component of acceleration is reduced by combining equations: F x = mg sin θ -  mg cos θ

70. 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:

71.  You are asked to find the acceleration.  You know the mass, friction force, and angle.  Use relationships: a = F ÷ m and F x = mg sinθ.  Calculate the x component of the skier’s weight:  F x = (50 kg)(9.8 m/s 2 ) × (sin 20 o ) = 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/s 2 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?

72. 7.3 The vector form of Newton’s 2 nd 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 = (a x, a y, a z ) m/s 2.

73. 7.3 The vector form of Newton’s 2 nd 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.

74.  You are asked to find the acceleration of the satellite.  You know the mass, forces, and assume no friction in space.  Use relationships: F = net force and a = F ÷ m  Calculate the net force by adding components. F = (50, 0, 0) N  Calculate acceleration: a y = a z = 0 a x = 50 N ÷ 100 kg = 0.5 m/s 2  a = (0.5, 0, 0) m/s 2 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? F 1 = (0, 0, 50) N F 2 = (25, 0, –50) N F 3 = (25, 0, 0) N

75.  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