1. Physics Chapter 7 Forces and Motion in Two Dimensions

2. 7.1 Forces in Two Dimensions 7.2 Projectile Motion 7.3 Circular Motion And…Simple Harmonic Motion from Chapter 6!

3. Forces in Two Dimensions We have already studied a few forces in two dimensions When dealing with friction acting on an object (parallel), the normal force on the object (perpendicular) also plays a role in friction In this chapter we will be looking at forces on a object that are at angles other than 90°

4. Equilibrium and the Equilibrant Equilibrium—when the net forces on an object is equal to zero This means an object can be stationary or moving but it can’t be _________. When graphically adding vectors together, if a closed geometric shape is formed, there is no resultant (?) so net force is zero

5. Equilibrium and the Equilibrant If an object is not in equilibrium this means that: There is a net force on it When the vectors are added together, there is a resultant It is accelerating

6. Equilibrium and the Equilibrant If add a vector to the resultant that is equal and opposite to it, there will be no net force and the object will be in equilibrium This vector is called the equilibrant The equilibrant is the vector that puts an object in equilibrium It is equal and opposite to the resultant of the existing vectors

7. Creating Equilibrium Hanging a Sign A 168 N sign is equally supported by two ropes that make an angle between them of 135°. What is the tension in the rope? 168 N

8. Creating Equilibrium Hanging a Sign Is 168 N the sign’s weight, resultant force, and/or equilibrant? 168 N

9. Creating Equilibrium Hanging a Sign So both ropes together must pull up with a force of 168 N to put the sign into equilibrium 168 N

10. Creating Equilibrium Hanging a Sign So each ropes together must pull up with a force of 84 N (168  2) to put the sign into equilibrium 168 N 84 N

11. Creating Equilibrium Hanging a Sign So to find the tension in the rope (hypotenuse of a right triangle) use trig! cos  = a/h so h = a/cos  H = 84/cos 67.5 H = 219.5 N Why was  = 67.5° and not 135°? 84 N

12. Creating Equilibrium Try this….. Hanging a Sign A 150 N sign is equally supported by two ropes that make an angle between them of 96°. What is the tension in the rope? Answer: 112 N

13. Motion Along an Inclined Plane When an object sits on a flat surface, there are four forces that determine if there is a net force acting on the object. What are these forces? (five force equation)

14. Motion Along an Inclined Plane If the object is in equilibrium, what can you say about the relationship between F g, F n, F a and F f ?

15. Motion Along an Inclined Plane If the object is in equilibrium on a sloping surface, which of the following forces doesn’t change; F g, F n, F a and F f ?

16. Motion Along an Inclined Plane If the object is in equilibrium on a sloping surface, which of the following forces doesn’t depend on the angle of the slope; F g, F n, F a and F f ?

17. Motion Along an Inclined Plane The object weight (F g ) is always the same (regardless of the slope of the surface) and it always directed straight down (toward center of the earth).

18. Motion Along an Inclined Plane F n, F a, F f all change with the angle of the slope

19. Motion Along an Inclined Plane So lets look how F g, F a, F n, and F f are related

20. Motion Along an Inclined Plane When an object sits on an inclined plane it is being pulled into the plane and down the plane by the force of gravity FgFg

21. Motion Along an Inclined Plane F g is always the hypotenuse of the right force triangle formed FgFg

22. Motion Along an Inclined Plane The force with which the object is pulled into the plane is called the perpendicular force (F  ) FgFg FF

23. Motion Along an Inclined Plane The force with which the object is pulled down the plane is called the parallel force (F  ) FgFg FF F 

24. Motion Along an Inclined Plane These forces are perpendicular and parallel to what? FgFg FF F 

25. Motion Along an Inclined Plane We could solve for F  and F || using F g and angle  if we knew what  was. Angle  is always equal to the slope of the plane FgFg FF F  

26. Motion Along an Inclined Plane So for F  ; cos  = a/h Since a = F  and h = F g F  = F g cos  FgFg F  

27. Motion Along an Inclined Plane So for F  ; sin  = o/h Since o = F  and h = F g F  = F g sin  FgFg F  = F g cos  F  = F g sin  

28. Motion Along an Inclined Plane So if gravity pulls the object down into the plane with a force F , which force counteracts it? FgFg F  = F g cos  F  = F g sin  

29. Motion Along an Inclined Plane So F , is equal and opposite F n FgFg F  = F g cos  F  = F g sin   FnFn

30. Motion Along an Inclined Plane So if gravity pulls the object down the plane with a force F , which force counteracts it? FgFg F  = F g cos  F  = F g sin  

31. Motion Along an Inclined Plane So F  is equal and opposite to F f If there is no motion (or acceleration)! FgFg F  = F g cos  F  = F g sin   FfFf

32. Motion Along an Inclined Plane So F , is equal and opposite F n and F  is equal and opposite to F f Only if there is no ________or_______.

33. Motion Along an Inclined Plane Example A trunk weighing 562 N is resting on a plane inclined 30°. Find the perpendicular and parallel components of its weight. Answers F  = - 487 N F  = - 281 N Why are they negative? What other forces are they equal to? Would the force change if the object was moving?

34. Motion Along an Inclined Plane Under what set of circumstances could the object be accelerating down the plane? F  > F f Slope too slippery—not enough friction to hold it We could be pushing or pulling the object What can we say about the net force on the object?

35. Motion Along an Inclined Plane What is the Five Forces Equation? F net = F a + F f + F g + F n Lets see how it can be modified to deal with inclined planes

36. Motion Along an Inclined Plane F net = F a + F f + F g + F n Does F n determine if the object will accelerate down the plane? So we get rid of it F net = F a + F f + F g

37. Motion Along an Inclined Plane F net = F a + F f + F g What are the two components of weight for an object on the plane? We replace weight with these two F net = F a + F f + F  + F 

38. Motion Along an Inclined Plane F net = F a + F f + F  + F  Does F  determine if the object accelerates down the plane? Get rid of it F net = F a + F f + F 

39. Motion Along an Inclined Plane F net = F a + F f + F  Can we still push or pull the object? Does friction still act on it? So this is the new Four Forces Equation for inclined planes F net = F a + F f + F 

40. Motion Along an Inclined Plane Example A trunk weighing 562 N is resting on a plane inclined 30°. Find the perpendicular and parallel components of its weight. If the force of friction is 200 N, find the trunk’s acceleration rate. Answers F  = - 487 N F  = - 281 N a = -1.41 m/s 2

41. Projectile Motion Projectile—a launched object that moves through air only under the force of gravity Ignoring air resistance! Trajectory—the path of a projectile through space

42. Projectile Motion A projectile has both horizontal and vertical components of its velocity These components are independent of each other This is because the force of gravity acts on the vertical component (causing acceleration) but not on the horizontal component (constant velocity). But one can equal zero!

43. Projectile Motion We will be learning how to solve three types of projectile motion problems: Dropped Objects Objects Thrown Horizontally Objects Launched at an Angle

44. A Dropped Object When an object is dropped from a height it will fall It has two components of its velocity, horizontal and vertical and both are equal to zero

45. A Dropped Object As it falls, it picks up speed (accelerates) in the vertical direction due to the force of gravity Acceleration due to gravity =? So when dropped any object will pick up negative vertical velocity at the rate of – 9.8 m/s for each second it falls

46. A Dropped Object As it falls, its horizontal speed stays constant (why?) So as an object falls its vertical speed changes but its horizontal speed doesn’t These two perpendicular components of speed are independent of each other, they have no effect on each other (this is true of all perpendicular vectors)

47. A Dropped Object Example A 2.5 kg stone is dropped from a cliff 44 m high. How long is it in the air? Answer 3.0 sec What is its velocity right before it hits the ground? Answer - 29.4 m/s

48. An Object Thrown Horizontally When an object is thrown horizontally, it has horizontal and vertical components to its velocity The horizontal velocity is constant during the entire time its in the air (why?) The vertical velocity starts as zero but increases as it falls just as if it were dropped! (why?) These two components are independent of each other

49. An Object Thrown Horizontally Example A 2.5 kg stone thrown horizontally at 15 m/s from a cliff 44 m high. How long is it in the air? What is its horizontal velocity right before it hits the ground? What is its vertical velocity right before it hits the ground? How far from the cliff does it land?

50. An Object Thrown Horizontally Example A 2.5 kg stone thrown horizontally at 15 m/s from a cliff 44 m high. How long is it in the air? Answer 3.0 sec

51. An Object Thrown Horizontally Example A 2.5 kg stone thrown horizontally at 15 m/s from a cliff 44 m high. How long is it in the air? 3.0 sec What is its horizontal velocity right before it hits the ground? Answer 15 m/s What is its vertical velocity right before it hits the ground? Answer - 29.4 m/s

52. An Object Thrown Horizontally Example A 2.5 kg stone thrown horizontally at 15 m/s from a cliff 44 m high. How long is it in the air?3.0 sec What is its horizontal velocity right before it hits the ground? 15 m/s What is its vertical velocity right before it hits the ground? - 29.4 m/s How far from the cliff does it land? Answer 45 m

53. Objects Launched at an Angle When a projectile is launched at an angle, there is a nonzero vertical and horizontal component to the initial velocity

54. Objects Launched at an Angle The trajectory of the projectile is called a parabola The amount of time the projectile is in the air (prior to landing) is called its flight time or hang time The horizontal distance it travels (prior to landing) is called it range The maximum vertical height is called the maximum height

55. Objects Launched at an Angle Flight time or hang time, range and maximum height all depend on the initial vertical and horizontal velocity And the initial vertical and horizontal components of velocity depend on the initial velocity and the angle at which the object is launched

56. Objects Launched at an Angle A projectile is launched at an angle  with a velocity, v o. vovo 

57. Objects Launched at an Angle At the beginning of its trajectory, it has an initial vertical component of its velocity (v v ) and an initial horizontal component of its velocity (v h ) vovo  v vhvh

58. Objects Launched at an Angle Can we use trig to solve for the initial vertical component of its velocity (v v ) and initial horizontal component of its velocity (v h )? vovo  v vhvh

59. Objects Launched at an Angle Initial vertical component of its velocity (v v ) v v = v o sin  vovo  v vhvh

60. Objects Launched at an Angle Initial horizontal component of its velocity (v h )? v h = v o cos  vovo  v vhvh

61. Objects Launched at an Angle What can you tell me about these perpendicular components of initial velocity? vovo  v v = v o sin  v h = v o cos 

62. Objects Launched at an Angle Lets make a table of what we know of what we know about the horizontal and vertical components of the projectile's motion vovo  v v = v o sin  v h = v o cos 

63. Objects Launched at an Angle Motionhorizontalvertical vovo v o cos  v o sin  vSameZero-at top of path aNone- 9.8 m/s 2

64. Objects Launched at an Angle Lets see how we find the hang time, range and maximum height for this type of projectile motion problem vovo  v v = v o sin  v h = v o cos 

65. Objects Launched at an Angle A soccer ball is kicked at an initial velocity of 4.47 m/s at a 66°. Find its hang time, range and maximum height.

66. Objects Launched at an Angle A soccer ball is kicked at an initial velocity of 4.47 m/s at a 66°. Find its hang time, range and maximum height. Step #1: draw a diagram (vector) vovo  v v = v o sin  v h = v o cos 

67. Objects Launched at an Angle A soccer ball is kicked at an initial velocity of 4.47 m/s at a 66°. Find its hang time, range and maximum height. Step #2: Find the initial horizontal and vertical components of the initial velocity vovo  v v = v o sin  = 4.47 sin 66 = 4.08 m/s v h = v o cos  = 4.47 cos 66 = 1.82 m/s

68. Objects Launched at an Angle A soccer ball is kicked at an initial velocity of 4.47 m/s at a 66°. Find its hang time, range and maximum height. Step #3: Make a table of the horizontal and vertical components of the projectile’s motion vovo  v v = v o sin  = 4.47 sin 66 = 4.08 m/s v h = v o cos  = 4.47 cos 66 = 1.82 m/s

69. Objects Launched at an Angle Step #3: Make a table of the horizontal and vertical components of the projectile’s motion vovo  v v = v o sin  = 4.47 sin 66 = 4.08 m/s v h = v o cos  = 4.47 cos 66 = 1.82 m/s Motionhorizontalvertical vovo 1.82 m/s4.08 m/s v1.82 m/s0 m/s at top aNone-9.8 m/s 2

70. Objects Launched at an Angle Step #4: Use the vertical components of the projectile’s motion to solve for its hang time v = v o + at 0 = 4.08 + -9.8t t 1/2 = 0.416 s (time to top) t t = 0.833 s (hang time) Motionhorizontalvertical vovo 1.82 m/s4.08 m/s v1.82 m/s0 m/s at top aNone-9.8 m/s 2

71. Objects Launched at an Angle Step #5: Use the vertical components of the projectile’s motion to solve for its maximum height d = d o + ½(v + v o )t d = 0 + ½ (4.08 + 0)(0.416) d = 0.849 m Motionhorizontalvertical vovo 1.82 m/s4.08 m/s v1.82 m/s0 m/s at top aNone-9.8 m/s 2

72. Objects Launched at an Angle Step #6: Use the horizontal components of the projectile’s motion to solve for its range d = d o + ½(v + v o )t d = 0 + ½ (1.82 + 1.82)(0.833) d = 1.52 m Motionhorizontalvertical vovo 1.82 m/s4.08 m/s v1.82 m/s0 m/s at top aNone-9.8 m/s 2

73. Objects Launched at an Angle A football is kicked at an initial velocity of 27 m/s at a 30°. Find its hang time, range and maximum height Hang time 2.76 s Range 64.6 m Maximum height. 9.27 m

74. Projectile Motion Summary: Dropped Objects Objects Thrown Horizontally Objects Launched at an Angle

75. Circular Motion Uniform circular motion (UCM)— when an object is moving in a circle at constant speed Is an object in uniform circular motion accelerating? (why?) What is uniform in UCM? What is changing in UCM?

76. Circular Motion During uniform circular motion an object is accelerating because the direction (not magnitude) of its velocity is changing.

77. Circular Motion During uniform circular motion an object is accelerating because the direction (not magnitude) of its velocity is changing.

78. Circular Motion So to accelerate an object we must apply a ________ to it and its direction is ______.

79. Circular Motion During UCM, the velocity's direction is tangential to its circular path and its acceleration’s direction is toward the center

80. Circular Motion “Centripetal”—center seeking Centripetal acceleration (a c )—acceleration that causes an object to move in a circular path Its direction is toward the center of the circular path Centripetal force (F c )—force that causes an object to move in a circular path Its direction is toward the center of the circular path

81. Circular Motion Velocity = rate of change in position during a time interval V =  d/t When an object travels a circular path, this is just the circumference of the circle (2  r) So for UCM; v = (2  r)/T T = period of revolution (sec/rev) Time it takes object to complete circular path

82. Circular Motion acceleration = rate of change in velocity during a time interval a =  v/t But for UCM; a c = v 2 /r r = radius of circular path Other possible equations?

83. Circular Motion force = cause a change in velocity (acceleration) F = ma so for UCM; F c = ma c = mv 2 /r r = radius of circular path Other possible equations?

84. Circular Motion Example: A 13-g rubber stopper is attached to a 0.93 m string. If the stopper is swung in a circular path at a rate of one revolution in 1.18 s, find the tension in the string (F c ) Answer 0.343 N

85. Circular Motion Summary: Uniform circular motion What is it? What changes and what doesn’t?

86. Circular Motion Summary: Circular velocity What is it? Direction? Equation?

87. Circular Motion Summary: Centripetal acceleration (a c ) What is it? Direction? Equation?

88. Circular Motion Summary: Centripetal Force (F c) What is it? Direction? Equation?

89. Simple Harmonic Motion A type of periodic motion Periodic motion—motion that repeats itself over and over again over same path Simple Harmonic Motion (SHM)— periodic motion in which the force that will restore the object to equilibrium is directly proportional to its displacement

90. Simple Harmonic Motion Examples: playground swing, pendulum, vibrating spring, guitar string

91. Simple Harmonic Motion Two quantities describe simple harmonic motion: Period Amplitude

92. Simple Harmonic Motion Period (T)—the amount of time that is needed to complete one complete cycle of motion Units: seconds (per cycle) Amplitude

93. Simple Harmonic Motion Amplitude—the maximum distance the object moves from equilibrium The larger the amplitude of vibration (displacement) the more force will need to be applied

94. Simple Harmonic Motion Lets look at some types of objects undergoing simple harmonic motion: Mass on a spring Pendulum

95. Mass on a spring When a mass oscillates on a spring we can change many things and see what effect it has on the oscillation rate (period) Lets look at: Size of mass Size of spring Stiffness of spring Amplitude of vibration

96. Mass on a spring Effects on period Size of mass Bigger mass means larger period Size of spring Bigger spring means larger period Stiffness of spring Stiffer spring means smaller period Amplitude of vibration No effect on period

97. Pendulum When a pendulum swings we can change many things and see what effect it has on the oscillation rate (remember your first lab!) Lets look at: Size of mass (bob) Length of pendulum Amplitude of vibration Does anyone remember which one changes the period?

98. Pendulum Effects on period Size of mass No effect Size of pendulum Longer string means larger period Amplitude of vibration No effect on period

99. Pendulum There is an equation that can be used to calculate the period of any pendulum T = 2  l/g Where: l = length of pendulum (meters) g = acceleration due to gravity (-9.8 m/s 2 )

100. Pendulum Example: What is the length of a pendulum that has a period of 2.25 s? Answer: 1.26 m