PHYSICS. UNIT 5 Forces in Equilibrium PERIODS 2 & 4. Mr. Largo

Transcription

1 PHYSICS UNIT 5 Forces in Equilibrium PERIODS 2 & 4 Mr. Largo

2 INTRODUCTION TO FORCES AND FORCE DIAGRAMS

3 FORCE DEFINITION: Force is a vector quantity that characterizes how hard (magnitude) and in which direction an external object pushes or pulls on the system object.

4 INTERACTIONS: PUSH or PULL System: Interaction: Objects: Physics: car Pushing Person & car Person exerts a force on the car.

5 INTERACTIONS: PUSH or PULL System: Interaction: Objects: Physics: dog Pulling Dog & Leash Leash exerts a force on the dog

6 FORCES: SOPHISTICATED LANGUAGE Forces are EXERTED... NOT applied, NOT acted upon. Objects do not have forces. A force requires TWO objects. Objects exert forces on other objects.

7

8 FORCE Forces are vector physical quantities: Magnitude Direction. F x F y Up or down Forward (left or right)

9 FORCE: SUBSCRIPTS F M on S The symbol for force has subscripts identifying the external object that exerts the force and the system object on which the force is exerted.

10 FORCE: SUBSCRIPTS System: Sled Interaction: Megan exerts a force on the sled. F M on S Force that Megan (M) exerts on the sled (S)

11 TWO KINDS OF FORCE Contact forces: Objects are in contact for the interaction (touching each other). Book on table Cellphone on hand Car on parking lot Colored pencils on box

12 TWO KINDS OF FORCE Field Forces: Non-contact force in which a field (created by an external object) exerts a force on an object inside the field. Gravitational Field (objects attracted to planet Earth) Magnetic Field (magnet attracting a paper clip) Electric Field (pieces of paper attracted to plastic after being rubbed)

13 GRAVITATIONAL FORCE

14 PLANET EARTH INTERACTS WITH US! Earth is an object, a huge rock with a huge mass! This massive rock creates a gravitational field that exerts a force on every single object on its surface. The closer you are to the center, the stronger that this force is. The direction of this force is always towards the center of Earth.

15 FORCE THAT EARTH EXERTS ON AN OBJECT Earth exerts a force on each and every single object on Earth. This force is exerted towards the center of Earth and it is what keeps our feet on the ground. Say NO to gravity Earth is an object, ONLY OBJECTS can exert forces on other objects. GRAVITY IS NOT AN OBJECT..... Say NO to GRAVITY!!!

16 FORCE THAT EARTH EXERTS ON AN OBJECT Represented by: F E on O F EonO is a kind of Field force exerted inside the gravitational field of Earth. F EonO is ALWAYS present whether the object is moving, falling, resting on the ground or being lifted. Gravitational Force is an acceptable name for F EonO The direction of F EonO is always down (toward the center).

17 FORCE DIAGRAM

18 FORCE DIAGRAM A FORCE DIAGRAM is a graphical representation used by physicists to analyze the forces exerted on objects by other objects.

19 Drawing force diagrams 1. Sketch the situation. 2. Circle the system. 3. Identify external interactions. 4. Place a dot at the side of the sketch representing the system object. 5. Draw force arrows to represent the external interactions. 6. Label the forces with a subscript containing two elements.

20 An SUV is parked at Madison High School parking lot 1. Sketch 3. Force Diagram (FD) F GonS Earth 2. Objects Interacting with the system SUV I am 4x4 Earth Ground (field force) (contact force) F EonS

21 EQUILIBRIUM The SUV is in equilibrium: Its acceleration is a = 0 m s 2 Its velocity is v = 0 m s (velocity is constant) F x = 0 Newtons F y = 0 Newtons

22 WHITEBOARD Bowling ball on the ground 1. Sketch 3. Force Diagram (FD) F GonBB Earth 2. Objects Interacting with the system Bowling Ball Earth Ground (field force) (contact force) F EonBB

23 WHITEBOARD Bowling ball on a table 1. Sketch 3. Force Diagram (FD) F TonBB Earth 2. Objects Interacting with the system Bowling Ball Earth Table F EonBB

24 WHITEBOARD Bowling ball hanging from a string 1. Sketch 3. Force Diagram (FD) F SonBB Earth 2. Objects Interacting with the system Bowling Ball Earth String F EonBB

25 WHITEBOARD Bowling ball hanging from a spring 1. Sketch 3. Force Diagram (FD) F SonBB Earth 2. Objects Interacting with the system Bowling Ball Earth Spring F EonBB

26 WHITEBOARD Bowling ball with a book on top 1. Sketch 3. Force Diagram (FD) F GonBB 2. Objects Interacting with the system Earth Book Ground Earth F BonBB Bowling Ball F EonBB

27 WHITEBOARD Bowling ball is rolling at constant velocity across a frictionless frozen lake 1. Sketch 3. Force Diagram (FD) F GonBB Earth 2. Objects Interacting with the system Bowling Ball Earth Ground F EonBB

28 WHITEBOARD Bowling ball in free fall 1. Sketch 3. Force Diagram (FD) Earth 2. Objects Interacting with the system Bowling Ball Earth F EonBB

29 GRAVITATIONAL FORCE

30 F E on O [N] GRAPH MADE IN LAB F E on O vs mass m [kg]

31 Graph: F Earth on O vs mass Shape of the graph: Diagonal straight line SLOPE Symbol : Name: Units: Meaning: g Gravitational constant m s 2 Acceleration caused by the Force exerted by Earth

32 F E on O [N] Slope = g Gravitational constant m [kg]

33 MATH MODEL MATH CLASS y = m x + b PHYSICS F EonO = g m

34 F E on O [N] Slope = g Gravitational constant m [kg]

35 I exert a force on any object on my surface!!! F EonO = g m

36 F EonO = g m Force exerted by Earth [N] Newtons Gravitational constant [m/s 2 ] Meters per second squared Mass [kg] kilograms

37 Force exerted by Earth [N] Newtons F EonO g m Gravitational constant [m/s 2 ] Meters per second squared Mass [kg] kilograms

38 F EonO = g m metric System [N] Newtons [m/s 2 ] Meters per second squared [kg] kilograms imperial System [lb] Pound [ft/s 2 ] feet per second squared [slug] slug

39 Mass m Mass is a property of subatomic particles (electron, proton, neutron). It determines the amount of matter that makes up an object. Your mass will be the same, no matter if you are on Earth, the Moon, Saturn, Jupiter, etc.

40 gravitational constant g Constant acceleration caused by the gravitational force. It depends on the size and mass of the planet. g Earth = 9.8 m s 2 g Moon = 1.6 m s 2 g Jupiter = m s 2

41 Gravitational Force F EonO Field force exerted between two objects that have mass. The force depends on the distance separation between the two objects and their masses. Objects only feel the force exerted by another big (planet size) objects. F JonO = g Jupiter m F EonO = g Earth m

42 FORCE EARTH vs. WEIGHT GRAVITATIONAL FORCE (on Earth) Force that Earth exerts on an object (F EonO ) Vector: magnitude and direction Weight (F W ) Scalar: only magnitude (No Direction)

43 FORCE OF EARTH vs. WEIGHT After working out, Mr. L 2 Stands on the scale to measure his weight. F W = 980 N Weight is a scalar: only magnitude (No Direction) F EonL 2 = 980 N F EonL2 is a Vector: magnitude and direction

44 GRAVITATIONAL FORCE F E on o = g m F E on o g m

45 WHITEBOARD Pumpkin on a table 1. Sketch 3. Force Diagram (FD) F TonP Earth 2. Objects Interacting with the system Pumpkin Earth Table F EonP

46 The mass of the pumpkin is 8 kg m = g = 8 kg m/s 2 F EonP = F TonP = F W = N N 78.4 N

47 The same pumpkin is taken to the moon m = g = 8 kg m/s 2 F MonP = F TonP = F W = N N 12.8 N

48 The weight of the same pumpkin in Pluto is 4.96 N m = 8 kg g = m/s 2 F PonP = F TonP = F W = N N 4.96 N

49 The mass of the new iphone X is kg m = g = kg m/s 2 F EoniP = F ToniP = F W = N N 1.71 N

50 The same iphone X is taken to the moon m = g = kg m/s 2 F MoniP = F ToniP = F W = N N 0.28 N

51 The weight of the same iphone X in Venus is 1.54 N m = g = kg m/s 2 F VoniP = F ToniP = F W = N N 1.54 N

52 The weight of a girl on Saturn is N. The gravitational constant of Saturn is m/s 2 EARTH MOON SATURN m 60 kg 60 kg 60 kg g m/s m/s m/s 2 F EonG N - 96 N N F GonG N + 96 N N F W 588 N 96 N N

53 HOW CAN YOU LOSE WEIGHT WITHOUT LOSING MASS? Go to a planet (or size like object) with lower gravitational constant than Earth g = -9.8 m/s 2. Moon: g = -1.6 m/s 2 Pluto: g = m/s 2 Venus: g = m/s 2

54 HOW CAN YOU LOSE MASS? (without a diet) When a person loses a limb.

55 FORCE THAT A SPRING EXERTS ON AN OBJECT

56 Initial length of spring L Elongation Final length of spring

57 Earth FORCE DIAGRAM Earth (down) Spring (up) F SonO object F EonO

58 F S on O [N] L [m]

59 MATH MODEL MATH CLASS y = m x + b PHYSICS F SonO = k L

60 F S on O [N] Slope = k Spring constant L [m]

61 Graph: F Spring on O vs Elongation Shape of the graph: Diagonal straight line SLOPE Symbol : Name: Units: Meaning: k Spring constant N m Force (in Newtons) required to stretch/compress the spring (1 meter)

62 A diagonal straight line can have ONLY one slope! Slope = Spring constant Springs can have ONLY one spring constant! Spring constant tells you how easy/difficult it is to stretch/compress a spring

63 F SonO = k L Force exerted by a Spring [N] Newtons Spring constant [N/m] Newtons per Meters Elongation [m] meters

64 Force exerted by a Spring [N] Newtons F SonO k L Spring constant [N/m] Newtons per Meters Elongation [m] meters

65 FORCE DIAGRAM F SonO =k L Earth (down) Spring (up) Object Earth F EonO = g m

66 Robert Hooke ( ) HOOKE S LAW F SonO =k L Hooke's law: The Force (F SonO ) needed to extend or compress a spring by some distance is proportional to its elongation ( L) and the spring constant (k).

67 Robert Hooke ( ) Iris diaphragm The universal joint The respirator. He ground lenses and assembled his own microscopes for studying life on a small scale. Coined the word "cell" for the basic structure of biological tissues. Greatly improved clock design with the balance spring. Understanding of how fossils are formed by the intrusion of "petrifying water." Explained the geological uplift as the mechanism whereby fossil sea life came to rest on mountaintops. Developed the correct theory of combustion. Helped Robert Boyle in working out the relationship of gas pressure to volume (Boyle s Law). Great draughtsman and designer of experiments. Played a leading role in the rebuilding of London after the Great Fire of 1666.

68 FORCE EXERTED BY A SPRING F S on o = k L F S on o k L

69 A PHYSICS CLASS USES A SPRING WITH k = 24.5 N/m

70 PREDICTION: The mass at the bottom of the spring is 1.5 kg. Complete all physical quantities

71 - 9.8 m/s 2 g = 1.5 kg Mass = N F EonO = F EonO = g m 24.5 N/m k = 0.6 m L = L = F SonO k N (Equilibrium) F SonO =

72 SAME SPRING USED IN CLASS PREDICTION: The Force exerted by the spring is N. Complete all physical quantities

73 - 9.8 m/s 2 g = 3.5 kg Mass = m = F EonO g N (Equilibrium) F EonO = 24.5 N/m k = 1.4 m L = L = F SonO k N F SonO =

74 SAME SPRING USED IN CLASS PREDICTION: The elongation of the spring is 1.12 m. Complete all physical quantities

75 - 9.8 m/s 2 g = 2.8 kg Mass = m = F EonO g N F EonO = 24.5 N/m k = 1.12 m L = (Equilibrium) N F SonO = F SonO = k L

76 SAME SPRING USED IN CLASS AFTERNOON CLASS k = N/m

77 SAME SPRING USED IN CLASS PREDICTION: The mass at the bottom of the spring is 1.5 kg. Complete all physical quantities

78 - 9.8 m/s 2 g = 1.5 kg Mass = N F EonO = F EonO = g m N/m k = 0.52 m L = L = F SonO k N (Equilibrium) F SonO =

79 SAME SPRING USED IN CLASS PREDICTION: The Force exerted by the spring is N. Complete all physical quantities

80 - 9.8 m/s 2 g = 3.5 kg Mass = m = F EonO g N (Equilibrium) F EonO = N/m k = 1.22 m L = L = F SonO k N F SonO =

81 SAME SPRING USED IN CLASS PREDICTION: The elongation of the spring is 1.12 m. Complete all physical quantities

82 - 9.8 m/s 2 g = 3.22 kg Mass = m = F EonO g N F EonO = N/m k = 1.12 m L = (Equilibrium) N F SonO = F SonO = k L

83 EXAMPLE #1 Alex places a 1.50 kg salmon on the scale at the fish market. L The spring of the scale has a spring constant 175 N/m. Identify objects that interact with the fish. Draw a force diagram. Find all Physical quantities.

84 Earth FORCE DIAGRAM L Earth (down) Spring (up) F SonF Fish F EonF

85 - 9.8 m/s 2 g = 1.5 kg Mass = F EonF = g m F EonF = 9.8 m 1.5 kg s2 F EonF = 14.7 N N F EonF = 175 N/m k = m L = N (Equilibrium) F SonF = L = L = F SonF k N 175 N/m L = m

86 EXAMPLE #2 A 0.8 kg mass hangs at the bottom of a spring. The spring elongates m. L Identify all objects that interact with the mass. Draw a force diagram. Find all Physical quantities.

87 Earth FORCE DIAGRAM Earth (down) F SonM Spring (up) L Mass F EonM

88 - 9.8 m/s 2 g = 0.8 kg Mass = F EonM = g m F EonM = 9.8 m 0.8 kg s2 F EonM = 7.84 N N F Eonm = 140 N/m k = m L = N (Equilibrium) F Sonm = k = F SonM L k = 7.84 N m k = 140 N/m

89 EXAMPLE #3 Mr. Hooke measures the weight of his luggage before going away for the holiday. Mr. Hooke does some online research and he finds the spring constant to be 6440 N/m, he then measures m as the elongation of the spring Identify all objects that interact with the luggage. Draw a force diagram. Find all Physical quantities.

90 Earth FORCE DIAGRAM Earth (down) Spring (up) F SonL Luggage F EonL

91 - 9.8 m/s 2 g = 23 kg Mass = F SonL = k L F SonL = 6440 N m m F SonL = N N F Eonm = 6440 N/m k = m L = N (Equilibrium) F Sonm = m = m = F EonL g N 9.8 m s 2 m = 23 kg

92 FORCE THAT A SURFACE EXERTS ON AN OBJECT

93 Box SLIDES with constant velocity. Why is it important to know that? EQUILIBRIUM!!!

94 Surfaces exert TWO forces on an object: Normal Force: Perpendicular to the surface (90 ). Force of kinetic friction: Parallel to the surface (0 ). Direction opposite to the direction of the motion of the object. Object must be sliding.

95 Earth (down) FORCE DIAGRAM Finger (right) F NonB Surface - Normal (up) Surface kinetic friction (left) box F KFonB F FonB Earth F EonB

96 45 FKFonO FNonO

97 MATH MODEL MATH CLASS y = m x + b PHYSICS F KFonO = k F NonO

98 45 FKFonO Slope = k Coefficient of kinetic friction FNonO

99 Graph: F KF on O vs F N on O Shape of the graph: Diagonal straight line SLOPE Symbol : Name: Units: Meaning: k Coefficient of kinetic friction No units Ratio between the forces exerted by a surface: F KF on O and F N on O.

100 FORCE EXERTED BY A SURFACE F KF on o = μ K F N on o F KF on o k FN on o

101 The coefficient of kinetic friction ( k ) depends on TWO surfaces. Slope = coefficient of kinetic friction Coefficient of kinetic friction stays constant between TWO surfaces

102 FORCE OF KINETIC FRICTION (FKFonO = µk FNonO ) Depends on the mass: If you change mass, then FEonO changes If you change FEonO, then FNonO changes If you change FNonO, then FKFonO changes Depends on the surfaces: If you change any of the surfaces, then µk changes. If you change µk, then FKFonO changes

103 COEFFICIENT OF KINETIC FRICTION ( k ) Depends on two surfaces. It is NOT a property of ONE surface! µk is a positive number between 0 and 1 µk is NOT the force of kinetic friction. It does not have units

104 Force of kinetic friction: Created at the microscopic level, due to imperfections on BOTH surfaces.

105 F KFonO = μ k F NonO Force of kinetic friction: Force exerted by a surface, parallel to the surface [N] Newtons Coefficient of kinetic friction: Ratio of two forces: Force of kinetic friction and Normal force. [No units] Normal Force: Force exerted by a surface, perpendicular to the surface [N] Newtons

106 Force of kinetic friction [N] Newtons F KFonO k F NonO Coefficient of kinetic friction [No units] Normal Force [N] Newtons

107 FORCE DIAGRAM F NonB box F KFonB = μf kkfonb F NonB F FonB Earth F EonB F EonB = g m

108 Force of KINETIC friction: Object must be sliding Force of STATIC friction: Object is rotating or walking

109 Look a sliding box!

110 A moose stands on top of the crate. What will happen to these quantities? Force Earth Increases Normal Force Increases Force Finger Increases Force kinetic Friction Increases Coefficient of kinetic friction Stays the same

111 EXAMPLE #1 Jackie exerts a horizontal force on the 400 kg sled. The coefficient of kinetic friction between the sled and the ground is The sled moves with constant velocity. Identify objects that interact with the crate. Draw a force diagram. Find all Physical quantities.

112 Earth (down) FORCE DIAGRAM Jackie (right) Surface Normal (up) Surface friction (left) F NonS sled F KFonS F JonS F EonS Earth

113 g = m/s 2 F EonS = g m 400 kg Mass = N F EonS = F NonS = N 0.21 k = F FKonS = N N F JonS = F EonS = 9.8 m 400 kg s2 F EonS = 3920 N (Equilibrium) (Equilibrium) F KFonS = μ K F NonS F KFonS = N (negative because force is to the left) F KFonS = N

114 EXAMPLE #2 Oscar exerts a force of N pulling on a rope attached to a stubborn donkey. The mass of an average donkey is 500 kg. The force exerted by Oscar is completely horizontal and the donkey slides with constant velocity. Identify objects that interact with the donkey. Draw a force diagram. Find all Physical quantities.

115 FORCE DIAGRAM Earth (down) Rope (right) Surface Normal (up) Surface friction (left) F NonD donkey F KFonD F RonD F EonD Earth

116 g = m/s 2 F EonD = g m 500 kg Mass = N F EonD = F EonD = 9.8 m 500 kg s2 F EonD = 4900 N F NonD = N 0.35 k = F KFonD = N (Equilibrium) N F RonD = (Equilibrium) μ K = F FKonD F NnD μ K = 1715 N 4900 N μ K = 0.35

117 EXAMPLE #3 Identify objects that interact with the crate. Draw a force diagram. Find all Physical quantities. Exerting a force of N, Mr. Largo moves a crate filled with lab equipment with constant speed along G hall. The coefficient of kinetic friction between the plastic crate and the floor is 0.18.

118 Earth Earth (down) Mr. L 2 (left) FORCE DIAGRAM Surface Normal (up) Surface friction (right) F NonC crate F MLonC F KFonC F EonC

119 g = m/s kg Mass = N (Equilibrium) F EonC = F NonC = N k = 0.18 F FKonC = N (Equilibrium) N F MLonC = F NonC = F FKonC μ k F EonC = N 0.18 F NonC = N m = F EonC g 1960 N m = 9.8 m s 2 m = 200 kg

120 FORCE EXERTED BY A STRING (TENSION FORCE)

121 WHITEBOARD A rescue hero hangs at the bottom of the string. Sketch Identify objects that interact with the Rescue hero. Draw a force diagram. Find the magnitude and direction of all forces exerted on the rescue hero. mrh = kg

122 Earth (down) String (up) F SonRH Rescue Hero Earth F EonRH

123 Find the force that Earth exerts on the Rescue Hero F EonRH = g m F EonRH = 9.8 m kg s2 F EonRH = N F SonRH = N (equilibrium)

124 Force Exerted by Earth: F EonRH = N Force Exerted by String (Tension Force): F SonRH = N Weight of Rescue Hero (No direction): F W = N

125 TENSION FORCES The Force exerted by a rope, cable, string, thread, etc. The direction of the tension force is always along the rope, cable, string, thread, etc.

126 TENSION FORCES The length of the rope does not determine the magnitude ofthe tension force. It is possible to have a very short rope exerting a large tension force, or a very long rope exerting a small tension force. In a force diagram the length of the arrow corresponds to the magnitude of the force, not the length of the physical rope.

127 EQUILIBRIUM All forces exerted on the object are balanced: F X = 0 N F Y = 0 N Objects moves with constant velocity (including 0 m/s). The acceleration is 0 m/s 2 The velocity is constant The net Force is 0 N

128 WHITEBOARD A 5 kg bowling ball hangs off a rope. Identify objects that interact with the bowling ball. Draw a force diagram. Find the magnitude and direction of all forces exerted on the bowling ball.

129 Earth (down) String (up) F SonBB = +49 N bowling ball F EonBB = -49 N

130 Find the forces exerted on the bowling ball F EonBB = g m F EonBB = 9.8 m s 2 5 kg F EonBB = 49 N F SonBB = +49 N (equilibrium)

131 TWO STRINGS! A rescue hero hangs at the bottom of TWO strings. Draw a force diagram.

132 m RH = kg = 55

133 Earth (down) String 1 (left) String 2 (at an angle) F S2onRH F S1onRH Rescue Hero F EonRH

134 FORCE EXERTED BY EARTH ON RESCUE HERO F EonRH = g m F EonRH = 9.8 m kg s2 F EonRH = N

135 X-COMPONENT Y-COMPONENT RESULTANT F S1 on RH 0 N (HORIZONTAL) F S2 on RH N F E on RH F (sum of forces) 0 N (EARTH DOES NOT PUSH YOU SIDEWAYS) 0 N (EQUILIBRIUM) F EonRH=gm N 0 N (EQUILIBRIUM)

136 F S2onRH = 55 (F S2onRH ) x (F S2onRH) y = N From vectors unit: R x = R cosθ R y = R sinθ

137 R y = R sinθ = R sin sin55 = R sin N = R

138 R x = R cosθ R x = cos55 R x = N

139 X-COMPONENT Y-COMPONENT RESULTANT F S1 on RH N 0 N (HORIZONTAL) F S2 on RH N N F E on RH F (sum of forces) 0 N (EARTH DOES NOT PUSH YOU SIDEWAYS) 0 N (EQUILIBRIUM) F EonRH=gm N 0 N (EQUILIBRIUM)

140 X-COMPONENT Y-COMPONENT RESULTANT (Pythagorean Theorem) F S1 on RH N 0 N (HORIZONTAL) N F S2 on RH N N N F E on RH F 0 N (EARTH DOES NOT PUSH YOU SIDEWAYS) 0 N (EQUILIBRIUM) F EonRH=gm N N 0 N (EQUILIBRIUM)

141 WHITEBOARD A rescue hero hangs at the bottom of TWO strings. Draw a force diagram. Find the magnitude and direction of all forces exerted on the rescue hero. Predict the mass of the hero

142

143 F T2 on H F T1 on H Hero F E on H

144 (F T1 on H ) y F T1 on H = 3.5 N (F T1 on H ) y = F T1 on H sin F T1 on H ) y = 3.5 sin 20 (F T1 on H ) y = N = 20 (F T1 on H ) x Hero (F T1 on H ) x = F T1 on H cos (F T1 on H ) x = 3.5 cos 20 (F T1 on H ) x = N Negative: Force is to the left

145 F T2 on H = 7.0 N (F T2 on H ) y (F T2 on H ) y = F T2 on H sin (F T2 on H ) y = 7.0 sin 62 (F T2 on H ) y = N ero = 62 (F T2 on H ) x (F T2 on H ) x = F T2 on H cos (F T2 on H ) x = 7.0 cos 62 (F T2 on H ) x = N

146 F T1onO F T2onO F EonO F x y RESULT N N 3.5 N N N 7.0 N 0 N N 7.38 N 0 N 0 N Pythagorean Theorem Equilibrium

147 m = F EonRH g m = 7.38 N 9.8 m s 2 m = 0.75 kg