Forces I. Newtons Laws

Transcription

1 Forces I Newtons Laws

2 Kinematics The study of how objects move

3 Dynamics The study of why objects move

4 Newton s Laws and Forces What is force? What are they?

5 Force A push or a pull Symbol is F Unit is N (Newton) One newton is the force necessary to cause a one kilogram mass to accelerate at the rate of 1m/s 2 1N=1 kg m/s 2

6 FORCE A force is any influence that can change the velocity of a body. Forces can act either through the physical contact of two objects (contact forces: push or pull) or at a distance (field forces: magnetic force, gravitational force). Contact Forces Frictional Force Tensional Force Normal Force Air Resistance Force Applied Force Spring Force Action-at-a-Distance Forces Gravitational Force Electrical Force Magnetic Force

7 What do we mean by balanced and unbalanced forces? Balanced forces are EQUAL and OPPOSITE in direction Unbalanced forces have a greater force in one direction The forces on the book are unbalanced

8 If all the forces are balanced, we say the object is in EQUILIBRIUM

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10 A net force is the vector sum of the forces acting on an object

11 What are the net forces? F NET = Σ forces 0 N These are free-body diagrams

12 Free Body Diagram Practice 1. A book is at rest on a tabletop. Diagram the forces acting on the book. 2. A picture is hanging from the ceiling by two ropes. Diagram the forces acting on the picture. He was not injured in the experiment. 3. An egg is free-falling from a nest in a tree. Neglect air resistance. Diagram the forces acting on the egg as it is falling. 4. A rightward force is applied to a book in order to move it across a desk with a rightward acceleration. Consider frictional forces. Neglect air resistance. Diagram the forces acting on the book. 5. A rightward force is applied to a book in order to move it across a desk at constant velocity. Consider frictional forces. Neglect air resistance. Diagram the forces acting on the book. 6. A college student rests a backpack upon his shoulder. The pack is suspended motionless by one strap from one shoulder. Diagram the vertical forces acting on the backpack. 7. A skydiver is descending with a constant velocity. Consider air resistance. Diagram the forces acting upon the skydiver.

13 Newton Three Laws of Motion His laws explain why objects move (or don t move) as they do

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15 Newton's First Law of Motion (Law of Inertia) or [Law of Balanced Forces] Newton's First Law of Motion says: an object at rest stays at rest and an object in motion stays in motion unless the object is acted on by an unbalanced force.

16 Things keep doing what they are doing! Constant Velocity It can be Zero or it can be greater than zero..it just can t change!!!!!!

17 What is required to change the An unbalanced force constant velocity?

18 Diagram from the Physics Classroom

19 Diagram from the Physics Classroom

20 Other Examples to Consider: Blood rushes from your head to your feet while quickly stopping when riding on a descending elevator. The head of a hammer can be tightened onto the wooden handle by banging the bottom of the handle against a hard surface.

21 Inertia and Mass Inertia: the resistance an object has to a change in its state of motion The more mass the more inertia As mass increases the more an object resists change in its motion Mass (kg) is the measurement of Inertia

22 Newton s First Law

23 Remember an object in constant motion has balanced forces working on it. This is the first law described mathmatically ΣF = 0 Where Σ (sigma) is sum F is force

24 FREE BODY DIAGRAM:Identify the missing force marked with a (?) in the diagram below. The car is moving at a constant velocity of 25 m/s eastward. Air Friction 20N Car Force pushing car? 15N Ground Friction Ground Friction

25 FREE BODY DIAGRAM:Identify the missing force marked with a (?) in the diagram below. The car is accelerating eastward with a net force of 50 N. Air Friction 20N Car Force pushing car? 15N Ground Friction Ground Friction

26 Assign Fun With Forces Forces WS I A Work an example on top and on bottom

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28 . Newton's Second Law of Motion (Law of Acceleration) [Law of Unbalanced Forces]

29 SECOND LAW OF MOTION According to Newton's Second Law of Motion, the net force acting on a body equals the product of the mass and the acceleration of the body. The direction of the force is the same as that of the acceleration. In equation form: F = ma F = force applied to an object [N = Newtons = kg m/s 2 ] m = object s mass [kg] a = acceleration [m/s 2 ]

30 What is the relationship between Force, Mass, and Acceleration? According to Newton's second law of motion, if mass is constant what is the relationship between force and acceleration? Direct According to Newton's second law of motion, if force is constant what is the relationship between mass and acceleration? Inverse

31 What happens to acceleration of an object if force is doubled? (mass is constant) Force Tripled? Cut in half? What happens to acceleration if force is constant but the mass is doubled? Mass Tripled? Mass cut in half?

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33 2 nd Law: F = m a EX A A 2000 N net force acts horizontally to the right on a 50 kg object. Draw the free body diagram. Calculate the acceleration. F m a 2000N 50kg a a 40m / s 2

34 Are weight and mass the same thing?

35 WEIGHT (of a body or object) aka Force of Weight The gravitational force with which the Earth attracts the body. Causes it to accelerated downward with the acceleration of gravity g. Vector quantity Measured in Newtons Varies with its location near the Earth (or other astronomical body) Force Weight = mass x gravity F W = m x g Newton = kg x m/s 2 1 N = 1 kgm/s 2

36 Mass A scalar quantity Same everywhere in the universe. It doesn t change Measured in kg

37 EX B: What is the weight of a person whose mass at sea level is 72 kg? Draw the free body diagram. g = 9.80 m/s 2 m = 72 kg F = mg F w = mg F w = (72 kg)(9.80 m/s 2 ) F w = 706 kg m/s 2 F w = 706 N

38 Ex C: What is the mass of a box (in grams) if it weighs 625N? Answer to the nearest tenth. Draw the free body diagram. F = ma m = F/a m = 625N/9.8 m/s2 m = 63.8 kg m = grams

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40 What is the third law of motion For every action force there is an equal and opposite reaction force Two different objects and two different forces (equal but opposite) Forces always occur in pairs Also known as The Law of Force Pairs Are the effects of the force the same? NO They are not always acting on the same amount of mass Examples: Bug hitting windshield, recoil of shot gun, pushing off of on a raft

41 Third Law Mathmatically F = F Which can be written as: m 1 a 1 = m 2 a 2

42 Forces occurs in pairs

43 What forces cause the swimmer to move forward? Her push on the wall? Or The wall s push on her?

44 Ex D A 2500 kg car hits a kg bug with a force of 500N. According to the third law of motion, the bug hits the car with how many Newton's of force? Why isn t the car damaged? Calculate the acceleration of the bug due to the force. Calculate the acceleration of the car due to the force.

45 A 2500 kg car hits a kg bug with a force of 500N. According to the third law of motion, the bug hits the car with how many Newton's of force? Why isn t the car damaged? Calculate the acceleration of the bug due to the force. Calculate the acceleration of the car due to the force. acceleration of the bug due to impact: a = F/m a = 500N/.001 kg = 500,000 m/s 2! acceleration of the car due to impact : a = F/m a = 500N/2500kg = 0.2 m/s 2

46 Third Law of Motion The girl pushes the boy with 10N of force. How much does the boy push back with?

47 Force and Kinematics

48 Net Force (F net ) F net in simple situations (only 1 force implied): F net = ma F net and acceleration are always in the same direction How is net force on an object determined with multiple? It is the sum of all forces F net = Σ F Remember force is a vector, therefore it has direction (+,-, N,S etc)

49 Net force ( F) = ma The second law of motion is the key to understanding the behavior of moving bodies since it links cause (force) and effect (acceleration) in a definite way.

50 Example E A 2000 N net force (horizontally to the right) acts on a 50 kg object. Draw the free body diagram. Calculate the acceleration. If the object started from rest, how far would it move in 5 seconds? F m a 2000N 50kg a a 40m / s 2

51 Example E A 2000 N net force (horizontally to the right) acts on a 50 kg object. Draw the free body diagram. Calculate the acceleration. If the object started from rest, how far would it move in 5 seconds? d = v i t +.5at 2 d = (.5)(40 m/s 2 )(5sec) 2 d = 500m Could we determine the velocity at 5 seconds? How? a = (v f v i )/t v f = v i + at v f = 0 m/s + (40 m/s 2 )(5sec) = 200 m/s

52 Ex F A 1000 kg car goes from 10 to 20 m/s in 5 s. What force is acting on it? m = 1000 kg v i = 10 m/s v f = 20 m/s t = 5 s a = (v f v i )/t a = (20m/s-10 m/s) / 5s a = 2 m/s 2 F = ma F = (1000 kg) (2 m/s 2 ) = 2000 N

53 Ex G-I as time permits

54 Ex G A 60-g tennis ball approaches a racket at 15 m/s, is in contact with the racket for s, and then rebounds at 20 m/s. Find the average force exerted by the racket. How can we draw this?

55 Ex G A 60-g tennis ball approaches a racket at 15 m/s, is in contact with the racket for s, and then rebounds at 20 m/s. Find the average force exerted by the racket. m = 0.06 kg v i = 15 m/s t = s v f = - 20 m/s a = (v f v i )/t a = (-20m/s-15 m/s) / 0.005s a = m/s 2 F = ma F = (0.06 kg) (-7000 m/s 2 ) = -420 N

56 Ex H The brakes of a 1000-kg car exert 3000 N. a. How long will it take the car to come to a stop from a velocity of 30 m/s? How can we draw this?

57 Ex H The brakes of a 1000-kg car exert 3000 N. a. How long will it take the car to come to a stop from a velocity of 30 m/s? m = 1000 kg F = N v i = 30 m/s v f = 0 m/s t a v F m f a 3000 = - 3 m/s v i = 10 s b. How far will the car travel during this time? d = v i t+.5at 2 = 30(10)+.5 (-3)(10) 2 = 150 m If you got 450m you did not use -3 m/s 2

58 Ex H The brakes of a 1000-kg car exert 3000 N. a. How long will it take the car to come to a stop from a velocity of 30 m/s? m = 1000 kg F = N v i = 30 m/s v f = 0 m/s t a v F m f a 3000 = - 3 m/s v i = 10 s b. How far will the car travel during this time? d = v i t+.5at 2 = 30(10)+.5 (-3)(10) 2 = 150 m If you got 450m you did not use -3 m/s 2

59 Ex I A net horizontal force of 4000 N is applied to a car at rest whose weight is 10,000 N. What will the car's speed be after 8 s? F A = 4000 N F w = 10,000 N t = 8s m F a w g = kg a F m 4000 = 3.92 m/s v f = v i + at = (8) = m/s

60 Forces Part II Friction

61 Friction Is a force that opposes motion.

62 When surfaces are pressed together, we can identify four forces

63 Friction Force: F F Force opposing motion Measured in Newtons (N)

64 Applied Force: F A The push or pull applied to the object Measured in Newtons (N)

65 F w Force of Weight or Gravity (Mass in kg) (Acceleration due to Gravity) Kg x 9.8 m/s 2 Measured in Newtons (N)

66 Normal Force: F N Usually a 3 rd Law reaction to gravity, that is equal and opposite of Force of Weight (F w ) Perpendicular to the surface. Measured in Newtons (N) F N is NOT F NET Don t confuse them just because they begin with N!

67 For Examples J-M we will look at the same object under different conditions Ex J: F NET No F F Ex K: F F Ex L: F NET Ex M: F NET No F NET & F F & F F

68 A 50kg object is moving horizontally at a constant velocity. Is there acceleration? Is there a net force? F N F F F A F W

69 Ex J No Friction F net = ma A 50 kg object experiences an applied force of 400 N, what is the F net? (Disregard friction) What is the acceleration?: F net = F A F net = Net Force, results in acceleration a = F net /m = (400N )/ (50 kg) = 8 m/s 2

70 Example J F N F A F W

71 Ex K Friction and Constant velocity. Is there a net force? F net = 0 F A + - F F = ma or F A + - F F = F net A 50 kg object moves at a constant velocity when acted upon by an applied force of 400 N. What is the F F? What is the F net? What is the acceleration?: F F = F A F net = F A + - F F = 0 N If F net = 0 then acceleration =? 0

72 How does friction affect net force? F net = ma ma = F A + F F Where: F net = Net Force F F = Friction Force F A = Applied Force You are actually subtracting F F from F A, since F f is in the opposite direction Friction will reduce the net force

73 Example K Friction and constant velocity F N F F F A F W

74 Ex K What would have to change to have acceleration?

75 Do all surfaces provide the same amount of friction? How is this described? F N F F F A F W

76 F F = Force of Friction What does it depend on? Depends on the surface of the materials. Depends on how tightly the surfaces are pressed together.

77 Coefficient of Friction The coefficient of friction is a measure of how difficult it is to slide a material of one kind over another; the coefficient of friction applies to a pair of materials, and not simply to one object by itself Coefficient of Friction Reference Table - Engineer's Handbook

78 When Surfaces are Pressed Together Coefficent of Friction µ can be calculated It is a ratio of F F and F N F F F N F F F N

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80 Coefficient of friction What is it equal to? What is the unit for Coefficient of friction? If Coefficient of friction is small, what does that mean about the F F? If Coefficient of friction is large, what does that mean about the F F? The higher the coefficient of friction, the more difficult to slide

81 What forces cause the swimmer to move forward? Her push on the wall? Or The wall s push on her?

82 I call these FAWN problems

83 Example L A 400 N force is applied horizontally to a 50kg object. Calculate the acceleration of the object if = 0.3. F N F F F A F W

84 Example Cont d. F A = 400 N F g = m g (50kg)(9.8m/s 2 ) F g = 490 N F N = 490 N also F F = (0.3)(490N) = 147 N

85 Example cont d. F N = 490N F F = 147N F A = 400N F W = 490N

86 To solve for Acceleration must calculate Net Force F NET = ma = F A + - F F F NET = ma= 400N 147N ma = 253N

87 Now use Net Force and mass of object in F=ma formula a = F/m a = 253N/50kg a =5.06 m/s 2

88 Ex M Putting it all together A 50 kg object accelerates horizontally at 0.2 m/s 2 from rest for 5 seconds. If the coefficient of friction is 0.01, what is the F net? What is the F F? What is the F A?How far does it travel in 5 seconds?: F net = (50 kg) (0.2 m/s 2 ) = 10 N F F = μ F N F N = (50kg)(9.8 m/s 2 ) = 490 N F F = (0.01) (490 N) =4.9N F A = F net - F F F A = 10 N N = 14.9 N

89 Ex M A 50 kg object accelerates horizontally at 0.2 m/s 2 from rest for 5 seconds. If the coefficient of friction is 0.01, what is the F net? What is the F F? What is the F A?How far does it travel in 5 seconds?: F net = (50 kg) (0.2 m/s 2 ) = 10 N Set up DVVAT d =? v i = 0 m/s v f =? a =0.2 m/s 2 d = (.5)(.2)(5 )2 d = 2.5m

90 What does a small coefficient of friction imply about FF?

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92 TENSION aka F T is the magnitude of the pulling force exerted by a string, cable, chain, or similar object on another object. It is the opposite of compression. It is a response force That is to say, if one pulls on the rope, the rope fights back by resisting being stretched Ropes, strings, and cables can only pull. They cannot push because they bend. is measured in newtons is always measured parallel to the string on which it applies.

93 What does the rope provide? A lift (vertical force) and a pull (horizontal force) If there was no angle, would there be any vertical force? No If the angle was at 90, how would that affect the force components? Force would only be in the vertical plane How would you calculate the horizontal and vertical force components if the angle of the rope with the floor was 57 and the Force of tension (F A ) in the rope was 400 N?

94 Ex N. This crate is be pulled with a rope across a friction based horizontal surface at a constant velocity. The rope exerts a tension of 400 N at an angle of 57. What is the coefficient of friction? 50 kg 57

95 A box is pulled into motion with a rope across a horizontal surface. The rope makes an angle of 57 to the floor. The Force of tension (F A ) in the rope is 400N 400 N F AY 50 kg 57º F AX A. Determine F AX or F Horiz = cos (57) (400N) = N B. Determine F AY or F Vert = sin (57) (400N) = N

96 I work in a circle Determine F W Determine F AX Determine F AY Determine F N The F AY supplies part of the upward force. The total upward force is F AY + F N and together these are equal but opposite the F W. F N = F W F AY Determine F F Determine coefficient of friction

97 F AY N 50 kg N F AX N Determine F w = (50 kg)(9.8 m/s 2 ) = 490 N Determine F N = F W - F AY = 490 N N = N

98 F AY N 50 kg N F AX N Determine F F F F = F ax (Constant Velocity) F F = N Determine µ= F F / F FN = N/ N = 1.41

99 Force III Statics

100 TENSION aka F T is the magnitude of the pulling force exerted by a string, cable, chain, or similar object on another object. Ropes, strings, and cables can only pull. They cannot push because they bend. is measured in newtons is always measured parallel to the string on which it applies.

101 Example O: What is the tension in the cable supporting the sign? Mass of sign 5.86 kg

102 Example P: What is the tension in each cable supporting the sign? Mass of sign 5.86 kg

103 Example Q: What is the tension in each cable supporting the sign? Angle A and Angle B = 35 o Mass of sign 5.86 kg

104 Example Q What do we know? Object not moving, therefore F W = F T F W = 57.4 N F T is shared between the two cables. F T represents the vertical component Do you see triangles? Angle A and Angle B = 35 o Mass of sign 5.86 kg

105 Example Q Solution Object not moving, therefore F W = F T F W = 57.4 N F T is shared between the two cables. F T represents the vertical component Do you see triangles? Angle A and Angle B = 35 o Mass of sign 5.86 kg

106 Solution 35 F T 28.7 N S = O/H F T = 28.7 N/ sin 35 F T = 50 N How would this change as angle changes? Try with 60º, 15º Angle A and Angle B = 35 o Mass of sign 5.86 kg

107 A word about angle location: F T N 35º is in reference to the horizontal 55º is in reference to the vertical

108 Ex R A 7.5 g sign is suspended by two equal strings. The angle of the string to the vertical is 35. What is the tension in each string?

109 A 7.5 g sign is suspended by two equal strings. The angle of the string to the vertical is 35. What is the tension in each string? N