2. N EWTON ’ S L AWS OF M OTION LG: TSW be able to analyze situations in which a particle remains at rest or moves due to influences of forces.

3. The space shuttle Endeavor lifts off for an 11-day mission in space. All of Newton’s laws of motion - the law of inertia, action- reaction, and the acceleration produced by a resultant force - are exhibited during this lift-off. Credit: NASA Marshall Space Flight Center (NASA- MSFC). NASA

4. SECOND LAW OF MOTION If there is a net force acting on an object, the object will have an acceleration and the object’s velocity will change. Newton's Second Law of Motion states that for a particular force, the acceleration of an object is directly proportional to the net force and inversely proportional to the mass of the object. The direction of the force is the same as that of the acceleration. In equation form:

5. SECOND LAW OF MOTION In the SI system, the unit for force is the newton (N): A newton is that net force which, when applied to a 1-kg mass, gives it an acceleration of 1 m/s 2.

7. Acceleration and Force With Zero Friction Forces Pushing the cart with twice the force produces twice the acceleration. Three times the force triples the acceleration.

8. Acceleration and Mass Again With Zero Friction F F a a/2 Pushing two carts with same force F produces one-half the acceleration. The acceleration varies inversely with the mass.

9. THIRD LAW OF MOTION According to Newton's third law of motion, when one body exerts a force on another body, the second body exerts on the first an equal force in opposite direction. For every action force, there must be an equal and opposite reaction force. The Third Law of Motion applies to two different forces on two different objects: " For every action force, there must be an equal and opposite reaction force. " Action and reaction forces never balance out because they act on different objects.

10. Newton’s Third Law Third Law: For every action force, there must be an equal and opposite reaction force. Forces occur in pairs. Action Reaction Action Reaction

11. Acting and Reacting Forces Use the words by and on to study action/reaction forces below as they relate to the hand and the bar: The action force is exerted by the _____ on the _____. The reaction force is exerted by the _____ on the _____. bar handsbar hands Action Reaction

12. MASS The property that a body has of resisting any change in its state of rest or of uniform motion is called inertia. The inertia of a body is related to the amount of matter it contains. A quantitative measure of inertia is mass. The unit of mass is the kilogram (kg).

13. WEIGHT (F G ) Weight (a vector quantity) is different from mass (a scalar quantity). The weight of a body varies with its location near the Earth (or other astronomical body), whereas its mass is the same everywhere in the universe. The weight of a body is the force that causes it to be accelerated downward with the acceleration of gravity g. F G = mg Units: Newtons (N) g = acceleration due to gravity = 9.81 m/s 2

15. FORCE An object that experiences a push or a pull has a force exerted on it. Notice that it is the object that is considered. The object is called the system. The world around the object that exerts forces on it is called the environment. system

16. FORCE 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).

17. Type of Force and its Symbol Description of Force Direction of Force Applied Force An applied force is a force that is applied to an object by another object or by a person. If a person is pushing a desk across the room, then there is an applied force acting upon the desk. The applied force is the force exerted on the desk by the person. FAFA In the direction of the pull or push.

18. Type of Force and its Symbol Description of Force Direction of Force Normal Force Perpendicular to the surface FNFN The normal force is the support force exerted upon an object that is in contact with another stable object. For example, if a book is resting upon a surface, then the surface is exerting an upward force upon the book in order to support the weight of the book. The normal force is always perpendicular to the surface

19. THE NORMAL FORCE A normal force is a force exerted by one surface on another in a direction perpendicular to the surface of contact. Note: The gravitational force and the normal force are not an action-reaction pair.

20. Type of Force and its Symbol Description of Force Direction of Force Friction Force The friction force is the force exerted by a surface as an object moves across it or makes an effort to move across it. The friction force opposes the motion of the object. For example, if a book moves across the surface of a desk, the desk exerts a friction force in the direction opposite to the motion of the book. Opposite to the motion of the object F

21. Type of Force and its Symbol Description of Force Direction of Force Air Resistance Force Air resistance is a special type of frictional force that acts upon objects as they travel through the air. Like all frictional forces, the force of air resistance always opposes the motion of the object. This force will frequently be ignored due to its negligible magnitude. It is most noticeable for objects that travel at high speeds (e.g., a skydiver or a downhill skier) or for objects with large surface areas. Opposite to the motion of the object FDFD

22. Type of Force and its Symbol Description of Force Direction of Force Tensional Force Tension is the force that is transmitted through a string, rope, or wire when it is pulled tight by forces acting at each end. The tensional force is directed along the wire and pulls equally on the objects on either end of the wire. In the direction of the pull FTFT

23. Type of Force and its Symbol Description of Force Direction of Force Gravitational Force (also known as Weight) The force of gravity is the force with which the earth, moon, or other massive body attracts an object towards itself. By definition, this is the weight of the object. All objects upon earth experience a force of gravity that is directed "downward" towards the center of the earth. The force of gravity on an object on earth is always equal to the weight of the object. Straight downward FgFg

24. FORCES HAVE AGENTS Each force has a specific identifiable, immediate cause called agent. You should be able to name the agent of each force, for example the force of the desk or your hand on your book. The agent can be animate such as a person, or inanimate such as a desk, floor or a magnet. The agent for the force of gravity is Earth's mass. If you can't name an agent, the force doesn't exist. agent

25. Directions: “Book resting on a Table” - Choose a coordinate system defining the positive direction of motion. - Replace the object by a dot and locate it in the center of the coordinate system. - Draw arrows to represent the forces acting on the system. A free-body-diagram (FBD) is a vector diagram that shows all the forces that act on an object whose motion is being studied.

26. E XAMPLE OF F REE B ODY D IAGRAM 30 0 60 0 50 NA A B B FgFg 30 0 60 0 BxBx ByBy AxAx AyAy 1. Draw and label a sketch. 2. Draw and label vector force diagram. (FBD) 3. Label x and y components opposite and adjacent to angles.

27. FNFN FGFG

28. FNFN FGFG

29. FNFN FGFG F

30. FGFG FNFN F GX F GY

31. FGFG FNFN F

32. FTFT FGFG

33. F T2 FGFG F T1

34. F T2 FGFG F T1

35. FGFG FNFN FTFT

36. FGFG F N1 F N2

37. FNFN FGFG F FAFA

38. FNFN FGFG F FAFA θ

39. FTFT FGFG

40. FNFN FGFG F FAFA θ

41. FGFG

42. FDFD FGFG

43. FGFG 17.The ball has been punted by a football player.

44. 4.1 A 5.0-kg object is to be given an upward acceleration of 0.3 m/s 2 by a rope pulling straight upward on it. What must be the tension in the rope? FTFT FGFG a (+) m = 5 kg a = 0.3 m/s 2 ΣF y = F T - F G = ma F T = m(a + g) = 5(0.3 + 9.8) = 50.5 N 5 kg

45. 4.2 A cord passing over a frictionless pulley has a 7.0 kg mass hanging from one end and a 9.0-kg mass hanging from the other. (This arrangement is called Atwood's machine). a. Find the acceleration of the masses. FTFT FTFT F G1 F G2 a (+) m 1 = 7 kg m 2 = 9 kg 7kg 9 kg

46. FTFT FTFT F G1 F G2 a (+) ΣF = F T - F G1 - F T + F G2 = m TOTAL a = 1.22 m/s 2

47. FTFT FTFT F G1 F G2 a (+) b. Find the tension of the cord Using either side of the pulley yields the same answer! F T – F G1 = m 1 a F T = m 1 a + F G1 = m 1 (a + g) = 7(1.22 + 9.8) = 77.1 N

48. APPARENT WEIGHT The actual weight of a body is the gravitational force that acts on it. The body's apparent weight is the force the body exerts on whatever it rests on. Apparent weight can be thought of as the reading on a scale a body is placed on. scale

49. 4.3 What will a spring scale read for the weight of a 75 kg man in an elevator that moves at a constant velocity: m = 75 kg F G = 75(9.8) = 735 N

50. ΣF y = F N - F G = 0 F N = F G = 735 N FNFN FGFG a. With constant upward speed of 5 m/s m = 75 kg F G = 735 N b. With constant downward speed of 5 m/s F N = 735 N N1L

51. ΣF = F N - F G = ma F N = ma + F G = 75 (2.45) + 735 = 919 N FNFN FGFG c. Going up with an acceleration of 0.25 g a = 2.45 m/s 2

52. ΣF = F G - F N = ma F N = F G - ma = 735 - 75 (2.45) = 551 N FNFN FGFG d. Going down with an acceleration of 0.25 g a = 2.45 m/s 2

53. For free fall the only force acting is F G so F N = 0 N FGFG e. In free fall?

54. “Weightlessness” More properly, this effect is called apparent weightlessness, because the gravitational force still exists. It can be experienced on Earth, but only briefly:

56. ARISTOTLE studied motion and divided it into two types: natural motion and violent motion. Natural motion: up or down. Objects would seek their natural resting places. Natural for heavy things to fall and for very light things to rise. Violent motion: imposed motion. A result of forces.

57. GALILEO demolish the notion that a force is necessary to keep an object moving. He argued that ONLY when friction is present, is a force needed to keep an object moving. In the absence of air resistance (drag) both objects will fall at the same time.

58. A ball rolling down the incline rolls up the opposite incline and reaches its initial height. As the angle of the upward incline is reduced, the ball rolls a greater distance before reaching its initial height. How far will the ball roll along the horizontal?

60. FIRST LAW OF MOTION According to Newton's First Law of Motion: " If no net force acts on it, a body at rest remains at rest and a body in motion remains in motion at constant speed in a straight line." Isaac Newton (1642-1727)

61. NEWTON'S FIRST LAW OF MOTION "An object at rest tends to stay at rest and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force." There are two parts to this statement: - one which predicts the behavior of stationary objects and - the other which predicts the behavior of moving objects.

63. The behavior of all objects can be described by saying that objects tend to "keep on doing what they're doing" (unless acted upon by an unbalanced force). If at rest, they will continue in this same state of rest. If in motion with an eastward velocity of 5 m/s, they will continue in this same state of motion (5 m/s, East).

64. It is the natural tendency of objects to resist changes in their state of motion. 1. This tendency to resist changes in their state of motion is described as inertia. 2. Inertia is the resistance an object has to a change in its state of motion. The elephant at rest tends to remain at rest.

66. Tablecloth trick: Too little force, too little time to overcome "inertia" of tableware.

67. If the car were to abruptly stop and the seat belts were not being worn, then the passengers in motion would continue in motion. Now perhaps you will be convinced of the need to wear your seat belt. Remember it's the law - the law of inertia.

68. If the truck were to abruptly stop and the straps were no longer functioning, then the ladder in motion would continue in motion.

69. If the motorcycle were to abruptly stop, then the rider in motion would continue in motion. The rider would likely be propelled from the motorcycle and be hurled into the air.

70. FIRST CONDITION FOR EQUILIBRIUM A body is in translational equilibrium if and only if the vector sum of the forces acting upon it is zero. Σ F x = 0Σ F y = 0

71. 4.4 A block of weight 50. N hangs from a cord that is knotted to two other cords, A and B fastened to the ceiling. If cord B makes an angle of 60.˚ with the ceiling and cord A forms a 30.° angle, draw the free body diagram of the knot and find the tensions A and B. AB 50 N ΣF x = B cos 60º - A cos 30º = 0 = 1.73 A ΣF y = B sin 60º + A sin 30º - 50 = 0 1.73 A sin 60º + A sin 30º = 50 1.5 A + 0.5 A = 50 A = 25 N B = 1.73 (25) = 43.3 N A = 25 N B = 43.3 N 60º 30º AxAx AyAy ByBy BxBx

72. 4.5 A 200 N block rests on a frictionless inclined plane of slope angle 30º. A cord attached to the block passes over a frictionless pulley at the top of the plane and is attached to a second block. What is the weight of the second block if the system is in equilibrium? FNFN 200 N F G2 FTFT FTFT x y θ ΣF x = F T - 200 sin 30º = 0 F T = 200 sin 30º = 100 N ΣF y = F T - F G = 0 F T = F G2 F G2 = 100 N

73. FRICTION: STATIC AND KINETIC FRICTION Frictional forces act to oppose relative motion between surfaces that are in contact. Such forces act parallel to the surfaces. Static friction occurs between surfaces at rest relative to each other. When an increasing force is applied to a book resting on a table, for instance, the force of static friction at first increases as well to prevent motion. In a given situation, static friction has a certain maximum value called starting friction. When the force applied to the book is greater than the starting friction, the book begins to move across the table.

74. The origin of friction: on a microscopic scale, most surfaces are rough.

75. FRICTION: STATIC AND KINETIC FRICTION The kinetic friction (or sliding friction) that occurs afterward is usually less than the starting friction, so less force is needed to keep the book moving than to start it moving.

76. COEFFICIENT OF FRICTION The frictional force between two surfaces depends on the normal force F N pressing them together and on the natures of the surfaces. The latter factor is expressed quantitatively in the coefficient of friction  (mu) whose value depends on the materials in contact. The frictional force is experimentally found to be: Static friction Kinetic friction

78. 4.6 A horizontal force of 140 N is needed to pull a 60.0 kg box across the horizontal floor at constant speed. What is the coefficient of friction between floor and box? F a = 140 N m = 60 kg ΣF y = F N – F G = 0 F N = F G =mg = 60(9.8) = 588 N FNFN FGFG FfFf FaFa ΣF x = F a – F f = 0 F a = F f = μF N = 0.24

79. 4.7 A 70-kg box is slid along the floor by a horizontal 400-N force. Find the acceleration of the box if the value of the coefficient of friction between the box and the floor is 0.50. m = 70 kg F a = 400 N μ = 0.5 ΣF y = F N – F G = 0 F N = F G =mg = 70(9.8) = 686 N ΣF x = F a – F f = ma F f = μF N = (0.5)(686) = 343 N FNFN FaFa FGFG FfFf

80. m = 70 kg F a = 400 N μ = 0.5 ΣF x = F a – F f = ma = 0.81 m/s 2 FNFN FaFa FGFG FfFf

81. 4.8 A 70-kg box is pulled by a rope with a 400-N force at an angle of 30  to the horizontal. Find the acceleration of the box if the coefficient of friction is 0.50 m = 70 kg F a = 400 N, 30  μ = 0.5 FGFG FaFa FfFf FNFN F ax = 400 cos 30  = 346.4 N F ay = 400 sin 30  = 200 N ΣF y = F N + F ay - F g = 0 F N = F G - F ay = 70(9.8) - 200 = 486 N F f = μF N = (0.5)(486) = 243 N

82. FGFG FaFa FfFf FNFN ΣF x = F ax – F f = ma = 1.47 m/s 2 m = 70 kg F a = 400 N, 30  μ = 0.5

83. FNFN FfFf FGFG F Gy F Gx θ 4.9 A box sits on an incline that makes an angle of 30˚ with the horizontal. Find the acceleration of the box down the incline if the coefficient of friction is 0.30. θ = 30  μ = 0.3 ΣF y = F N - F Gy = 0 F N = F Gy ΣF x = F Gx – F f = ma F f = μF N ΣF x = F Gx – μ F Gy = ma

84. FNFN FfFf FGFG F Gy F Gx θ ΣFx = F Gx – μ F Gy = ma mg sin θ - μ mg cos θ = ma = 2.35 m/s 2

85. 4.10 Two blocks m 1 (300 g) and m 2 (500 g), are pushed by a force F. If the coefficient of friction is 0.40. a. What must be the value of F if the blocks are to have an acceleration of 200 cm/s 2 ? m 1 = 0.3 kg m 2 = 0.5 kg μ = 0.4 a = 2 m/s 2 F G1 = F N1 F G2 = F N2 F G1 F G2 F N1 F N2 F F f2 F f1 ΣF x = F - F f1 - F f2 = m T a F = m T a + F f1 + F f2 = m T a + μ F NT = m T a + μ m T g = 0.8(2) + (0.4) 0.8 (9.8) = 4.7 N

86. b. How large a force does m 1 then exert on m 2 ? m 2 alone ΣF x = m 2 a F 12 - F f2 = m 2 a F 12 = F f2 + m 2 a = μ m 2 g + m 2 a = 0.4(0.5) (9.8) + 0.5 (2) = 2.96 N F f2 F 12 F N2 F G2

87. 4.11 An object m A = 25 kg rests on a tabletop. A rope attached to it passes over a light frictionless pulley and is attached to a mass m B = 15 kg. The coefficient of friction between the table and block A is 0.20. a. What is the acceleration of the 25 kg block? m A = 25 kg m B = 15 kg μ = 0.2 F N = F GA = 25(9.8) = 245 N F fA = μF N = 0.2 (245) = 49 N FNFN F GA FfFf FTFT FTFT F GB 25 kg 15 kg

88. FNFN F GA FfFf FTFT FTFT F GB ΣF = F T - F fA + F GB - F T = m T a = 2.45 m/s 2 m A = 25 kg m B = 15 kg μ = 0.2

89. FNFN F GA FfFf FTFT FTFT F GB ΣF = F T - F f = m A a F T = m A a + F f = 25 (2.45) + 49 = 110.25 N b. What is the tension on the string?

90. 4.12 Blocks 1 and 2 of masses m l and m 2, respectively, are connected by a light string, as shown above. These blocks are further connected to a block of mass M by another light string that passes over a pulley of negligible mass and friction. Blocks l and 2 move with a constant velocity v down the inclined plane, which makes an angle  with the horizontal. The kinetic frictional force on block 1 is f and that on block 2 is 2f.

91. a. On the figure below, draw and label all the forces on block m l. FGFG FNFN FTFT FfFf

92. Express your answers to each of the following in terms of m l, m 2, g, , and f. b. Determine the coefficient of kinetic friction between the inclined plane and block 1.

93. c. Determine the value of the suspended mass M that allows blocks 1 and 2 to move with constant velocity down the plane. Mg m 1 g m 2 g f 2f FTFT FTFT FTFT F N1 F N2 FTFT

94. m 1 g f F N1 d. The string between blocks 1 and 2 is now cut. Determine the acceleration of block 1 while it is on the inclined plane.

95. To swing open a door, you exert a force. The doorknob is near the outer edge of the door. You exert the force on the doorknob at right angles to the door, away from the hinges. To get the most effect from the least force, you exert the force as far from the axis of rotation (imaginary line through the hinges) as possible.

96. TORQUE Torque is a measure of a force's ability to rotate an object.

97. T ORQUE IS D ETERMINED BY T HREE F ACTORS : 1. The magnitude of the applied force. 2. The direction of the applied force. 3. The location of the applied force. 1. The magnitude of the applied force. 2. The direction of the applied force. 3. The location of the applied force. 20 N Magnitude of force 40 N The 40-N force produces twice the torque as does the 20-N force. Each of the 20-N forces has a different torque due to the direction of force. 20 N Direction of Force 20 N   Location of force The forces nearer the end of the wrench have greater torques. 20 N

98. The perpendicular distance from the axis of rotation to the line of force is called the lever arm of that force. It is the lever arm that determines the effectiveness of a given force in causing rotational motion. If the line of action of a force passes through the axis of rotation (A) the lever arm is zero.

99. U NITS FOR T ORQUE Torque Depends on the magnitude of the applied force and on the length of the lever arm, according to the following equation. r is measured perpendicular to the line of action of the force F Torque Depends on the magnitude of the applied force and on the length of the lever arm, according to the following equation. r is measured perpendicular to the line of action of the force F  = Fr Units: N  m 6 cm 40 N  = (40 N)(0.60 m) = 24.0 N  m

100. Applying a Torque

101. Sign Convention: (+)Torque will be positive if F tends to produce counterclockwise rotation. (-)Torque will be negative if F tends to produce clockwise rotation.

102. ROTATIONAL EQUILIBRIUM An object is in rotational equilibrium when the sum of the forces and torques acting on it is zero. First Condition of Equilibrium: Σ F x = 0 andΣ F y = 0 (translational equilibrium) Second Condition of Equilibrium: Στ = 0 (rotational equilibrium) By choosing the axis of rotation at the point of application of an unknown force, problems may be simplified.

103. CENTER OF MASS The terms "center of mass" and "center of gravity" are used synonymously in a uniform gravity field to represent the unique point in an object or system that can be used to describe the system's response to external forces and torques. The concept of the center of mass is that of an average of the masses factored by their distances from a reference point. In one plane, that is like the balancing of a seesaw about a pivot point with respect to the torques produced.

104. C ENTER OF G RAVITY The center of gravity of an object is the point at which all the weight of an object might be considered as acting for purposes of treating forces and torques that affect the object. The single support force has line of action that passes through the c. g. in any orientation.

105. E XAMPLES OF C ENTER OF G RAVITY Note: C. of G. is not always inside material.

106. 4.13 A 300 N girl and a 400 N boy stand on a 16 m platform supported by posts A and B as shown. The platform itself weighs 200 N. What are the forces exerted by the supports on the platform?

107. = 183. 3 N B = 900 - 183.3 = 716.7 N ΣF = 0 A + B - 300 - 200 - 400 = 0 A + B = 900 N Selecting B as the hinge Στ B = 0 -A(12) +300(10) +200(4) - 400(4) = 0 - A12 + 3000 + 800 - 1600 = 0

108. = 716.7 N A = 900 - 716.7 = 183. 3 N Selecting A as the hinge Στ A = 0 - 300(2) - 200(8) + B(12) - 400(16) = 0 - 600 - 1600 + B12 - 6400 = 0

109. 4.14 A uniform beam of negligible weight is held up by two supports A and B. Given the distances and forces listed, find the forces exerted by the supports. ΣF = 0 A - 60 - 40 + B = 0 A + B = 100 Στ A = 0 = - 60 (3) - 40(9) + B(11) = 0 A B 60 N40 N 3 m6 m2 m = 49.1 N A = 100 - 49.1 = 50.9 N

110. SUMMARY Newton’s Second Law: A net force produces an acceleration in the direction of the force that is directly proportional to the force and inversely proportional to the mass.

111. S UMMARY : P ROCEDURE Read and write data with units.Read and write data with units. Draw free-body diagram for each object.Draw free-body diagram for each object. Choose x or y-axis along motion and choose direction of motion as positive.Choose x or y-axis along motion and choose direction of motion as positive. Write Newton’s law for both axes:Write Newton’s law for both axes:  F x = m x  F y = m y  F x = m a x  F y = m a y Solve for unknown quantities.Solve for unknown quantities. Read and write data with units.Read and write data with units. Draw free-body diagram for each object.Draw free-body diagram for each object. Choose x or y-axis along motion and choose direction of motion as positive.Choose x or y-axis along motion and choose direction of motion as positive. Write Newton’s law for both axes:Write Newton’s law for both axes:  F x = m x  F y = m y  F x = m a x  F y = m a y Solve for unknown quantities.Solve for unknown quantities. N = (kg)(m/s 2 )

112. The space shuttle Endeavor lifts off for an 11-day mission in space. All of Newton’s laws of motion - the law of inertia, action- reaction, and the acceleration produced by a resultant force - are exhibited during this lift-off. Identify them! Credit: NASA Marshall Space Flight Center (NASA-MSFC). NASA