1. Chapter 4- Forces and Motion

2. List some examples of force
FOCUS QUESTION(S)
What is force?
List some examples of force

3. What are Newton’s three laws of motion?
FOCUS QUESTION(S)
What are Newton’s three laws of motion?

4. Objective(s) 1. Explain how force affects the motion of an object.
2. Distinguish between contact forces and field forces.
3. Interpret and construct free-body diagrams.

5. is the most volcanically active body in the solar system.
Think about the following questions: What is this object? Where is it? Why does it look like that?
Erupting Volcano!!
IO is a moon of Jupiter
Competing forces between Jupiter and the other Galilean moons cause the center of Io compress and melt. Consequently Io
is the most volcanically active body in the solar system.

6. Other examples of forces

7. What is a force?
IPC definition: A push or a pull exerted on some object
Better definition: Force represents the interaction of an object with its environment
The Unit for Force is a Newton

8. Two major types of forces
Contact Forces: Result from physical contact between two objects
Examples: Pushing a cart, Pulling suitcase
Field Forces: Forces that do not involve physical contact
Examples: Gravity, Electric/Magnetic Force

9. Force is a vector! (yay more vectors )
The effect of a force depends on magnitude and direction

10. Force Diagrams (p. 126)
Force Diagram: A diagram that shows all the forces acting in a situation

11. Free Body Diagrams p.127
Free Body Diagrams (FBDs) isolate an object and show only the forces acting on it
FBDs are essential! They are not optional! You need to draw them to get most problems correct!

12. How to draw a free body diagram
Situation: A tow truck is pulling a car
(p. 127)
We want to draw a FBD for the car only.

13. Steps for drawing your FBD
Step 1: Draw a shape representing the car (keep it simple)
Step 2: Starting at the center of the object, Draw and label all the external forces acting on the object
Force of Tow Truck on Car=
5800 N

14. Add force of gravity Force of Tow Truck on Car= 5800 N
Gravitational force (Weight of car)=
14700 N
Force of Tow Truck on Car=
5800 N

15. Add force of the road on the car (Called the Normal Force)
Gravitational force (Weight of car)=
14700 N
Force of Tow Truck on Car=
5800 N
Normal Force =
13690 N

16. Finally add the force of friction acting on the car
Gravitational force (Weight of car)=
14700 N
Force of Tow Truck on Car=
5800 N
Normal Force =
13690 N
Force of Friction= 775 N

17. A Free Body Diagram of a Football Being KickedFg
Fkick

18. A person is pushed forward with a force of 185 N
A person is pushed forward with a force of 185 N. The weight of the person is 500 N, the floor exerts a force of 500 N up. The friction force is 30 N.
FN= 500 N
Fg= 500 N
Fapp= 185 N
Ff= 30 N

19. Forces you will need Symbol of Force Description Fg
Gravitational Force is the Weight of the Object (equal to mass x g= mg) FN
Normal Force= Force acting perpendicular to surface of contact Ff
Frictional Force- Opposes applied force; acts in direction opposite of motion
Fapp
Applied Force

20. Sample Problem p. 128 #3
Draw a free body diagram of a football being kicked. Assume that the only forces acting on the ball are the force of gravity and the force exerted by the kicker.

21. Practice Problem
A large, square box of exercise equipment sits on a storeroom floor. A rope is tied around the box. Assume that if the box moves along the floor, there is a backward force that resists its motion.
Suppose that the box remains at rest. In the space provided, draw a free body diagram for the box. Label each force involved in the diagram.

22. A large, square box of exercise equipment sits on a storeroom floor
A large, square box of exercise equipment sits on a storeroom floor. A rope is tied around the box. Assume that if the box moves along the floor, there is a backward force that resists its motion.
Suppose a warehouse worker moves the box by pulling the rope to the right horizontal to the ground. In the space provided, draw a free-body diagram for the box. Label each force involved in the diagram

23. A large, square box of exercise equipment sits on a storeroom floor
A large, square box of exercise equipment sits on a storeroom floor. A rope is tied around the box. Assume that if the box moves along the floor, there is a backward force that resists its motion.
Suppose the warehouse worker moves the box by pulling the rope to the right at a 50° angle to the ground. In the space provided, draw a free body diagram for the box. Label each force involved in the diagram

24. FOCUS QUESTION(S) Explain how force affects the motion of an object.
2. Distinguish between contact forces and field forces.
A Chair is pushed forward with a force of 185N. The gravitational force of the earth on the chair is 155N downward, and the floor exerts a force of 155N upward on the chair. Draw a free-body diagram showing the forces acting on the chair.

25. FBD Worksheet

26. FOCUS QUESTION(S)
Two soccer players kick a ball at the same instant. One player kicks with a force of 65 N to the north, while the other player kicks with a force of 88 N to the east. In what direction does the ball travel?

27. The muscle responsible for closing the mouth is the strongest muscle in the human body. It can exert a force greater than that exerted by a man lifting a mass of 400 kg. Richard Hoffman of Florida recorded the force of biting at 4.33 x 103 N. If each force shown in the diagram below has a magnitude equal to the force of Hoffman’s bite, determine the net force.

28. FOCUS QUESTION(S)
The gravitational force of earth on a cake on a plate is 8.9 N. The plate exerts a force of 11.0 N. Draw a free- body diagrams of the cake.

29. Newton’s first Law
SECTION 2

30. Objective(s)
1. Explain the relationship between the motion of an object and the net external force acting on it.
2. Determine the net external force on an object.
3. Calculate the force required to bring an object into equilibrium.

31. Newton’s 1st Law of Motion
The Law of Inertia
An object at rest remains at rest, and an object in motion continues in motion with constant velocity (constant speed in straight line) unless the object experiences a net external force
The tendency of an object not to accelerate is called inertia

32. Acceleration
The net external force (Fnet) is the vector sum of all the forces acting on an object
If an object accelerates (changes speed or direction) then a net external force must be acting upon it

33. Equilibrium
If an object is at rest (v=0) or moving at constant velocity, then according to Newton’s First Law, Fnet =0
When Fnet =0, the object is said to be in equilibrium

34. How do we use this information? Sample Problem p. 133 #2
A crate is pulled to the right with a force of 82.0 N, to the left with a force of 115 N, upward with a force of 565 N and downward with a force of 236 N.
A. Find the net external force in the x direction
B. Find the net external force in the y direction
C. Find the magnitude and direction of the net external force on the crate.

35. Step 1: Draw a FBD Fup = 565 N Fright = 82 N Fleft = 115 N
Fdown = 236 N

36. Find the vector sum of forces
A. 82 N + (-115 N )= -33 N
B. 565 N + (-236 N) = 329 N
C. Find the resultant of the two vectors from part a and b.
33 N
329 N
R = 331 N at 84.3 North of West

37. Newton’s 1st Law Review Newton’s 1st Law:
When Fnet=0, an object is in equilibrium and will stay at rest or stay in motion
In other words, if the net external force acting on an object is zero, then the acceleration of that object is zero

38. Practice Problems Practice 4A page 133 Classwork
Section 4.2 review page 135 Homework.

39. FOCUS QUESTION(S) What is Newton 2nd and 3rd Law of Motion?
The net force exerted by a woodpecker’s head when its beak strikes a tree can be as large as 4.90 N, assuming that the bird’s head has a mass of 50.0 g. Assume that two different muscles pull the woodpecker’s head forward and downward, exerting a net force of 4.90 N. If the forces exerted by the muscles are at right angles to each other and the muscle that pulls the woodpecker’s head downward exerts a force of 1.70 N, what is the magnitude of the force exerted by the other muscle? Draw a free-body diagram showing the forces acting on the woodpecker’s head.

40. Chapter 4
Section 3

41. Objective(s)
1. Describe the acceleration of an object in terms of its mass and the net external force acting on it.
2. Predict the direction and magnitude of the acceleration caused by a known net external force.
3. Identify action-reaction pairs.
4. Explain why action-reaction pairs do not result in equilibrium.

42. Newton’s 2nd Law (p.137)
The acceleration of an object is directly proportional to the net external force acting on the object and inversely proportional to the object’s mass

43. Example p. 138 # 4
A 2.0 kg otter starts from rest at the top of a muddy incline 85 cm long and slides down to the bottom in 0.50 s. What net external force acts on the otter along the incline?

44. Solving the problem To calculate Fnet, we need m and a M=2.0 kg
What is a?
Vi= 0 m/s, t=0.50 s,
displacement=85 cm=.85 m
Welcome back kinematic equations! 

46. Newtons’ 3rd Law Forces always exist in pairs
For every action there is an equal and opposite reaction

47. Action- Reaction Pairs
Some action-reaction pairs:

48. Although the forces are the same, the accelerations will not be unless the objects have the same mass.

49. Practice
Practice page 138 HW section 4.3 review page 140

50. FOCUS QUESTION(S)

51. FOCUS QUESTION(S)
A cabinet initially at rest on a horizontal surface requires a 115 N horizontal force to set it in motion. If the coefficient of static friction between the cabinet and the floor is 0.38,what is the normal force exerted on the cabinet? What is the mass of the cabinet?

52. FOCUS QUESTION(S)
Two students reach for a jar of mustard at the same time. One student pulls to the left with a force of 13.2 N, while the other student pulls to the right with a force of 12.9 N. If the jar has a net acceleration of 0.44 m/s2 to the left, what is the mass of the jar?

53. FOCUS QUESTION(S)
Are mass and weight the same? Why or why not?
Explain the difference between static and kinetic friction.

54. CHAPTER 4
Section 4

55. Objective(s)
1. Explain the difference between mass and weight. 2. Find the direction and magnitude of the normal force. 3. Describe air resistance as a form of friction. 4. Use coefficients of friction to calculate frictional force.

56. Everyday Forces Weight= Fg = mg
Normal Force= FN= Is always perpendicular to the surface.
Friction Force= Ff
Opposes applied force
There are two types of friction: static and kinetic

57. Static Friction
Force of Static Friction (Fs) is a resistive force that keeps objects stationary
As long as an object is at rest:
Fs = -Fapp

58. Kinetic Friction
Kinetic Friction (Fk) is the frictional force on an object in motion
Fnet= Fapp- Fk

59. Coefficients of Friction
The coefficient of friction (μ) is the ratio of the frictional force to the normal force
Coefficient of kinetic Friction
Coefficient of Static Friction

60. Sample Problem p. 145 #2
A 25 kg chair initially at rest on a horizontal floor requires a 365 N horizontal force to set it in motion. Once the char is in motion, a 327 N horizontal force keeps it moving at a constant velocity.
A. Find coefficient of static friction
B. Find coefficient of kinetic friction

61. Coefficient of Static Friction
In order to get the chair moving, it was necessary to apply 365 N of force to overcome static friction. Therefore Fs = 365 N.
The normal force is equal to the weight of the chair (9.81 x 25= 245 N)

62. Coefficient of Kinetic Friction
The problem states that the chair is moving with constant velocity, which means Fnet=0. Therefore, Fapp must equal -Fk.
Fapplied= 327 N
Fk= 327 N

63. Solve for Coefficient of Kinetic Friction

64. FOCUS QUESTION(S)
A 20.0 kg trunk is pushed across the floor of a moving van by a horizontal force. If the coefficient of kinetic friction between the trunk and the floor is 0.255,what is the magnitude of the frictional force opposing the applied force?

65. FOCUS QUESTION(S)
A cabinet initially at rest on a horizontal surface requires a 115 N horizontal force to set it in motion. If the coefficient of static friction between the cabinet and the floor is 0.38,what is the normal force exerted on the cabinet? What is the mass of the cabinet?

66. FOCUS QUESTION(S)
A box of books weighing 319 N is shoved across the floor by a force of 485 N exerted downward at an angle of 35° below the horizontal. Draw the free body diagram for this situation.

67. Forces at an angle
A woman is pulling a box to the right at an angle of 30 above the horizontal. The box is moving at a constant velocity. Draw a free body diagram for the situation.

68. FBD F app,y F app,x FN= Normal Force Fapp= Applied Force
Ff= Friction Force
Fg=Weight

69. What is Fnet?
Since the suitcase is moving with constant velocity, Fnet=0.
That means the forces in the x direction have to cancel out and the forces in y direction have to cancel out
Fk = Fapp,x
FN + Fapp,y = Fg
NOTICE THAT NORMAL FORCE DOES NOT EQUAL WEIGHT IN THIS SITUATION

70. Let’s do an example. P. 154 #42
A 925 N crate is being pushed across a level floor by a force F of 325 N at 25° above the horizontal. The coefficient of kinetic friction is Find the magnitude of the acceleration of the crate.

71. What do we need to know? So we need mass and Fnet.
We have weight (925 N). So what is mass?
How to find Fnet?
Find vector sum of forces acting on crate.

72. FBD F app,y F app,x FN= Normal Force Fapp= 325 N Ff= Friction Force
Fg=Weight=925 N

73. Finding Fnet,y Is box accelerating in y direction?
No. Therefore Fnet in y direction is 0
So FN + Fapp,y = Fg
So FN = Fg- Fapp,y= 925 N- 325sin(25)
FN= N

74. Finding Fnet,x Is box accelerating in x direction?
Yes. Therefore Fnet,x is not 0
Fnet,x= Fapp,x – Ff
Fapp,x = Fappcos(25)=294.6 N
Use coefficient of friction to find Ff
Ff=μFN=(0.25)(787N)=197 N

75. Finish the Problem Fnet,x = 294 N – 197 N= 97 N
So now we know that the Fnet on the box is 97 N since Fnet,y is 0

76. Another example. P. 154 #54 part a
A box of books weighing 319 N is shoved across the floor by a force of 485 N exerted downward at an angle of 35° below the horizontal.
If μk between the floor and the box is 0.57, how long does it take to move the box 4.00 m starting from rest?

77. DRAW FBD FN
Fapp,x Ff
Fapp,y
Fg=319 N
Fapp= 485 N

78. Find Fnet Is box accelerating in y direction?
No. Therefore Fnet in y direction is 0
So FN = Fapp,y + Fg
So FN = 485sin(35) N= 598 N

79. Fnet,x Is box accelerating in x direction?
Yes. Therefore Fnet,x is not 0
Fnet,x= Fapp,x – Ff
Fapp,x = 485cos(35)= N
Use coefficient of friction to find Ff
Ff=μFN=(0.57)(598)=341 N
Fnet, x = = N

80. So now we know that the Fnet on the box is
57.29 N since Fnet,y is 0
Weight of box is 319 N.
Find mass by dividing by 9.81
m= kg

81. Finish the problem
We want to know how long it takes for the box to move 4.00 m.
Find vf so that you can solve for t
Solve for t

82. Forces on An Incline
A block slides down a ramp that is inclined at 30° to the horizontal. Write an expression for the normal force and the net force acting on the box.

83. Draw a Free Body DiagramFF FN θ
Fg,x
Fg,y Fg θ

84. Closer look at gravity triangle.
Solve for Fg,y and Fg,x Fg
Fg,x
Fg,y θ

85. Coordinate system for inclined planes
Y axis
X axis

86. Fnet in the y direction
When a mass is sliding down an inclined plane, it is not moving in the y direction.
Therefore Fnet,y =0 and all the forces in the y direction cancel out.

87. Forces In the y-direction
So what are the forces acting in the y direction?
Look at your FBD
We have normal force and Fg,y
Since they have to cancel out…
FN= mgcos(θ)

88. Forces in the x direction
What is the force that makes the object slide down the inclined plane?
Gravity…but only in the x direction

89. Remember that Vectors can be moved parallel to themselves!!FF FN Fg θ
Fg,y θ
Fg,x

90. Forces in the x direction
So what are the forces acting in the x direction?
Friction Force (Ff) and Gravitational Force (Fg,x)
If the box is in equlibrium
Fg,x = Ff
If the box is accelerating
Fnet= Fg,x - Ff

91. What if there is an additional applied force?
Example: a box is being pushed up an inclined plane…
Fapp FN
Fg,x Ff
Fg,y θ

92. In that case… FN= mgcosθ Fnet = Fapp- Fg,x – Ff
If the object is in equilibrium then
Fapp= Fg,x + Ff

93. An Example p. 153 #40
A 5.4 kg bag of groceries is in equilibrium on an incline of 15° angle. Find the magnitude of the normal force on the bag.

94. Draw a FBD Ff FN
Fg,x
Fg,y Fg θ

95. Solve the Problem The block is in equilibrium so…
Fnet=0
Fg,y= FN=mgcosθ=(5.4kg)(9.81)cos(15)
FN=51 N
Additionally, what is the force of friction acting on the block?

96. Find Force of Friction Fnet= 0 Fg,x= Ff= mgsinθ=5.4(9.81)sin(15)
Ff= 13.7N

97. Example p. 147 #3
A 75 kg box slides down a 25.0° ramp with an acceleration of 3.60 m/s2.
Find the μk between the box and the ramp
What acceleration would a 175 kg box have on this ramp?

98. FBD Ff FN
Fg,x
Fg,y Fg θ

99. What is Fnet? They give mass and acceleration
So Fnet= ma= 75kg x 3.60 m/s2
Fnet= 270 N
FN= mgcosθ
Fnet= Fg,x – Ff=mgsinθ - Ff

100. Solve for Ff Fnet= Fg,x – Ff=mgsinθ – Ff Ff= mgsinθ – Fnet
Ff = 75kg(9.8)sin(25) – 270 N
Ff = N

101. Finish the Problem
We are trying to solve for μk

102. CLASS PRACTICE
Practice 4D p.147
HW 4.4 Section Review p. 149
#1-5

103. GET OUT YOUR HW