1. Chapter 4- Forces and Motion

2. 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.

3. Other examples of forces

4. 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

5. 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

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

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

8. 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!

9. 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.

10. 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

11. 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

12. 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

13. 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

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

15. 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

16. 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

17. 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.

18. 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

19. 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

20. 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

21. 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.

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

23. 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

24. 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

25. 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

26. 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?

27. 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! 

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

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

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

32. 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

33. 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

34. Kinetic Friction
Kinetic Friction (Fk) is the frictional force on an object in motion

35. 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

36. 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

37. 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)

38. 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

39. Solve for Coefficient of Kinetic Friction

40. 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.

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

42. 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

43. 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 an angle of 25 above the horizontal. The coefficient of kinetic friction is Find the magnitude of the acceleration of the crate.

44. 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.

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

46. 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

47. 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

48. 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

49. 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?

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

51. 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

52. 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

53. 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

54. 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

55. 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.

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

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

58. Coordinate system for inclined planes
Y axis
X axis

59. 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.

60. 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)

61. 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

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

63. 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

64. 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

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

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

67. Draw a FBD Ff FN
Fg,x
Fg,y Fg θ

68. 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?

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

70. 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?

71. FBD Ff FN
Fg,x
Fg,y Fg θ

72. 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

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

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