1. Newton’s Laws of Motion
Chapter 4 1

2. Force 2

3. FORCE
Newton stated that the change in velocity of an object is caused by FORCES.
When the velocity of an object is constant, or if the object is at rest, it is said to be in equilibrium. 3

4. FORCE
Contact forces: forces that result from physical contact between two objects.
Examples: friction, a push or a pull
Field forces: forces that can act at a distance.
Examples: magnetism, gravity 4

5. FORCE DIAGRAMS Force is a vector.
Force diagrams – show forces vectors as arrows
Free body diagrams – shows only the forces acting on a single object 5

6. Newton’s First Law 6

7. Newton’s First Law (Law of Inertia)
A body at rest will remain at rest, a body in motion will remain in motion, traveling with a constant velocity in a straight line, unless an unbalanced force acts on it.
INERTIA = a measure of a body’s ability to resist changes in velocity.
INERTIA ≈ MASS
(the greater the mass of a body, the less it will accelerate under the action of an applied force) 7

8. 8

9. 9

10. It’s easier to push a Volkswagen than a Mack Truck
But, once it’s moving, it’s harder to stop the truck 10

11. That’s because the Mack Truck has more mass, and thereby more inertia.11

12. 12

13. 13

14. Newton’s Second and Third Laws15

15. Newton’s Second Law F = ma
The acceleration of an object is directly proportional to the resultant force acting on it and inversely proportional to its mass.
The direction of the acceleration is in the direction of the resultant force.
The SI unit of force is the NEWTON.
1N = 1 kg•m/s2
Weight: w = mg 16

16. Any acceleration requires a force.
Acceleration can be… Q 17

17. Newton’s Third Law
If two bodies interact, the force exerted on a body 1 by body 2 is equal in magnitude, but opposite in direction to the force exerted on body 2 by 1.
OR….
For every action there is an equal and opposite reaction. 20

18. Action Reaction Pairs Ball being thrown Car hitting wall
Elevator pulled up by cable 21

19. Implicit in these laws are the following ideas.
If no unbalanced force acts on a body, its acceleration must be zero.
An unbalanced force is one whose vector sum does not equal zero. 22

20. Implicit in these laws are the following ideas.
If an unbalanced force acts on a body, it must accelerate. It will continue to accelerate for as long as the force(s) are unbalanced. 23

21. Implicit in these laws are the following ideas.
If a body has no acceleration, the vector sum of all the forces acting on it must be zero. 24

22. What is the vector sum of all the forces acting on this point?
Zero 25

23. A 1 kg rock is thrown at 10 m/s straight upward
A 1 kg rock is thrown at 10 m/s straight upward. Neglecting air resistance, what is the net force that acts on it when it is half way to the top of its path?
N /2 N N /4 N 26

24. Everyday Forces 27

25. Weight vs Mass
The weight of an object is the force of gravity acting on the mass of that object.
Measured in pounds in the USA
Measured in Newtons elsewhere (and in this class)
Weight = mg or (mass) x (9.8m/s2)
On Earth 28

26. Normal Force
In this case, N = mg;
But not always
Normal force: the upward force acting on an object on a surface; the normal force is perpendicular to the surface on which the object is sitting. N
W = mg 29

27. Tension
Tension (T) is a force acting on a rope. When a rope is taught there is tension in the rope. 30

28. Friction More on friction later
Friction (Ff) is a resistive force that opposes motion.
More on friction later 31

29. How to draw a FBD
All the forces on one object are drawn as vectors (arrows)
The tail of the arrow starts on the object and point off in the direction of the force.
If the force is at an angle, include the angle.

30. Example: A car hits a wall. Draw the FBD of the car
Normal Force (N)
Weight (mg)
Force from wall (Fwall)

31. Multi-Object Force Diagrams
Draw all objects as boxes.
Create a table of interactions within the system.
Using the table of interactions, draw the force pairs acting on all objects.

32. Example 1: Car hits wall
Car
Wall
Earth
Wall
Car
Earth + + + + + +

33. Example 2: Elephant vs. Man
Rope
Earth
Man
Eleph
Rope
Man
Earth - + + + + + - + + + + +

34. + - + + + + - + + + + + Example 3: Hanging Box Rope Box Earth Ceiling+ - + + + + - + + + + +

35. How to solve problems with Newton’s Laws
Draw the picture or Free Body Diagram
Label ALL forces (in x and y direction)
Pick +X (or +y) to be in the direction of acceleration
Sum the forces in each direction
Solve

36. Example 1: A car engine will push a 2500 kg car forward with 800
Example 1: A car engine will push a 2500 kg car forward with 800. N of force on flat ground. If the air resistance is 300. N, what is the acceleration of the car? N X Y
Fcar
Fair w N
SFx = max
800.
300.N
= (2500kg)ax
800.N
-300.
500. = (2500kg)ax mg
-mg
ax = 0.20m/s2 N

37. SFy = may T X Y T - 8800 = (m)(1.30) T T - 8800 = (898kg)(1.30)
Example: What is the tension in the cable of an elevator with a weight of 8800 N that ascends with an acceleration of 1.30 m/s2?
SFy = may T X Y
T = (m)(1.30) T
T = (898kg)(1.30)
T = 9,967 N
T = 1.0 x 104 N
-8800
W=mg
W=8800N
m = W/g
= 8800/9.8
= 898 kg 40

38. X Y SFx = max 84cos35o = 68.81 84sin35o = 48.81 68.81 – Fr = 0 N -mg
A coach pulls a 35 kg bag of soccer balls across the field with a force of 84 N directed at an angle of 35o above the horizontal. If the bag travels at a constant velocity, (a) how strong are the resistive forces from the field on the bag and (b) how strong is the normal force on the bag? N X Y
SFx = max
84 N
35o
84cos35o
= 68.81
84sin35o
= 48.81
68.81 – Fr = 0
Fr = ? N
-mg
- Fr
Fr = N mg
Fr = 69 N

39. SFy = may 84cos35o = 69.63 84sin35o = 48.18 N -mg - Fr X Y
A coach pulls a 35 kg bag of soccer balls across the field with a force of 84 N directed at an angle of 35o above the horizontal. If the bag travels at a constant velocity, (a) how strong are the resistive forces from the field on the bag and (b) how strong is the normal force on the bag?
SFy = may N
84cos35o
= 69.63
84sin35o
= 48.18 N
-mg
- Fr X Y
N - mg = 0
84 N
35o
Fr = ?
N – (35)(9.8) = 0
N = 0 mg
N = N
N = 290 N

40. Example: A child holds a sled at rest on a frictionless snow covered hill. If the sled weighs 100. N, find the force the child must exert on the rope and the force the hill exerts on the sled. n T
W=100N
40
40 43

41. Example: A child holds a sled at rest on a frictionless snow-covered hill. If the sled weighs 100. N, find the force the child must exert on the rope and the force the hill exerts on the sled.
40
W=100N T n
X Y
SFx = max
0 N
T – = 0
T 0
T = = 64.3 N
- 100sin40o =
- 100cos40o =
SFy = may
N – = 0
N = = 76.6 N

42. Example: A traffic light weighing 100
Example: A traffic light weighing 100. N hangs from a cable tied to two other cables fastened to a support. The upper cables make angles of 37 and 53 with the horizontal. Find the tension in each of the three cables.
37
53 T3 T1 T2 T3 T1 T2
37
53
T3=W W 45

43. X y T1 T2 T3 -T1cos37 T1sin37 T2cos53 T2sin53 -100 N
Example: A traffic light weighing 100. N hangs from a cable tied to two other cables fastened to a support. The upper cables make angles of 37 and 53 with the horizontal. Find the tension in each of the three cables. T3 X y T1 T2 T3 T1 T2
-T1cos37
T1sin37
37
53
T2cos53
T2sin53
T3=W
-100 N W 46

44. Example: A traffic light weighing 100
Example: A traffic light weighing 100. N hangs from a cable tied to two other cables fastened to a support. The upper cables make angles of 37 and 53 with the horizontal. Find the tension in each of the three cables.
Fx = 0 = T2cos53-T1cos37 x y T1 T2 T3
T2cos53 = T1cos37
-T1cos37
T1sin37
T2 = T1(cos37/cos53)
T2cos53
T2sin53
T2 = 1.33T1
-100 N
Fy = 0 = T1sin37+T2sin53-100
100 = T1sin37+1.33T1sin53
100 = T1(.602)+T1(1.06)
T1= 60.2 N 47

45. Example: A traffic light weighing 100
Example: A traffic light weighing 100. N hangs from a cable tied to two other cables fastened to a support. The upper cables make angles of 37 and 53 with the horizontal. Find the tension in each of the three cables.
T1= 60.2 N
T2 = 1.33T1
T2 = 1.33(60 N)
T2= 80.1 N
T3 = W = 100N
T3= 100 N = 1.00 x 102 N 48

46. Forces of Friction
The term friction refers to the resistive forces that arise to oppose the motion of a body past another with which it is in contact. 52

47. Forces of Friction
Kinetic friction (sliding friction) is the frictional resistance a body in motion experiences.
Static friction is the frictional resistance a stationary body must overcome in order to be set in motion. 53

48. Ff=μN Where μ = coefficient of friction N = normal force
The magnitude of μ depends on the nature of the surfaces in contact
NOTE: friction does NOT depend on the area of contact! 54

49. n T=90 N 90.0cos30o = 90.0sin30o = Fkf - Ff N W=mg -mg
Example: You need to move a box of books into your dormitory room. To do so, you attach a rope to the box and pull on it with a force of 90.0 N at an angle of 30.0. The box of books has a mass of 20.0 kg, and the coefficient of friction between the bottom of the box and the hallway surface is Find the acceleration of the box. n
X Y
T=90 N
90.0cos30o =
90.0sin30o =
Fkf
30
- Ff N
W=mg
-mg 55

50. Fx = max = 90cos30-75.5 Fy = 0 = n + 90sin30-(20)(9.8)
Example: You need to move a box of books into your dormitory room. To do so, you attach a rope to the box and pull on it with a force of 90.0 N at an angle of 30.0. The box of books has a mass of 20.0 kg, and the coefficient of friction between the bottom of the box and the hallway surface is Find the acceleration of the box.
Fx = max = 90cos30-75.5
(20)(ax) = 90cos30-75.5
Fy = 0 = n + 90sin30-(20)(9.8)
ax = m/s2 n
n = 151 N
T=90 N
Fkf = μn = (0.5)(151 N)
Fkf
Fkf = 75.5 N (to the left)
30
W=mg 56

51. Example: A hockey puck is given an initial speed of 20
Example: A hockey puck is given an initial speed of 20.0 m/s on a frozen pond. The puck remains on the ice and slides 120. m before coming to rest. Determine the coeficient of friction between the puck and the ice. n
d=120 m
Fkf
W=mg 57

52. d=120 m n vi = 20 m/s vf2 = vi2 + 2ad Fkf vf = 0 m/s
Example: A hockey puck is given an initial speed of 20.0 m/s on a frozen pond. The puck remains on the ice and slides 120. m before coming to rest. Determine the coeficient of friction between the puck and the ice.
d=120 m n
vi = 20 m/s
vf2 = vi2 + 2ad
Fkf
vf = 0 m/s
0 = (20)2 + 2(a)(120)
d = 120 m
a = m/s2
W=mg 58

53. Fy = 0 = n - W Fx = max = -Fkf n n = W = mg
Example: A hockey puck is given an initial speed of 20.0 m/s on a frozen pond. The puck remains on the ice and slides 120. m before coming to rest. Determine the coeficient of friction between the puck and the ice.
Fy = 0 = n - W
Fx = max = -Fkf n
n = W = mg
(m)(-1.67) = -(9.8)(μ)(m)
Thus, Fkf = μn =μmg
μ = 0.170
Fkf
Fkf = μmg = 9.8μm
W=mg
d=120 m 59

54. N 4.00 kg object: Fx = ma = T - Fkf Fkf T T Fy = 0 = n – mg m1g m2g
Example: A 4.00-kg object is connected to a 7.00-kg object by a light string that passes over a frictionless pulley. The coefficient of sliding friction between the 4.00-kg object and the surface is Find the acceleration of the two objects and the tension in the string. N
4.00 kg object:
Fx = ma = T - Fkf
Fkf T
4 kg
4a = T - Fkf T
Fy = 0 = n – mg
m1g
n = (4.00 kg)(9.8m/s2)
7 kg
n = 39.2 N
m2g 60

55. Example: A 4. 00-kg object is connected to a 7
Example: A 4.00-kg object is connected to a 7.00-kg object by a light string that passes over a frictionless pulley. The coefficient of sliding friction between the 4.00-kg object and the surface is Find the acceleration of the two objects and the tension in the string. N
Fkf T
4 kg
a = 5.16 m/s2 T
m1g
7 kg
m2g

56. N Fkf T T m1g m2g a = 5.16 m/s2 T = 32.5 N 4 kg 7 kg
Example: A 4.00-kg object is connected to a 7.00-kg object by a light string that passes over a frictionless pulley. The coefficient of sliding friction between the 4.00-kg object and the surface is Find the acceleration of the two objects and the tension in the string. N
a = 5.16 m/s2
Fkf T
4 kg
T = 32.5 N T
m1g
7 kg
m2g

57. Multiple Objects Hooked Together
Treat it as a whole system together.
Add all the masses up and use that as the total mass.
Then just look at the forces on the outside.

58. 10.0 kg 10.0 kg 10.0 kg 150 N 25 N 25 N 25 N SF = mtotala
Example: Three boxes are tied together by light strings as shown. They are then pulled along a horizontal floor with a force of 150 N directed parallel to the floor. If each box has a mass of 10.0 kg and experiences a frictional force of 25 N, what is the total acceleration of the system?
10.0 kg
10.0 kg
10.0 kg
150 N
25 N
25 N
25 N
SF = mtotala
150 – 25 – = (30.0)a
a = 2.5 m/s2

59. Air resistance & Terminal Velocity68

60. Air resistance or Drag
Air resistance is similar to friction in that it opposes the direction of motion.
Air resistance is the resistance of the air that the object is falling through.
When the object is moving slowly, it is not moving through much air so the air resistance is small.
The faster the object moves, the more air it travels through and the larger the air resistance is. 69

61. Air Resistance
Weight (mg) 70

62. balanced, acceleration
When the forces are
balanced, acceleration
stops.
Air Resistance
Weight (mg) 71

63. Air Resistance Terminal velocity weight
The forces are balanced. That is, the vector sum is equal to zero.
weight 72

64. Free Fall Motion 73

65. Common Misconceptions
About Forces 74

66. Common misconceptions about forces
When a ball has been thrown, the force of the hand that threw it remains on it for awhile.
NO! The force of the hand is a contact force; therefore, once contact is broken, the force is no longer exerted. 75

67. Common misconceptions about forces
Even if no force acts on a moving object, it will eventually stop.
Weight and mass are two names for the same thing.
Air does not exert a force.
NO. Air exerts a huge force, but because it is balanced on all sides, it usually exerts no net force unless an object is moving. 76