1. Introduction to Newton’s Laws Newton’s First Law.

2. What is Force? A force is a push or pull on an object.
Forces cause an object to accelerate…
To speed up
To slow down
To change direction

3. Newton’s First Law
A body in motion stays in motion at constant velocity and a body at rest stays at rest unless acted upon by a net external force.
This law is commonly referred to as the Law of Inertia.

4. The First Law is Counterintuitive

5. Implications of Newton’s 1st Law
If there is zero net force on a body, it cannot accelerate, and therefore must move at constant velocity, which means
it cannot turn,
it cannot speed up,
it cannot slow down.

6. What is Zero Net Force? SF = 0 The table pushes up on the book. FT
A book rests on a table.
Physics
Gravity pulls down on the book. FG
Even though there are forces on the book, they are balanced.
Therefore, there is no net force on the book.
SF = 0

7. Diagrams
Draw a force diagram and a free body diagram for a book sitting on a table.
Physics N W
Force Diagram N W
Free Body Diagram

8. Sample Problem
A monkey hangs by its tail from a tree branch. Draw a force diagram representing all forces on the monkey FT FG

9. Sample Problem
Now the monkey hangs by both hands from two vines. Each of the monkey’s arms are at a 45o from the vertical. Draw a force diagram representing all forces on the monkey.
Fa1
Fa2 FG

10. Mass and Inertia
Chemists like to define mass as the amount of “stuff” or “matter” a substance has.
Physicists define mass as inertia, which is the ability of a body to resist acceleration by a net force.

11. Sample Problem
A heavy block hangs from a string attached to a rod. An identical string hangs down from the bottom of the block. Which string breaks
when the lower string is pulled with a slowly increasing force?
when the lower string is pulled with a quick jerk?
Top string breaks due to its greater force.
Bottom string breaks because block has lots of inertia and resists acceleration. Pulling force doesn’t reach top string.

13. Newton’s Second Law

14. Newton’s Second Law
A body accelerates when acted upon by a net external force.
The acceleration is proportional to the net force and is in the direction which the net force acts.

15. Newton’s Second Law ∑F = ma
where ∑F is the net force measured in Newtons (N)
m is mass (kg)
a is acceleration (m/s2)

16. Units of force Newton (SI system)
1 N = 1 kg m /s2
1 N is the force required to accelerate a 1 kg mass at a rate of 1 m/s2
Pound (British system)
1 lb = 1 slug ft /s2

17. Working 2nd Law Problems
Draw a force or free body diagram.
Set up 2nd Law equations in each dimension.
SFx = max and/or SFy = may
Identify numerical data.
x-problem and/or y-problem
Substitute numbers into equations.
“plug-n-chug”
Solve the equations.

18. Sample Problem
In a grocery store, you push a 14.5-kg cart with a force of 12.0 N. If the cart starts at rest, how far does it move in 3.00 seconds?
m = 14.5 kg, SF = 12 N, t = 3 s, vo = 0 m/s
x = ?
x = vot + ½at2
Need to find ‘a’, use SF = ma
12 = (14.5)a a= 12/14.5 = .83 m/s2
x = (0)(3) + ½(.83)(3)2
x = N

19. Sample Problem
A catcher stops a 41 m/s pitch in his glove, bringing it to rest in 0.15 m. If the force exerted by the catcher is 803 N, what is the mass of the ball?
vo = 41.1 m/s, v = 0 m/s,
x = 0.15 m, SF = 803 N
m = ?
m = SF/a need to find ‘a’, use v2 = vo2 +2ax
m = 803/(5603)
m = .14 kg

20. Sample Problem
A 747 jetliner lands and begins to slow to a stop as it moves along the runway. If its mass is 3.50 x 105 kg, its speed is 27.0 m/s, and the net braking force is 4.30 x 105 N
a) what is its speed 7.50 s later?
m = 350,000 kg, vo = 27 m/s, SF= x105 N
t = 7.5 s
v = ?
v = vo + at need to find ‘a’, use SF = ma
= (350000)a a = m/s2
v = 27 + ( )7.5
v = m/s

21. b) How far has it traveled in this time. x =. x = vot + ½at2 x = 27
b) How far has it traveled in this time? x = ? x = vot + ½at2 x = 27.5(7.5) + ½(-1.23)(7.5)2 x = – = m

23. Newton’s Third Law

24. Newton’s Third Law
For every action there exists an equal and opposite reaction.
If A exerts a force F on B, then B exerts a force of -F on A.

25. Examples of Newton’s 3rd Law
Copyright James Walker, “Physics”, 1st edition

26. Sample Problem You rest an empty glass on a table.
a) How many forces act upon the glass?
b) Identify these forces with a free body diagram.
c) Are these forces equal and opposite?
d) Are these forces an action-reaction pair?

27. Sample Problem
A force of magnitude 7.50 N pushes three boxes with masses m1 = 1.30 kg, m2 = 3.20 kg, and m3 = 4.90 kg as shown. Find the contact force between (a) boxes 1 and 2 and (b) between boxes 2 and 3.
m1 = 1.3 kg, m2 = 3.2 kg, m3 = 4.9 kg, SF = 7.5 N
Find total ‘a’ for all boxes. mT = 9.4 kg, using SF = ma
a=.798 m/s2
Copyright James Walker, “Physics”, 1st edition

28. Looking at box 1 only, there is only 1 force acting on it, from box 2.
F2 on 1
SF = F2 on 1 = m1a = 1.3(.798) = N

29. b. ) Looking at box 3 only, there are 2 forces acting on it
b.) Looking at box 3 only, there are 2 forces acting on it. Box 2 on 3 and the applied force.
Fapp
F2 on 3
SF = Fapp - F2 on 3 = m3a F2 on 3 = Fapp - m3a = 7.5 – 4.9( .798 ) = 3.58 N

30. Mass and Weight

31. Mass and Weight
Many people think mass and weight are the same thing. They are not.
Mass is inertia, or resistance to acceleration.
Weight can be defined as the force due to gravitation attraction.
W = mg

32. Sample Problem
A man weighs 150 pounds on earth at sea level. Calculate his
a) mass in kg. (2.2 lb. = 1 kg) Convert.
b) weight in Newtons.
w = mg

33. Apparent weight
If an object subject to gravity is not in free fall, then there must be a reaction force to act in opposition to gravity.
We sometimes refer to this reaction force as apparent weight.

34. Elevator rides
When you are in an elevator, your actual weight (mg) never changes.
You feel lighter or heavier during the ride because your apparent weight increases when you are accelerating up, decreases when you are accelerating down, and is equal to your weight when you are not accelerating at all.

35. Going Up? v = 0 a = 0 v > 0 a > 0 v > 0 a = 0 v > 0W
Wapp
Ground
floor
Normal feeling
v = 0
a = 0 W
Wapp
Just
starting up
Heavy feeling
v > 0
a > 0 W
Wapp
Between
floors
Normal feeling
v > 0
a = 0 W
Wapp
Arriving at
top floor
Light feeling
v > 0
a < 0
Going Up?

36. Going Down? v = 0 a = 0 v < 0 a < 0 v < 0 a = 0 v < 0
Wapp
Top
floor
Normal feeling
v = 0
a = 0 W
Wapp
Beginning
descent
Light feeling
v < 0
a < 0 W
Wapp
Between
floors
Normal feeling
v < 0
a = 0 W
Wapp
Arriving at
Ground floor
Heavy feeling
v < 0
a > 0
Going Down?

37. Sample Problem
An 85-kg person is standing on a bathroom scale in an elevator. What is the person’s apparent weight
when the elevator accelerates upward at 2.0 m/s2?
ay = 2 m/s2, m = 85 kg, g = 9.8 m/s2
Wapp = ?
SF = Wapp – W = may Find W: W = mg
Wapp = W + may
Wapp = (9.8*85) + 85(2)
Wapp =

38. b) when the elevator is moving at constant velocity between floors?
Constant velocity  ay = 0 m/s2
SF = Wapp – W = may
Wapp = W =
c) when the elevator begins to slow at the top floor at 2.0 m/s2?
ay = -2 m/s2
Wapp = ?
SF = Wapp – W = may Find W: W = mg
Wapp = W + may
Wapp = (9.8*85) + 85(-2)
Wapp =

39. Sample Problem
A 5-kg salmon is hanging from a fish scale in an elevator. What is the salmon’s apparent weight when the elevator is
at rest?
m = 5 kg, g = 9.8 m/s2, a = 0 m/s2
Wapp – W = may
Wapp = mg + ma = m(g + a)
Wapp = 5( )
Wapp =
b) moving upward and slowing at 3.2 m/s2? a = -3.2 m/s2
Wapp = m(g + a)
Wapp = 5( )

40. c) moving downward and speeding up at 3. 2 m/s2. a = 3
c) moving downward and speeding up at 3.2 m/s2? a = 3.2 m/s2 Wapp = m(g + a) Wapp = 5( ) Wapp =

41. Newton’s Laws in 2D

42. Newton’s 2nd Law in 2-D
The situation is more complicated when forces act in more than one dimension.
You must still identify all forces and draw your force diagram.
You then resolve your problem into an x-problem and a y-problem (remember projectile motion????).

43. Forces in 2-D
Copyright James Walker, “Physics”, 1st edition

44. Forces in 2-D
Copyright James Walker, “Physics”, 1st edition

45. Forces in 2-D
Copyright James Walker, “Physics”, 1st edition

46. Forces in 2-D
Copyright James Walker, “Physics”, 1st edition

47. Forces in 2-D
Copyright James Walker, “Physics”, 1st edition

48. Sample Problem
You are moving a chair with a mass of 50 kg, with a force of 35 N at an angle of 24 degrees. A: draw a free body diagram. B. What is the normal force? C. how fast will the chair be moving after 4 seconds if friction is ignored?

49. Normal Force

50. Normal force
The normal force is a force that keeps one object from penetrating into another object.
The normal force is always perpendicular a surface.
The normal exactly cancels out the components of all applied forces that are perpendicular to a surface.

51. Normal force on flat surface
The normal force is equal to the weight of an object for objects resting on horizontal surfaces.
N = W = mg N mg

52. Normal force on ramp N mg N = mgcos  
The normal force is perpendicular to angled ramps as well. It’s always equal to the component of weight perpendicular to the surface.
N = mgcos N 
mgcos
mgsin mg 

53. Normal force not associated with weight.
A normal force can exist that is totally unrelated to the weight of an object.
applied force
friction
weight
normal
N = applied force

54. The normal force most often equals the weight of an object…

55. …but this is by no means always the case!

56. More on the Normal Force

57. Draw a free body diagram for the skier.

58. Sample problem
A 5.0-kg bag of potatoes sits on the bottom of a stationary shopping cart. Sketch a free-body diagram for the bag of potatoes. Now suppose the cart moves with a constant velocity. How does this affect the free-body diagram?

59. Sample problem
Find the normal force exerted on a 2.5-kg book resting on a surface inclined at 28o above the horizontal.
m = 2.5 kg, g = 9.8 m/s2,  = 28o
N = ?
N = mgcos
N = 2.5(9.8)cos(28)
b) If the angle of the incline is reduced, do you expect the normal force to increase, decrease, or stay the same?

60. Sample problem
A gardener mows a lawn with an old-fashioned push mower. The handle of the mower makes an angle of 320 with the surface of the lawn. If a 249 N force is applied along the handle of the 21-kg mower, what is the normal force exerted by the lawn on the mower?

61. Sample problem
Larry pushes a 200 kg block on a frictionless floor at a 45o angle below the horizontal with a force of 150 N while Moe pulls the same block horizontally with a force of 120 N.
a) Draw a free body diagram.
b) What is the acceleration of the block?
c) What is the normal force exerted on the block?