1. Newton’s Third Law

2. Sec. 6.3 Interaction Forces
Objectives
Explain the meaning of interaction pairs (third law pairs) of forces and how they are related by Newton’s third law
List the four fundamental forces and illustrate the environment in which each can be observed.
Explain the tension in ropes and strings in terms of Newton’s third law

3. Newton’s laws of motion
The law of inertia. An object in motion remains in motion with constant velocity if the net force on the object is zero.
Force and acceleration. If the net force acting on an object of mass m is F, then the acceleration of the object is a = F/m. Or, F = ma.
Action and reaction. For every action there is an equal but opposite reaction.
Here again are Newton’s three laws of motion.
{READ}
= The third law is a statement about forces rather than a statement about motion. But it is called a law of motion because this property of forces affects the way mechanical systems move.
The words “action” and “reaction” are Newton’s original terminology. These terms are really archaic. In fact, this phraseology may be downright misleading.
What did Newton mean by “action” and “reaction”?
{APPEAR last}
Action means a force. Reaction means another force. So in any interaction there are two forces. Which one is called the action and which one is called the reaction is very arbitrary.
The archaic terminology of “action” and “reaction” suggests, wrongly, that one occurs first and then the other occurs. On the contrary, the two forces occur simultaneously. Either one can be considered the reaction force of the other.
= We need to restate Newton’s third law in a modern and more clear way.
Action means force.

4. Newton’s 3rd Law
For every action force there is an equal and opposite reaction force
(You cannot touch without being touched)

5. Action-reaction Pair
If object A exerts a force on object B, then object B exerts an equal and opposite force on object A. The pair of forces (due to one interaction), is called an action/reaction pair.
NOTA BENE:
The action/reaction pair will never appear in the same free body diagram.

6. Summary: Interaction forces (‘Third Law Pairs’)
Two forces that are in opposite directions and have equal magnitude
They never act on the same object!
(Therefore never in same FBD!)
General Form is:
FA on B = -FB on A

7. For every action there is an equal but opposite reaction.
Newton’s third law
Whenever one object exerts a force on another object, the second object exerts an equal but opposite force on the first object.
force of A on B
force of B on A
Forces always occur like this, in pairs.
{READ first line}
But here is a more modern statement…
{READ purple}
The figure shows attractive forces between two objects, as a specific illustration. The blue object A exerts a force on the red object B and B exerts an equal (but opposite) force on A. In this example, the objects attract each other. It would be the same for repulsion.
{READ: Forces always occur like this, in pairs. Etc}
DEMO [1/4]
Two spring balances; need two volunteers – a big hulk and a little coed. Each pulls on one of the spring balances and they both register the same force. Ask the girl to hold still while the big guy pulls harder. But the forces are always measured to be equal!
We will see that this is very hard to accept! It is just not common sense. That is why it took a great genius like Newton to figure it out.

8. Reaction: road pushes on tire Action: tire pushes on road

9. When you walk or run, what forces occur?
At constant velocity the horizontal force is 0 and you continue to move because of inertia.
To accelerate, you push backward against the floor; the reaction force, which is a friction force exerted by the floor on your foot, pushes you forward.
This reaction force may be hard to visualize, but imagine what would happen if you were on a frictionless surface – can’t accelerate!
{READ it all}
This idea that the Earth pushes you forward when you walk (or, rather, when you accelerate while walking) is really hard to comprehend. I guess the reason it is so difficult to see is that the Earth does not move, at least not perceptibly. How can you be pushed forward by something that does not move?
DEMO
A meter stick. (1) Stand it up and let it fall. Because of the frictional force, the center of mass moves forward. (2) Now put some soapy water on the bench and stand up the meter stick. When it falls, it falls straight down.

10. Recoil force
Force accelerating bullet
a = F/m
a = F/m

11. A bug and a car collide. Which experiences the greater force?
(b) car
(c) neither, they both experience the same magnitude of force
(c) neither, they both experience the same magnitude of force

12. Consider hitting a baseball with a bat
Consider hitting a baseball with a bat. If we call the force applied to the ball by the bat the action force, identify the reaction force.
(a) the force applied to the bat by the hands
(b) the force applied to the bat by the ball
(c) the force the ball carries with it in flight
(d) the centrifugal force in the swing
(b) the force applied to the bat by the ball

13. Which vehicle exerts a greater force ― the tow truck or the car?
Newton’s third law concerns “interactions” --- the forces when two objects are exerting forces on one another.
{READ}
Most people will answer this question incorrectly.
Do you think the answer is obvious? Then you probably have the wrong answer.
We’ll see what Newton’s third law says about it.

14. A puzzle:
The truck pulls to the right. According to Newton’s third law, the car pulls to the left with an equal force. So how can they start moving, or accelerate?
Part 4 – Puzzles (15 minutes)
Newton’s third law is not plain common sense. On the contrary, it can be very puzzling – it’s a mystery of the physical world.
{READ}
Start with the car – {appear the force on the car}.
Now, what about the truck?
{Appear the two forces on the truck.}
The idea that the road exerts a force on the tires of the truck, to cause the two vehicles to accelerate forward, is hard to visualize; because the road is at rest. But it is obviously correct: What if the truck were located on a patch of very slippery ice; then the vehicles would not be able to accelerate forward. The force of the road would be zero in that case.
Resolution: Consider each part separately, and don’t forget that other forces are also acting.

15. Playing catch with a medicine ball
A throws the ball and B catches it.  four forces
When A throws the ball he exerts a force on the ball (toward the right) and the ball exerts a force on him so he recoils (toward the left).
► Newton’s third law for the throw
DEMO [2/4]
= Playing catch with a medicine ball.
Need two volunteers – people with good coordination and pretty strong and not too large.
When B catches the ball he exerts a force on the ball (toward the left to stop it) and the ball exerts a force on him so he is knocked back (toward the right).
► Newton’s third law for the catch

16. Example – A collision
Let’s work through an example that depends on Newton’s third law.
{READ}
Maybe the final answer seems obvious. But the detailed explanation shows how Newton’s third law is involved.

17. The force exerted by the ball on the toe (reaction) is equal to the force exerted by the toe on the ball.
{READ}
Is this common sense? You can see the big force on the ball by the fact that the ball is significantly deformed at the instant of the kick. This is not an under-inflated ball, or a ball with no air pressure. It is a normally pressurized football. But in the instant of the kick it is very much deformed by the force of the foot. (You can’t see this with your eye – your eye is too slow.)
Now, at that instant there is an equal force (in the other direction) exerted by the ball on the toe!
In this, and all the examples, the action-reaction forces are equal. [lead in to next slide]
Really hard to accept!

18. A puzzle : Tug of War Which team will end up in the puddle?
String tension
String tension
Contact force
Contact force
Which team will end up in the puddle?
Here’s another puzzle.
{READ all}
The weightlifters win because of the frictional force of the ground, which is greater on the weightlifters than on the gymnasts.
The net force on the weightlifters is toward the left, because the friction of the ground is stronger than the tension of the rope.
The net force on the gymnasts is toward the left, because the friction of the ground is less than the tension of the rope.
By Newton’s third law the force of the rope is the same on both teams (equal but opposite). The frictional forces of the ground are different – that’s why the weightlifters win the tug of war.
Who would win if you put the weightlifters on slippery ice but the gymnasts on hard ground? Then the frictional force of the ground would put the weightlifters in the puddle.
But aren’t the forces equal but opposite !?
Resolution: Don’t forget that there are other forces acting.
Each team exerts a force on the Earth, so the Earth exerts a force on the team (3rd law!). The net force on either team is toward the left.

19. A puzzle
Horse and Cart
According to Newton, the cart pulls the horse with a force –A (to the left).
The horse pulls the cart with a force A (to the right).
{Blackboard:
The horse: F1 = - A + friction = m_horse a
The cart: F2 = +A = m_cart a
The accelerations are the same.
The directions of the net forces are the same, toward the right.
The magnitudes of the net forces are different.
The action/reaction pair are equal but opposite.
The idea that the Earth exerts a force on the horse’s hooves is hard to visualize, because the Earth is at rest. But it is obviously correct: What if the horse were standing on slippery ice; then he would not be able to pull the cart.
So how can they start moving, or accelerate?
Resolution: Consider each part separately, and don’t forget that there are other forces acting.

20. car. Which of the following statements is true?
A small car is pushing a larger truck that has a dead battery. The mass of the truck is larger than the mass of the
car. Which of the following statements is true?
The truck exerts a larger force on the car than the car exerts on the truck.
The truck exerts a force on the car but the car doesn’t exert a force on the truck.
The car exerts a force on the truck but the truck doesn’t exert a force on the car.
The car exerts a larger force on the truck than the truck exerts on the car.
The car exerts the same amount of force on the truck as the truck exerts on the car.
Answer: E 20

21. car. Which of the following statements is true?
A small car is pushing a larger truck that has a dead battery. The mass of the truck is larger than the mass of the
car. Which of the following statements is true?
The truck exerts a larger force on the car than the car exerts on the truck.
The truck exerts a force on the car but the car doesn’t exert a force on the truck.
The car exerts a force on the truck but the truck doesn’t exert a force on the car.
The car exerts a larger force on the truck than the truck exerts on the car.
The car exerts the same amount of force on the truck as the truck exerts on the car.
STT8.6 21

22. Newton’s Third Law Pairs

23. Example - analyzing interacting objects
A person pushes a large crate across a rough surface.
Identify the objects that are systems of interest
Draw free-body diagrams for each system of interest.
Identify all action/reaction pairs with a dashed line.

24. Forces involved in pushing a crate – FBD of person and crate

25. Propulsion Force
The force label fp shows that the static friction force on the person is acting as a propulsion force.
This is a force that a system with an internal source of energy uses to drive itself forward.

26. Propulsion forces

27. Free Body Diagrams – Exercises
Draw a freebody diagram of each object in the interacting system.
Show action/reaction pair with red/orange dotted lines.
Draw force vectors in another color.
Label vectors with standard symbols.
Label action/reaction pairs FAonB , FBonA for example.

28. A fishing line of negligible mass lifts a fish
upward at constant speed. The line and the fish are the system, the fishing pole is part of the environment. What, if anything, is wrong with the free-body diagrams?
Answer: C 28

29. A fishing line of negligible mass lifts a fish
upward at constant speed. The line and the fish are the system, the fishing pole is part of the environment. What, if anything, is wrong with the free-body diagrams?
The gravitational force and the tension force are incorrectly identified as an action/reaction pair. The correct action reaction pair is…?
Action/reaction pairs are never on the same free body diagram.
Mass of line considered negligible so no weight force necessary.
STT7.1 29

30. Boxes A and B are sliding to the right across a frictionless table
Boxes A and B are sliding to the right across a frictionless table. The hand H is slowing them down. The mass of A is larger than the mass of B. Rank in order, from largest to smallest, the horizontal forces on A, B, and H. Ignore forces on H from objects not shown in the picture.
Answer: C 30

31. Recognize any third law pairs?
Draw a FBD for each object: A, B, and H.
Only consider horizontal forces in this problem. A B C
FAB
FHB
FBH
FBA
Answer: C
Recognize any third law pairs? 31

32. Boxes A and B are sliding to the right across a frictionless table
Boxes A and B are sliding to the right across a frictionless table. The hand H is slowing them down. The mass of A is larger than the mass of B. Rank in order, from largest to smallest, the horizontal forces on A, B, and H.
STT8.3
FB on H = FH on B > FA on B = FB on A
from Newton’s 2nd and 3rd Laws 32

33. Problem-Solving Strategy: Interacting-Objects Problems

34. Challenge Problem:
Two strong magnets each weigh 2 N and are on opposite sides of the table. The table, by itself, has a weight of 20 N. The long range-range attractive force between the magnets keeps the lower magnet in place. The magnetic force on the lower magnet is 3 times its weight.
a. Draw a FBD for each magnet and table. Use dashed lines to connect all action/reaction pairs.
b. Find the magnitude of all forces in your FBD and list them in a table.

35. Challenge Problem ↓ (Wg) E,U ↓ (Wg) E,T ↓ (Wg) E,L ↑ N T,U ↑ N S,T
Upper Table Lower
↓ (Wg) E,U
↓ (Wg) E,T
↓ (Wg) E,L
↑ N T,U
↑ N S,T
↑ LM U,L
↓ LM L,U
↓ N U.T
↓ N T,L -
↑ N L,T

36. Find Third Law Pairs ↓ (Wg) E,U ↓ (Wg) E,T ↓ (Wg) E,L ↑ N T,U ↑ N S,T
Upper Table Lower
↓ (Wg) E,U
↓ (Wg) E,T
↓ (Wg) E,L
↑ N T,U
↑ N S,T
↑ LM U,L
↓ LM L,U
↓ N U.T
↓ N T,L -
↑ N L,T

37. FBDs for EOC 8

38. Challenge Problem: Answer

39. Ranking Task – Pushing blocks
Block 1 has a mass of m, block 2 has a mass of 2m, block 3 has a mass of 3m. The surface is frictionless.
Rank these blocks on the basis of the net force on each of them, from greatest to least. If the net force on each block is the same, state that explicitly.

40. Ranking Task – Pushing blocks
Answer:
Reason: ΣF = ma. Acceleration is equal for all blocks.

41. Ranking Problem Example
Block 1 has a mass of 1 kg, block 2 has a mass of 2 kg, block 3 has a mass of 3 kg. The surface is frictionless.
a. Draw a fbd for each block. Use dashed lines to connect all action/ reaction pairs.
b. How much force does the 2-kg block exert on the 3-kg block?
c. How much force does the 2-kg block exert on the 1-kg block?

42. Ranking Problem Example – Answer:
b. How much force does the 2-kg block exert on the 3-kg block? – 6N
c. How much force does the 2-kg block exert on the 1-kg block? – 10N

43. Newton’s Third Law

44. Acceleration constraint
An acceleration constraint is a well-defined relationship between the acceleration of 2 (or more) objects.
In the case shown, we can assume ac =aT = ax

45. Is there an acceleration constraint in this situation
Is there an acceleration constraint in this situation? If so, what is it? The pulley is considered to be massless and frictionless.

46. Answer: Acceleration constraint is: aA = -aB The actual signs may not be known until the problem is solved, but the relationship is known from the start.

47. Exercises 17

48. answers
a2kg = -.5a1kg

49. Newton’s Third Law

50. Page 163, 2nd Ed (Found in Chapter 5, 1st ed)

51. Interacting systems problem (EOC #35)
A rope attached to a 20 kg wooden sled pulls the sled up a 200 snow-covered hill. A 10 kg wooden box rides on top of the sled. If the tension in the rope steadily increases, at what value of tension will the box slip?

52. Interacting systems problem (EOC #35)
What are the objects of interest? What kind of axes for the FBD for each? Acceleration constraints? Draw FBDs, with 3rd law pairs connected with dashed lines.
Find the max tension in the rope, so the box does not slip.

53. Box: 3 forces
Sled: 6 forces
0.06

54. Return to sled equations with newfound booty.
I suggest starting with the equations for the sled, since the unknown of interest is found there.
Identify quantities in sled equations that you can find by solving box equations.
Solve box equations.
Return to sled equations with newfound booty.
Plug and chug.
0.06

55. Interacting systems problem (EOC #35)
A rope attached to a 20 kg wooden sled pulls the sled up a 200 snow-covered hill. A 10 kg wooden box rides on top of the sled. If the tension in the rope steadily increases, at what value of tension will the box slip?
Answer: 155 N.

56. The Massless String Approximation
A horizontal forces only fbd for the string:
TAonS
TBonS ●
ΣF = TBonS – TAonS = ma. If string is accelerating to the right
TBonS = TAonS + ma

57. The Massless String Approximation
Often in physics and engineering problems the mass of the string or rope is much less than the masses of the objects that it connects. In such cases, we can adopt the following massless string approximation:
This allows the objects A and B to be analyzed as if they exert forces directly on each other.

58. Pulleys
If we assume that the string is massless and the pulley is both massless and frictionless, no net force is needed to turn the pulley. TAonB and TBonA act “as if” they are an action/reaction pair, even though they are not acting in opposite directions.

59. Pulleys
In this case the Newton’s 3rd law action/reaction pair point in the same direction!
T 100kg on m
Tm on 100kg

60. All three 50 kg blocks are at rest
All three 50 kg blocks are at rest. Is the tension in rope 2 greater than, less than, or equal to the tension in rope 1?
Equal to
Greater than
Less than
Answer: A 60

61. All three 50 kg blocks are at rest
All three 50 kg blocks are at rest. Is the tension in rope 2 greater than, less than, or equal to the tension in rope 1?
Equal to
Greater than
Less than
STT8.4 61

62. In the (moving) figure to the right, is the tension in the string greater than, less than, or equal to the weight of block B?
Answer: C
Equal to
Greater than
Less than 62

63. In the figure to the right, is the tension in the string greater than, less than, or equal to the weight of block B?
Equal to
Greater than
Less than 63

64. Interacting systems problem (EOC #40)
A 4.0 kg box (m) is on a frictionless 350 incline. It is connected via a massless string over a massless, frictionless pulley to a hanging 2.0 kg mass (M). When the box is released:
Which way will it go George?
What is the tension in the string?
4.0 kg
350

65. Interacting systems problem (EOC #40)
a. Which way will it go? Even if you have no clue, follow the plan!
What are the objects of interest? What kind of axes for the FBD for each? Acceleration constraints?? Draw FBDs, with 3rd law pairs connected with dashed lines.
4.0 kg
350

66. Interacting systems problem (EOC #40)
How do you figure out which way the system will move, once m is released from rest?
massless string approx. allows us to join the tensions as an “as if” interaction pair

67. Interacting systems problem (EOC #40)

68. Interacting systems problem (EOC #40)
a = m/s2, T = 21 N.
Which way is the system moving?
How does the tension compare to the tension in the string while the box was being held?
Greater than, less than, equal to?

69. EOC # 33
The coefficient of static friction is 0.60 between the two blocks in the figure. The coefficient of kinetic friction between the lower block and the floor is Force F causes both blocks to slide 5 meters, starting from rest. Determine the minimum amount of time in which the motion can be completed without the upper block slipping.

70. EOC # 33
The coefficient of static friction is 0.60 between the two blocks in the figure. The coefficient of kinetic friction between the lower block and the floor is Force F causes both blocks to slide 5 meters, starting from rest. Determine the minimum amount of time in which the motion can be completed without the upper block slipping.

71. EOC # 33
amax = 3.27 m/s2
tmin = 1.75 s (time is important in this one)

72. EOC # 46
Find an expression for F, the magnitude of the horizontal force for which m1 does not slide up or down the wedge. This expression should be in terms of m1, m2 , θ, and any known constants, such as g. All surfaces are frictionless.

73. EOC # 46 .

74. EOC # 46 .
F = (m1 + m2) g tan θ

75. EXAMPLE 7.6 Comparing two tensions
QUESTION:

76. EXAMPLE 7.6 Comparing two tensions

77. EXAMPLE 7.6 Comparing two tensions

78. EXAMPLE 7.6 Comparing two tensions

79. EXAMPLE 7.6 Comparing two tensions

80. The spring force ---another example of 3rd law
Suppose the spring is stretched beyond its equilibrium length by a length x.
The force on the mass m1 is F1 = +kx. 
(k = Hooke’s constant)
The force on the mass m2 is F2 = -kx. 
( + means to the right; - means to the left.)
Part 3 - Spring force (10 minutes)
Another example of a force is the spring force.
{READ}
So the spring force obeys Newton’s third law.
The forces are equal but opposite.

81. Example
One end of a spring is attached to a wall. When the other end is pulled with a force of 50 N, the spring is stretched by 3 cm. What force would be required to stretch the spring by 5.5 cm?
Answer: 91.7 N
Let’s do an example problem.
{READ the problem; READ Hooke’s law}
EXAMPLE # 2
F = k x
From the information, k = F/x = (50 N) / (3 cm) = N/cm.
Now calculate the force for the larger displacement,
F = k x = (16.67 N/cm) * (5.5 cm) = 91.7 N.
DEMO [4/4]
A mass hanging from a spring. What happens as I add mass?
Hooke’s law: The strength of a spring force is proportional to the displacement (extension or compression). F = k x where k is called Hooke’s constant for the spring.

82. Forces obey Newton’s third law.
We’ll consider two examples:
The force of universal gravitation
The spring force
{READ}

83. Universal Gravitation --- an example of Newton’s third law
Part 2 – Universal Gravitation (20 minutes)
Newton’s third law is a statement about forces. Here is a specific example – gravity.
Newton’s Law of Universal Gravitation states that any two of masses in the universe attract each other. The Earth pulls an apple down to the ground; anyone could see that. The apple pulls the Earth upward toward the apple; it took a great genius like Newton to see that.
{READ slide; inverse square law; you’ll need the equation for CAPA; man and earth – gravity obeys the third law}
So this theory of universal gravitation was another remarkable discovery of Isaac Newton. But how could such a sweeping statement be accepted as a scientific truth?
(next slide!)

84. The Earth pulls the apple down (“action”).
The apple pulls the Earth up (“reaction”).
The two forces are equal (but opposite).

85. When does a scientific theory become accepted as true?
How could Newton’s theory of Universal Gravitation become accepted as scientific truth?
Well, for one thing, Newton explained the motion of the planets based on this theory. No one else could explain planetary motions by any competing theory. Newton explained certain other natural phenomena based on this theory – comets and the tides. No other theory could explain those phenomena. So Newton’s theory of universal gravitation became accepted as true in his time.
But this is a general question: When does a theory become a fact? Can a theory ever be accepted as a fact. Is the theory of evolution in biology a theory or a fact? Obviously it’s a theory. But when can you say the theory is true?
In physics, a theory can be tested in the laboratory, to test whether it is true. Can the Law of Universal Gravitation be tested in the laboratory?

86. Henry Cavendish, 1798 : first measurement of G
For a laboratory measurement, the gravitational force is really very weak.
The force on the 1 kg mass is +3.3 x N.
The force on the 5 kg mass is –3.3 x N.
( + means to the right, i.e., increasing x)
{READ the first line}
We tend to think of gravity as a strong force but that is because we experience the force exerted by an entire planet – Earth. But the force between laboratory masses is very weak.
{Read …}
{Blackboard: F = G m1 m2 /r2 equal but opposite}
The first measurement of Newton’s constant G was by Henry Cavendish in 1798, long after the death of Isaac Newton (71 years later). Why was the measurement so difficult? You can see from the value that the force is very weak.
The Cavendish experiment provides a direct laboratory test of Newtonian gravitation.
Henry Cavendish, 1798 : first measurement of G

87. What makes g?
As long as we are considering gravity, let’s take a little detour and learn something more about gravity. We’ll come back to Newton’s third law in a few minutes. But first, what makes g?
Newton’s law of universal gravitation is a fundamental theory.
How does it determine little g, the acceleration due to Earth’s gravity?
DEMO [3/4] Drop a ball
What makes g?
{READ}
9.81 m/s2

88. Weighing the Earth
Calculate the mass of the Earth.
The force of gravity on m is, by definition, its weight,
By Newton’s theory of universal gravitation,
Let’s do an example problem: Calculate the mass of the Earth.
How can we determine the mass of he Earth? You can’t put it on scales – it’s too big. So, how do we know the mass of the Earth? We know it from Cavendish’s measurement of G.
EXAMPLE # 1
Calculate the mass of the Earth.
Blackboard: GMm/R^2 = mg implies M = gR^2/G
Henry Cavendish did this very calculation, and deduced that the average density of the Earth is 5.5 times greater than the density of water, i.e., 5.5 g/cc.
This is interesting, because the density of various common rocks ranges from 2.5 to 4.5 g/cc. Evidently the Earth is not just made of rock. The density of iron is 7.87 g/cc. We can conclude that the Earth has a large iron core. We can’t go to the center of the earth, but we can know that the iron core exists from the calculation.
the mass of the Earth, relying on the Cavendish measurement

89. Quiz Question The planet is pulled toward the moon (and vice versa).
Calculate the gravitational force on the planet.