Unit 5: Newton’s Third Law of Motion
Did you know that when an apple falls from a tree, the whole Earth moves slightly up toward the falling apple? After completing this unit, you should be able to explain how forces are always created in pairs.
Learning Outcomes
By the end of this unit, you should be able to
- demonstrate how forces are the result of interactions.
- state Newton’s third law of motion.
- explain the difference between scalar and vector quantities and give examples of each.
eText Material
Reading Assignment
Read the following sections in Chapter 5 of the eText:
- 5.1: Forces and Interactions
- 5.2: Newton’s Third Law of Motion
- 5.3: Action and Reaction on Different Masses
- 5.4: Vectors and the Third Law
- 5.5: Summary of Newton’s Three Laws
Supplementary learning resources are available on the Mastering Physics learning platform.
Additional Reading
Newton’s Third Law of Motion
Of Newton’s laws of motion, the one most often quoted outside of physics is his third law of motion, which can be stated as follows:
To every action there is always an opposed equal reaction.[1]
Even though the statement appears to be quite simple, this law requires some thinking before you use it. The point to remember is that the action and reaction forces always act on different objects—never on the same one. If you apply a force on the wall, it reacts with an equal and opposite force on you. The wall experiences the action force, and you experience the reaction force. Therefore, an alternative statement of the third law is as follows:
Whenever one object exerts a force on a second object, the second object simultaneously exerts an equal and opposite force on the first.[2]
Figure 5.1: Action and reaction forces on a book at rest on a tabletop.
To see this concept in action, analyze what goes on when a book is placed on a tabletop, as shown in Figure 5.1. First, there is the force of gravity (or weight) $F_g = mg$, which acts on the book in the downward direction. When placed on the table, the book presses downward with equal force ($F = F_g$) on the tabletop, which in turn exerts an upward reaction force (called the normal force $N$) on the book. According to Newton’s third law, the two forces $F$ and $N$ form an action–reaction pair and have equal magnitudes (i.e., $F = N$), while the directions are exactly opposite as shown.
Note that $F_g$ and $N$ do not form an action–reaction pair because they both act on the same object (the book). Actually, the book’s weight forms an action–reaction pair with a force not shown in the diagram, which is the upward force of gravity the book applies on the Earth! (Newton’s universal law of gravitation will be covered in Unit 9.)
Video 5.1: Newton’s third law of motion.
Footnotes
[1] Hewitt, Paul G. (2021). Conceptual Physics (13th ed., p. 87) [Ebook]. Pearson.
[2] Hewitt, Paul G. (2021). Conceptual Physics (13th ed., p. 87) [Ebook]. Pearson.
Questions
The following questions are selected from the end of Chapter 5 of the eText. It is important to your learning that you try to answer each question independently before you read through the answer and explanation given.
For questions that ask you to explain or discuss your answer, the response revealed by the Answer button would earn you only partial marks on a quiz or exam in this course. Use the Answer to help you formulate a complete answer before you select the Explanation button to check your work.
Note that some questions have been adapted to suit the format of this course.
Chapter 5
Question 8
To produce a net force on a system, must there be an externally applied net force?
Answer
Yes.
Explanation
The net force on a system is the resultant of all the external forces applied on the system. According to the Newton’s third law, internal forces are generated in pairs, which act on different parts of the system and, therefore, cancel each other.
Chapter 5
Question 10
Earth pulls down on you with a gravitational force that you call your weight. Do you pull up on Earth with the same amount of force?
Answer
Yes.
Explanation
According to Newton’s third law, forces are created in pairs. So when you experience a downward force due to gravitational attraction, Earth experiences an equal force in the opposite direction. However, because of its relatively huge mass, Earth responds with unnoticeable upward acceleration toward you.
Chapter 5
Question 30
If you stand next to a wall on a frictionless skateboard and push the wall with a force of 30 N, how hard does the wall push on you? If your mass is 60 kg, show that your acceleration is $0.5\,\text{m}/\text{s}^2$.
Answer
According to Newton’s third law, the wall pushes back on you with a reaction force equal to 30 N. Since you stand on a frictionless skateboard, no other horizontal forces act on you. Thus, your acceleration, calculated using Newton’s second law, is given as \begin{align} a &= \frac{F}{m} \nonumber\\[6pt] &= \frac{30\,\text{N}}{60\,\text{kg}} \nonumber\\[6pt] &= 0.5\,\text{m}/\text{s}^2 \end{align}
Chapter 5
Question 31
Forces of 3.0 N and 4.0 N act at right angles on a 2.0 kg block. Show that the acceleration of the block is $2.5\,\text{m}/\text{s}^2$.
Diagram
Figure Q5.31
Answer
The vector representation of the resultant force on the block is drawn using the parallelogram rule. Referring to Figure Q5.31, you can determine the magnitude of the resultant (or net) force as follows: \begin{align} F_\text{net} &= \sqrt{(3.0\,\text{N})^2 + (4.0\,\text{N})^2} \nonumber\\[6pt] &= \sqrt{25\,\text{N}} \nonumber\\[6pt] &= 5.0\,\text{N} \end{align} So you can calculate the acceleration of the block as follows: \begin{align} a &= \frac{F_\text{net}}{m} \nonumber\\[6pt] &= \frac{5.0\,\text{N}}{2.0\,\text{kg}} \nonumber\\[6pt] &= 2.5\,\text{m}/\text{s}^2 \end{align}
Chapter 5
Question 38
Identify action and reaction forces for each of the following interactions:
- A hammer hits a nail.
- Earth gravity pulls down on a book.
- A helicopter or drone blade pushes air downward.
Answer
The hammer, Earth, and the helicopter or drone blade each receive an upward reaction force.
Explanation
- The hammer exerts a force on the nail; the nail reacts with an equal upward force on the hammer, which brings it to a halt.
- Gravitational force pulls the book down; the book reacts with an equal force that pulls Earth up.
- The rotating helicopter (or drone) blade pushes air down; the compressed air reacts with equal upward reaction force on the blade, which keeps the helicopter from falling.
Chapter 5
Question 45
You drop a rubber ball on the floor, and it bounces back up. What force acts on the ball to provide the bounce?
Answer
The upward reaction force exerted by the floor on the ball.
Explanation
When the ball pushes down on the floor during impact, the floor reacts with an equal upward force, which causes the ball to accelerate sharply in the upward direction. As a result, the ball slows down quickly, comes to a stop, and gains an upward velocity, which causes the ball to bounce back in the upward direction. All this occurs during the relatively short time interval while the ball is in contact with the floor.
Chapter 5
Question 54
When an athlete holds a barbell overhead, the reaction force is the weight of the barbell on his hand. How does this force vary when the barbell is accelerated upward? Accelerated downward?
Answer
The barbell feels heavier when accelerated upward and lighter when accelerated downward.
Diagram
Figure Q5.54
Explanation
To accelerate the barbell upward, the athlete pushes up with a force ($F$) greater than the force of gravity (the barbell’s weight). As a result, the downward reaction force exerted by the barbell on the athlete’s hand is greater than the barbell’s weight.
When the athlete pushes up with a force smaller than the force of gravity, the barbell accelerates downward. In this case, the downward reaction force exerted on the athlete’s hand is less than the barbell’s weight (see Figure Q5.54). This is why it is harder to push a barbell up than to bring it down.
Chapter 5
Question 55
You push a heavy car by hand. The car, in turn, pushes back with an opposite but equal force on you. Doesn’t this mean that the forces cancel one another, making acceleration impossible? Why or why not?
Answer
The two forces do not cancel one another because they act on different objects.
Explanation
The acceleration of an object is determined by the net force applied on the object. However, the forces mentioned in this question do not act on the same object. One force pushes the car forward, while the other force acts on your hand in the backward direction. Therefore, the acceleration of the car is possible.
Chapter 5
Question 56
The strong man will push the two initially stationary freight cars of equal mass apart before he himself drops straight to the ground. Is it possible for him to give either of the cars a greater speed than the other? Why or why not?
Figure Q5.56(a)
Answer
No, it is not possible.
Diagram
Figure Q5.56(b)
Explanation
The two forces exerted by the strong man on the freight cars (see Figure Q5.56(b)) are action–reaction forces and, therefore, are equal in magnitude. Since the two cars have equal masses, they will also have the same acceleration, resulting in equal speeds. Here, of course, you assume that both cars experience equal (or no) frictional forces.
Chapter 5
Question 58
A Mack truck and Honda Civic have a head-on collision. Which vehicle experiences the greater impact force? Which vehicle experiences the greater deceleration? Defend your answers.
Answer
The impact force is the same for both cars, and the Honda Civic experiences greater acceleration.
Explanation
According to Newton’s third law, the impact forces on the two vehicles are action–reaction forces and, therefore, are equal in magnitude. Based on Newton’s second law ($a = F/m$), the lighter Honda Civic experiences greater deceleration than the more massive Mack truck.
Chapter 5
Question 68
A stone is shown at rest on the ground. The vector shows the weight of the stone.
- Complete the vector diagram showing another vector that results in zero net force on the stone.
- What is the conventional name of the vector you have drawn?
Figure Q5.68(a)
Diagram
Figure Q5.68(b)
Answer
- While at rest on the surface of the ground, the stone pushes down on the ground with a force equal to the stone’s weight, while the ground reacts with an upward support force of equal magnitude. Figure Q5.68(b) shows the two forces that act on the stone while at rest.
- The upward support force is conventionally referred to as the “normal force.”
Chapter 5
Question 73
Here the stone is sliding down a friction-free incline.
- Identify the forces that act on it, and draw appropriate force vectors.
- By the parallelogram rule, construct the resultant force on the stone (carefully showing its direction parallel to the incline—the same direction as the stone’s acceleration).
Figure Q5.73(a)
Diagram
Figure Q5.73(b)
Answer
- The two forces that act on the stone are the force of gravity (or weight) $mg$ directed downward and the normal (or support) force $N$ directed perpendicular to the inclined surface.
- Using the parallelogram method, you draw the resultant-force vector parallel to the inclined plane, as shown in Figure Q5.73(b). Note that the stone accelerates down the incline in the same direction as the resultant force.
Chapter 5
Question 95
Two 100-N weights are attached to a spring scale as shown. Does the scale read 0, 100, or 200 N, or register some other reading? (Hint: Would it read any differently if one of the ropes were tied to the wall instead of to the hanging 100-N weight?)
Figure Q5.95
Answer
100 N
Explanation
The tension in each rope must be equal to 100 N to balance the weight attached to it. Therefore, the scale reads 100 N.
If one of the ropes were tied to the wall, the scale would not feel the difference, as each rope would continue to experience a tension of 100 N.
Exercises
Spend some time completing the following exercises to test your understanding of the main concepts in Chapter 5 and increase your efficiency in answering exam questions.
End-of-Chapter Practice Questions
Answer questions 5, 11, 13, 15, 17, 27, 29, 33, 35, 37, 47, 57, 61, 67, 69, 75, 81, and 91 in Chapter 5 of the eText. If you require assistance, please contact your tutor. The answers are provided at the end of the eText.
Quiz 1
Before moving on to Unit 6, complete Quiz 1, which covers Units 1–5. For more information and to take the quiz, see the Quizzes section on the course home page. This quiz is worth 10% of your course grade.