4.4 Newton's Third Law of Motion

Section Key Terms

Describing Newton’s Third Law of Motion

If you have ever stubbed your toe, you have noticed that although your toe initiates the impact, the surface that you stub it on exerts a force back on your toe. Although the first thought that crosses your mind is probably “ouch, that hurt” rather than “this is a great example of Newton’s third law,” both statements are true.

This is exactly what happens whenever one object exerts a force on another—each object experiences a force that is the same strength as the force acting on the other object but that acts in the opposite direction. Everyday experiences, such as stubbing a toe or throwing a ball, are all perfect examples of Newton’s third law in action.

Newton’s third law of motion states that whenever a first object exerts a force on a second object, the first object experiences a force equal in magnitude but opposite in direction to the force that it exerts.

Newton’s third law of motion tells us that forces always occur in pairs, and one object cannot exert a force on another without experiencing the same strength force in return. We sometimes refer to these force pairs as action-reaction pairs, where the force exerted is the action, and the force experienced in return is the reaction (although which is which depends on your point of view).

Newton’s third law is useful for figuring out which forces are external to a system. Recall that identifying external forces is important when setting up a problem, because the external forces must be added together to find the net force.

We can see Newton’s third law at work by looking at how people move about. Consider a swimmer pushing off from the side of a pool, as illustrated in Figure 4.8. She pushes against the pool wall with her feet and accelerates in the direction opposite to her push. The wall has thus exerted on the swimmer a force of equal magnitude but in the direction opposite that of her push. You might think that two forces of equal magnitude but that act in opposite directions would cancel, but they do not because they act on different systems.

In this case, there are two different systems that we could choose to investigate: the swimmer or the wall. If we choose the swimmer to be the system of interest, as in the figure, then ${\text{F}}_{\text{wall on feet}}$ is an external force on the swimmer and affects her motion. Because acceleration is in the same direction as the net external force, the swimmer moves in the direction of ${\text{F}}_{\text{wall on feet}}.$ Because the swimmer is our system (or object of interest) and not the wall, we do not need to consider the force ${\text{F}}_{\text{feet on wall}}$ because it originates from the swimmer rather than acting on the swimmer. Therefore, ${\text{F}}_{\text{feet on wall}}$ does not directly affect the motion of the system and does not cancel ${\text{F}}_{\text{wall on feet}}.$ Note that the swimmer pushes in the direction opposite to the direction in which she wants to move.

Figure 4.8 When the swimmer exerts a force $F_{\text{feet on wall}}$ on the wall, she accelerates in the direction opposite to that of her push. This means that the net external force on her is in the direction opposite to $F_{\text{feet on wall}}.$ This opposition is the result of Newton’s third law of motion, which dictates that the wall exerts a force $F_{\text{wall on feet}}$ on the swimmer that is equal in magnitude but that acts in the direction opposite to the force that the swimmer exerts on the wall.

Other examples of Newton’s third law are easy to find. As a teacher paces in front of a whiteboard, he exerts a force backward on the floor. The floor exerts a reaction force in the forward direction on the teacher that causes him to accelerate forward. Similarly, a car accelerates because the ground pushes forward on the car's wheels in reaction to the car's wheels pushing backward on the ground. You can see evidence of the wheels pushing backward when tires spin on a gravel road and throw rocks backward.

Another example is the force of a baseball as it makes contact with the bat. Helicopters create lift by pushing air down, creating an upward reaction force. Birds fly by exerting force on air in the direction opposite that in which they wish to fly. For example, the wings of a bird force air downward and backward in order to get lift and move forward. An octopus propels itself forward in the water by ejecting water backward through a funnel in its body, which is similar to how a jet ski is propelled. In these examples, the octopus or jet ski push the water backward, and the water, in turn, pushes the octopus or jet ski forward.

Applying Newton’s Third Law

Forces are classified and given names based on their source, how they are transmitted, or their effects. In previous sections, we discussed the forces called push, weight, and friction. In this section, applying Newton’s third law of motion will allow us to explore three more forces: the normal force, tension, and thrust. However, because we haven’t yet covered vectors in depth, we’ll only consider one-dimensional situations in this chapter. Another chapter will consider forces acting in two dimensions.

The gravitational force (or weight) acts on objects at all times and everywhere on Earth. We know from Newton’s second law that a net force produces an acceleration; so, why is everything not in a constant state of freefall toward the center of Earth? The answer is the normal force. The normal force is the outward force that a surface applies to an object perpendicular to the surface, and it prevents the object from penetrating it. In the case of an object at rest on a horizontal surface, it is the force needed to support the weight of that object. If an object on a flat surface is not accelerating, the net external force is zero, and the normal force has the same magnitude as the weight of the system but acts in the opposite direction. In equation form, we write that

Note that this equation is only true for a horizontal surface.

The word tension comes from the Latin word meaning to stretch. Tension is the force along the length of a flexible connector, such as a string, rope, chain, or cable. Regardless of the type of connector attached to the object of interest, one must remember that the connector can only pull (or exert tension) in the direction parallel to its length. Tension is a pull that acts parallel to the connector, and that acts in opposite directions at the two ends of the connector. This is possible because a flexible connector is simply a long series of action-reaction forces, except at the two ends where outside objects provide one member of the action-reaction forces.

Consider a person holding a mass on a rope, as shown in Figure 4.9.

Figure 4.9 When a perfectly flexible connector (one requiring no force to bend it) such as a rope transmits a force T, this force must be parallel to the length of the rope, as shown. The pull that such a flexible connector exerts is a tension. Note that the rope pulls with equal magnitude force but in opposite directions to the hand and to the mass (neglecting the weight of the rope). This is an example of Newton’s third law. The rope is the medium that transmits forces of equal magnitude between the two objects but that act in opposite directions.

Tension in the rope must equal the weight of the supported mass, as we can prove by using Newton’s second law. If the 5.00 kg mass in the figure is stationary, then its acceleration is zero, so $F_{\text{net}}=0.$ The only external forces acting on the mass are its weight W and the tension T supplied by the rope. Summing the external forces to find the net force, we obtain

where T and W are the magnitudes of the tension and weight, respectively, and their signs indicate direction, with up being positive. By substituting mg for F_net and rearranging the equation, the tension equals the weight of the supported mass, just as you would expect

For a 5.00-kg mass (neglecting the mass of the rope), we see that

Another example of Newton’s third law in action is thrust. Rockets move forward by expelling gas backward at a high velocity. This means that the rocket exerts a large force backward on the gas in the rocket combustion chamber, and the gas, in turn, exerts a large force forward on the rocket in response. This reaction force is called thrust.

Links To Physics

Math: Problem-Solving Strategy for Newton’s Laws of Motion

The basics of problem solving, presented earlier in this text, are followed here with specific strategies for applying Newton’s laws of motion. These techniques also reinforce concepts that are useful in many other areas of physics.

First, identify the physical principles involved. If the problem involves forces, then Newton’s laws of motion are involved, and it is important to draw a careful sketch of the situation. An example of a sketch is shown in Figure 4.10. Next, as in Figure 4.10, use vectors to represent all forces. Label the forces carefully, and make sure that their lengths are proportional to the magnitude of the forces and that the arrows point in the direction in which the forces act.

Figure 4.10 (a) A sketch of Tarzan hanging motionless from a vine. (b) Arrows are used to represent all forces. T is the tension exerted on Tarzan by the vine, $F_{\text{T}}$ is the force exerted on the vine by Tarzan, and W is Tarzan’s weight (i.e., the force exerted on Tarzan by Earth’s gravity). All other forces, such as a nudge of a breeze, are assumed to be negligible. (c) Suppose we are given Tarzan’s mass and asked to find the tension in the vine. We define the system of interest as shown and draw a free-body diagram, as shown in (d). $F_{\text{T}}$ is no longer shown because it does not act on the system of interest; rather, $F_{\text{T}}$ acts on the outside world. (d) The free-body diagram shows only the external forces acting on Tarzan. For these to sum to zero, we must have $T=W\text{.}$

Next, make a list of knowns and unknowns and assign variable names to the quantities given in the problem. Figure out which variables need to be calculated; these are the unknowns. Now carefully define the system: which objects are of interest for the problem. This decision is important, because Newton’s second law involves only external forces. Once the system is identified, it’s possible to see which forces are external and which are internal (see Figure 4.10).

If the system acts on an object outside the system, then you know that the outside object exerts a force of equal magnitude but in the opposite direction on the system.

A diagram showing the system of interest and all the external forces acting on it is called a free-body diagram. Only external forces are shown on free-body diagrams, not acceleration or velocity. Figure 4.10 shows a free-body diagram for the system of interest.

After drawing a free-body diagram, apply Newton’s second law to solve the problem. This is done in Figure 4.10 for the case of Tarzan hanging from a vine. When external forces are clearly identified in the free-body diagram, translate the forces into equation form and solve for the unknowns. Note that forces acting in opposite directions have opposite signs. By convention, forces acting downward or to the left are usually negative.

Grasp Check

If a problem has more than one system of interest, more than one free-body diagram is required to describe the external forces acting on the different systems.

1. True
2. False

Worked Example

An Accelerating Equipment Cart

A physics teacher pushes a cart of demonstration equipment to a classroom, as in Figure 4.11. Her mass is 65.0 kg, the cart’s mass is 12.0 kg, and the equipment’s mass is 7.0 kg. To push the cart forward, the teacher’s foot applies a force of 150 N in the opposite direction (backward) on the floor. Calculate the acceleration produced by the teacher. The force of friction, which opposes the motion, is 24.0 N.

Figure 4.11

Strategy

Because they accelerate together, we define the system to be the teacher, the cart, and the equipment. The teacher pushes backward with a force $F_{\text{foot}}$ of 150 N. According to Newton’s third law, the floor exerts a forward force $F_{\text{floor}}$ of 150 N on the system. Because all motion is horizontal, we can assume that no net force acts in the vertical direction, and the problem becomes one dimensional. As noted in the figure, the friction f opposes the motion and therefore acts opposite the direction of $F_{\text{floor}}.$

We should not include the forces $F_{\text{teacher}}$, $F_{\text{cart}}$, or $F_{\text{foot}}$ because these are exerted by the system, not on the system. We find the net external force by adding together the external forces acting on the system (see the free-body diagram in the figure) and then use Newton’s second law to find the acceleration.

Solution

Newton’s second law is

$$a=\frac{F_{\text{net}}}{m}\text{.}$$

4.21

The net external force on the system is the sum of the external forces: the force of the floor acting on the teacher, cart, and equipment (in the horizontal direction) and the force of friction. Because friction acts in the opposite direction, we assign it a negative value. Thus, for the net force, we obtain

$$F_{\text{net}}=F_{\text{floor}}-f=\text{150 N}-\text{24}\text{.0 N}=\text{126 N.}$$

4.22

The mass of the system is the sum of the mass of the teacher, cart, and equipment.

$$m\text{ = (65}\text{.0 + 12}\text{.0 + 7}\text{.0) kg = 84 kg}$$

4.23

Insert these values of net F and m into Newton’s second law to obtain the acceleration of the system.

$$\begin{array}{l}a=\frac{F_{\text{net}}}{m}\hfill \\ a=\frac{\text{126 N}}{\text{84 kg}}=\text{1}{\text{.5 m/s}}^2\hfill \end{array}$$

4.24

$$F_1<F_2$$

4.25

Discussion

None of the forces between components of the system, such as between the teacher’s hands and the cart, contribute to the net external force because they are internal to the system. Another way to look at this is to note that the forces between components of a system cancel because they are equal in magnitude and opposite in direction. For example, the force exerted by the teacher on the cart is of equal magnitude but in the opposite direction of the force exerted by the cart on the teacher. In this case, both forces act on the same system, so they cancel. Defining the system was crucial to solving this problem.