Chapter 4: Linear Kinetics, Force and Newton’s Laws of Motion
4.3 Newton’s Second Law
Authors: William Moebs, Samuel Ling, Jeff Sanny
Adapted by: Rob Pryce, Alix Blacklin
Learning Objectives
By the end of this section, you will be able to:
- Distinguish between external and internal forces
- Describe Newton’s second law of motion
- Explain the dependence of acceleration on net force and mass
Newton’s second law is closely related to his first law. It gives the cause-and-effect relationship between force and changes in motion. Newton’s second law is quantitative and is used extensively in biomechanics to calculate what happens in situations involving a force. Before we write down Newton’s second law as a simple equation, let’s look at some of the ideas we mentioned earlier.
Force and Acceleration
First, what do we mean by a change in motion? The answer is that a change in motion is equivalent to a change in velocity. A change in velocity means, by definition, that there is acceleration. Newton’s first law says that a net external force causes a change in motion; thus, we see that a net external force causes acceleration.
We defined external force in the previous section as a force acting on an object that originates from outside the object. Let’s consider this further. An intuitive notion of external is correct—it is outside the system of interest. For example, in Figure 4.10(a), the system of interest is the car plus the person within it. The two forces exerted by the two students are external forces. In contrast, an internal force acts between elements of the system or car. For example, the force the person in the car exerts to hang on to the steering wheel is an internal force. According to Newton’s First Law, only external forces affect the motion of an object. (The internal forces cancel each other out, as explained in the next section.)
From the example above, you can see that different forces exerted on the same mass produce different accelerations. In Figure 4.10(a), the two students push a car with a driver in it. Arrows representing all external forces are shown. The object (or system of interest) is the car and its driver. The weight [latex]\overset{\to }{w}[/latex] of the system and the support of the ground [latex]\overset{\to }{N}[/latex] are also shown for completeness and are assumed to cancel (because there was no vertical motion and no imbalance of forces in the vertical direction to create a change in motion). The vector [latex]\overset{\to }{f}[/latex] represents the friction acting on the car, and it acts to the left, opposing the motion of the car. (We discuss friction in more detail in the next chapter.)
In Figure 4.10 (b), all external forces acting on the system add together to produce the net force [latex]{\overset{\to }{F}}_{\text{net}}.[/latex] The free-body diagram shows all of the forces acting on the system of interest. The dot represents the center of mass of the system. Each force vector extends from this dot. Because there are two forces acting to the right, the vectors are shown collinearly.
Finally, in Figure 4.10 (c), a larger net external force produces a larger acceleration [latex](\overset{\to }{{a}^{\prime }}>\overset{\to }{a})[/latex] when the tow truck pulls the car. This is because the force of the tow truck is greater than the force produced by the two people, which is evident by the length of the force vectors in each of the free body diagrams.
It seems reasonable that acceleration would be directly proportional to and in the same direction as the net external force acting on a system. This assumption has been verified experimentally and is illustrated in Figure 4.10. To obtain an equation for Newton’s second law, we can first write the relationship of acceleration [latex]\overset{\to }{a}[/latex] and net external force [latex]{\overset{\to }{F}}_{\text{net}}[/latex] as the proportionality:
[latex]\overset{\to }{a}\propto {\overset{\to }{F}}_{\text{net}}[/latex]
where the symbol [latex]\propto[/latex] means “proportional to.” (Recall from the previous section that the net external force is the vector sum of all external forces and is sometimes indicated as [latex]\sum \overset{\to }{F}.[/latex]) This proportionality shows what we have said in words—acceleration is directly proportional to net external force. (In the context of human movement, it can be a tremendous simplification to disregard the numerous internal forces acting between objects within the system, such as muscular forces within the students’ bodies. Still, this simplification helps us solve some complex problems about motion of people or objects).
It also seems reasonable that acceleration should be inversely proportional to the mass of the system. In other words, the larger the mass (the inertia), the smaller the acceleration produced by a given force. As illustrated in Figure 4.11, the same net external force applied to a basketball produces a much smaller acceleration when it is applied to an SUV. The proportionality is written as
where m is the mass of the system and a is the magnitude of the acceleration. Experiments have shown that acceleration is exactly inversely proportional to mass, just as it is directly proportional to net external force.
Combining the two proportionalities above yields Newton’s second law.
Newton’s Second Law of Motion
The acceleration of a system is directly proportional to and in the same direction as the net external force acting on the system and is inversely proportion to its mass.
In equation form, Newton’s second law is written as:
Units of Force
[latex]F = \text{ma}[/latex] is used to define the units of force in terms of the three basic units for mass, length, and time.
The SI unit of force is called the newton (abbreviated N) and is the force needed to accelerate a 1-kg system at the rate of 1m/s2.
That is, since F=ma, 1 N=1 kg⋅m/s2.
Example: What Acceleration Can a Person Produce When Pushing a Lawn Mower?
Suppose that the net external force (push minus friction) exerted on a lawn mower is 51 N (about 11 lb.) parallel to the ground (Figure 4.12). The mass of the mower is 24 kg. What is its acceleration?
Strategy
This problem involves only motion in the horizontal direction; we are also given the net force, indicated by the single vector, but we can suppress the vector nature and concentrate on applying Newton’s second law. Since [latex]{F}_{\text{net}}[/latex] and m are given, the acceleration can be calculated directly from Newton’s second law as [latex]{F}_{\text{net}}=ma.[/latex]
Solution
The magnitude of the acceleration a is [latex]a={F}_{\text{net}}\text{/}m[/latex]. Entering known values gives
Substituting the unit of kilograms times meters per square second for newtons yields
Discussion
The direction of the acceleration is the same direction as that of the net force, which is parallel to the ground. Although there is no information given in this example about the individual external forces acting on the system, we can say something about their relative magnitudes. For example, the force exerted by the person pushing the mower must be greater than the friction opposing the motion (since we know the mower moved forward), and the vertical forces must cancel because no acceleration occurs in the vertical direction (the mower is moving only horizontally). The acceleration found is small enough to be reasonable for a person pushing a mower. Such an effort would not last too long, because the person’s top speed would soon be reached.
In the preceding example, we dealt with net force only for simplicity. However, several forces act on the lawn mower. The weight [latex]\overset{\to }{w}[/latex] (discussed in detail in Mass and Weight) pulls down on the mower, toward the center of Earth; this produces a contact force on the ground. The ground must exert an upward force on the lawn mower, known as the normal force [latex]\overset{\to }{N}[/latex], which we will also define later. These forces are balanced and therefore do not produce vertical acceleration. In the following examples, we will show multiple external forces acting on an object.
Example: Comparing forces using Newton’s Second Law: Which Force Is Bigger?
(a) The car shown in Figure 4.13 is moving at a constant speed. Which force is bigger, [latex]{\overset{\to }{F}}_{\text{engine}}[/latex] or [latex]{\overset{\to }{F}}_{\text{friction}}[/latex]? Explain.
(b) The same car is now accelerating to the right. Which force is bigger, [latex]{\overset{\to }{F}}_{\text{engine}}[/latex] or [latex]{\overset{\to }{F}}_{\text{friction}}?[/latex] Explain.
Strategy
We must consider Newton’s first and second laws to analyze the situation. We need to decide which law applies; this, in turn, will tell us about the relationship between the forces.
Solution
- The forces are equal. According to Newton’s first law, if the net force is zero, the velocity is constant.
- In this case, [latex]{\overset{\to }{F}}_{\text{engine}}[/latex] must be larger than [latex]{\overset{\to }{F}}_{\text{friction}}.[/latex] According to Newton’s second law, a net force is required to cause acceleration.
Discussion
These questions may seem trivial, but they are commonly answered incorrectly. For a car or any other object to move, it must be accelerated from rest to the desired speed; this requires that the engine force be greater than the friction force. Once the car is moving at constant velocity, the net force must be zero; otherwise, the car will accelerate (gain speed). To solve problems involving Newton’s laws, we must understand whether to apply Newton’s first law (where [latex]\sum \overset{\to }{F}=\overset{\to }{0}[/latex]) or Newton’s second law (where [latex]\sum \overset{\to }{F}[/latex] is not zero). This will be apparent as you see more examples and attempt to solve problems on your own.
Example: Solving for Force: What Rocket Thrust Accelerates This Sled?
Before manned space flights, rocket sleds were used to test aircraft, missile equipment, and physiological effects on human subjects at high speeds. They consisted of a platform that was mounted on one or two rails and propelled by several rockets.
Calculate the magnitude of force exerted by each rocket, called its thrust T, for the four-rocket propulsion system shown in Figure 4.14. The sled’s initial acceleration is [latex]49\,{\text{m/s}}^{2}[/latex], the mass of the system is 2100 kg, and the force of friction opposing the motion is 650 N.
Strategy
Although forces are acting both vertically and horizontally, we assume the vertical forces cancel because there is no vertical acceleration. This leaves us with only horizontal forces and a simpler one-dimensional problem. Directions are indicated with plus or minus signs, with right taken as the positive direction. See the free-body diagram in FIgure 4.14.
Solution
Since acceleration, mass, and the force of friction are given, we start with Newton’s second law and look for ways to find the thrust of the engines. We have defined the direction of the force and acceleration as acting “to the right,” so we need to consider only the magnitudes of these quantities in the calculations. Hence we begin with
where [latex]{F}_{\text{net}}[/latex] is the net force along the horizontal direction. We can see from the figure that the engine thrusts add, whereas friction opposes the thrust. In equation form, the net external force is
Substituting this into Newton’s second law gives us
Using a little algebra, we solve for the total thrust 4T:
Substituting known values yields
Therefore, the total thrust is
and the individual thrusts are
Discussion
The numbers are quite large, so the result might surprise you. Experiments such as this were performed in the early 1960s to test the limits of human endurance, and the setup was designed to protect human subjects in jet fighter emergency ejections. Speeds of 1000 km/h were obtained, with accelerations of 45 g’s. (Recall that g, acceleration due to gravity, is [latex]9.80\,{\text{m/s}}^{2}[/latex]. When we say that acceleration is 45 g’s, it is [latex]45\,×\,9.8\,{\text{m/s}}^{2},[/latex] which is approximately [latex]440\,{\text{m/s}}^{2}[/latex].) Although living subjects are not used anymore, land speeds of 10,000 km/h have been obtained with a rocket sled.
In this example, as in the preceding one, the system of interest is obvious. We see in later examples that choosing the system of interest is crucial—and the choice is not always obvious.
So, we can see that Newton’s second law is more than a definition; it is a relationship among acceleration, force, and mass. It can help us make predictions. Each of those physical quantities can be defined independently, so the second law tells us something basic and universal about nature.
Interactive
Explore forces and motion in the following simulations: when applying a simple force; when pulling a cart; or put an object on a ramp and see how it affects its motion.
Advanced Perspective: Newton’s Second Law and Momentum
Newton actually stated his second law in terms of momentum: “The instantaneous rate at which a body’s momentum changes is equal to the net force acting on the body.” (“Instantaneous rate” implies that the derivative is involved.) This can be given by the vector equation:
[latex]{\overset{\to }{F}}_{\text{net}}=\frac{d\overset{\to }{p}}{dt}.[/latex]
This means that Newton’s second law addresses the central question of motion: What causes a change in motion of an object? Momentum was described by Newton as “quantity of motion,” a way of combining both the velocity of an object and its mass. We devote a later chapter to the study of momentum.
For this explanation, it is sufficient to define momentum [latex]\overset{\to }{p}[/latex] as the product of the mass of the object m and its velocity [latex]\overset{\to }{v}[/latex]:
Since velocity is a vector, so is momentum.
It is easy to visualize momentum. A train moving at 10 m/s has more momentum than one that moves at 2 m/s. In everyday life, we speak of one sports team as “having momentum” when they score points against the opposing team.
If we substitute the two equations above, we obtain:
[latex]{\overset{\to }{F}}_{\text{net}}=\frac{d\overset{\to }{p}}{dt}=\frac{d(m\overset{\to }{v})}{dt}.[/latex]
When m is constant, we have:
[latex]{\overset{\to }{F}}_{\text{net}}=m\frac{d(\overset{\to }{v})}{dt}=m\overset{\to }{a}.[/latex]
Thus, we see that the momentum form of Newton’s second law reduces to the form given earlier in this section.