Learning Objectives

By the end of this section, you will be able to do the following:

Define net force, external force, and system
Understand Newton’s second law of motion
Apply Newton’s second law to determine the weight of an object

Newton’s second law of motion is closely related to Newton’s first law of motion. It mathematically states the cause and effect relationship between force and changes in motion. Newton’s second law of motion is more quantitative and is used extensively to calculate what happens in situations involving a force. Before we can write down Newton’s second law as a simple equation giving the exact relationship of force, mass, and acceleration, we need to sharpen some ideas that have already been mentioned.

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 an 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.

Another question immediately arises. What do we mean by an external force? An intuitive notion of external is correct—an external force acts from outside the system of interest. For example, in Figure 4.5(a) the system of interest is the wagon plus the child in it. The two forces exerted by the other children are external forces. An internal force acts between elements of the system. Again looking at Figure 4.5(a), the force the child in the wagon exerts to hang onto the wagon is an internal force between elements of the system of interest. Only external forces affect the motion of a system, according to Newton’s first law. The internal forces actually cancel, as we shall see in the next section. You must define the boundaries of the system before you can determine which forces are external. Sometimes the system is obvious, whereas other times identifying the boundaries of a system is more subtle. The concept of a system is fundamental to many areas of physics, as is the correct application of Newton’s laws. This concept will be revisited many times on our journey through physics.

When we describe the acceleration of a system, we are modeling the system as a single point which contains all of the mass of that system. The point we choose for this is the point about which the system’s mass is evenly distributed. For example, in a rigid object, this center of mass is the point where the object will stay balanced even if only supported at this point. For a sphere or disk made of homogenous material, this point is of course at the center. Similarly, for a rod made of homogenous material, the center of mass will be at the midpoint.

For the rider in the wagon in Figure 4.5, the center of mass is probably between the rider’s hips. Due to internal forces, the rider’s hand or hair may accelerate slightly differently, but it is the acceleration of the system’s center of mass that interests us. This is true whether the system is a vehicle carrying passengers, a bowl of grapes, or a planet. When we draw a free-body diagram of a system, we represent the system’s center of mass with a single point and use vectors to indicate the forces exerted on that center of mass (see Figure 4.5).

(a) A boy in a wagon is pushed by two girls toward the right. The force on the boy is represented by vector F one toward the right, and the force on the wagon is represented by vector F two in the same direction. Acceleration a is shown by a vector a toward the right and a friction force f is acting in the opposite direction, represented by a vector pointing toward the left. The weight W of the wagon is shown by a vector acting downward, and the normal force acting upward on the wagon is represented by a

Figure 4.5 Different forces exerted on the same mass produce different accelerations. (a) Two children push a wagon with a child in it. Arrows representing all external forces are shown. The system of interest is the wagon and its rider. The weight

w

of the system and the support of the ground

N

are also shown for completeness and are assumed to cancel. The vector

f

represents the friction acting on the wagon, and it acts to the left, opposing the motion of the wagon. (b) All of the external forces acting on the system add together to produce a net force,

F_{net}

. 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, we draw the vectors collinearly. (c) A larger net external force produces a larger acceleration (

a′ > a

) when an adult pushes the child.

Now, it seems reasonable that acceleration should be directly proportional to and in the same direction as the net (total) external force acting on a system. This assumption has been verified experimentally and is illustrated in Figure 4.5. In part (a), a smaller force causes a smaller acceleration than the larger force illustrated in part (c). For completeness, the vertical forces are also shown; they are assumed to cancel since there is no acceleration in the vertical direction. The vertical forces are the weight $w$ and the support of the ground $N$ , and the horizontal force $f$ represents the force of friction. These will be discussed in more detail in later sections. For now, we will define friction as a force that opposes the motion past each other of objects that are touching. Figure 4.5(b) shows how vectors representing the external forces add together to produce a net force, $F_{net}$ .

To obtain an equation for Newton’s second law, we first write the relationship of acceleration and net external force as the proportionality

4.1

a \propto F_{net},

where the symbol $\propto$ means proportional to, and $F_{net}$ is the net external force. The net external force is the vector sum of all external forces and can be determined graphically, using the head-to-tail method, or analytically, using components. The techniques are the same as for the addition of other vectors, and are covered in Two-Dimensional Kinematics. This proportionality states what we have said in words—acceleration is directly proportional to the net external force. Once the system of interest is chosen, it is important to identify the external forces and ignore the internal ones. It is a tremendous simplification not to have to consider the numerous internal forces acting between objects within the system, such as muscular forces within the child’s body, let alone the myriad of forces between atoms in the objects, but by doing so, we can easily solve some very complex problems with only minimal error due to our simplification.

Now, it also seems reasonable that acceleration should be inversely proportional to the mass of the system. In other words, the larger the mass or the inertia, the smaller the acceleration produced by a given force. And indeed, as illustrated in Figure 4.6, the same net external force applied to a car produces a much smaller acceleration than when applied to a basketball. The proportionality is written as

4.2

a \propto \frac{1}{m}

where $m$ is the mass of the system. Experiments have shown that acceleration is exactly inversely proportional to mass, just as it is exactly linearly proportional to the net external force.

(a) A basketball player pushes the ball with the force shown by a vector F toward the right and an acceleration a-one represented by an arrow toward the right. M sub one is the mass of the ball. (b) The same basketball player is pushing a car with the same force, represented by the vector F towards the right, resulting in an acceleration shown by a vector a toward the right. The mass of the car is m sub two. The acceleration in the second case, a sub two, is represented by a shorter arrow than in the firs

Figure 4.6 The same force exerted on systems of different masses produces different accelerations. (a) A basketball player pushes on a basketball to make a pass. The effect of gravity on the ball is ignored. (b) The same player exerts an identical force on a stalled SUV and produces a far smaller acceleration even if friction is negligible. (c) The free-body diagrams are identical, permitting direct comparison of the two situations. A series of patterns for the free-body diagram will emerge as you do more problems.

Both of these proportionalities have been experimentally verified repeatedly and consistently, for a broad range of systems and scales. Thus, it has been experimentally found that the acceleration of an object depends only on the net external force and the mass of the object. Combining the two proportionalities just given yields Newton's second law of motion.

Applying the Science Practices: Testing the Relationship Between Mass, Acceleration, and Force

Plan three simple experiments using objects you have at home to test relationships between mass, acceleration, and force.

(a) Design an experiment to test the relationship between mass and acceleration. What will be the independent variable in your experiment? What will be the dependent variable? What controls will you put in place to ensure force is constant?

(b) Design a similar experiment to test the relationship between mass and force. What will be the independent variable in your experiment? What will be the dependent variable? What controls will you put in place to ensure acceleration is constant?

(c) Design a similar experiment to test the relationship between force and acceleration. What will be the independent variable in your experiment? What will be the dependent variable? Will you have any trouble ensuring that the mass is constant?

What did you learn?

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 inversely proportional to its mass.

In equation form, Newton’s second law of motion is

4.3

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

This is often written in the more familiar form

4.4

F_{net} = m a .

When only the magnitude of force and acceleration are considered, this equation is simply

4.5

F_{net} = ma .

Although these last two equations are really the same, the first gives more insight into what Newton’s second law means. The law is a cause and effect relationship among three quantities that is not simply based on their definitions. The validity of the second law is completely based on experimental verification.

Applying the Science Practices: Systems and Free-Body Diagrams

First, consider a person on a sled sliding downhill. What is the system in this situation? Try to draw a free-body diagram describing this system, labeling all the forces and their directions. Which of the forces are internal? Which are external?

Next, consider a person on a sled being pushed along level ground by a friend. What is the system in this situation? Try to draw a free-body diagram describing this system, labelling all the forces and their directions. Which of the forces are internal? Which are external?

Note that zero net force is not the same as the absence of external forces. Two or more forces may be acting on a body, but if the vector sum of those forces is zero, then there will be zero net force acting on the body. In that case, if the object was initially at rest, it would remain at rest, even though multiple forces are acting on it. If the object initially was moving before the forces were applied, the magnitude and direction of the object’s velocity would remain unchanged as a result of the forces.

Units of Force

$F_{net} = m a$ 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 $1 {m/s}^{2} .$ That is, since $F_{net} = m a$ ,

4.6

1 N = 1 kg \cdot {m/s}^{2} .

While almost the entire world uses the newton for the unit of force, in the United States, the most familiar unit of force is the pound (lb), where 1 N = 0.225 lb.

Weight and the Gravitational Force

When an object is dropped, it accelerates toward the center of Earth. Newton’s second law states that a net force on an object is responsible for its acceleration. If air resistance is negligible, the net force on a falling object is the gravitational force, commonly called its weight $w$ . Weight can be denoted as a vector $w$ because it has a direction; down is, by definition, the direction of gravity, and hence weight is a downward force. The magnitude of weight is denoted as $w$ . Galileo was instrumental in showing that, in the absence of air resistance, all objects fall with the same acceleration $g$ . Using Galileo’s result and Newton’s second law, we can derive an equation for weight.

Consider an object with mass $m$ falling downward toward Earth. It experiences only the downward force of gravity, which has magnitude $w$ . Newton’s second law states that the magnitude of the net external force on an object is $F_{net} = ma$ .

Since the object experiences only the downward force of gravity, $F_{net} = w$ . We know that the acceleration of an object due to gravity is $g$ , or $a = g$ . Substituting these into Newton’s second law gives

Weight

This is the equation for weight—the gravitational force on a mass $m$

4.7

w = mg .

Since $g = 9.80 {m/s}^{2}$ on Earth, the weight of a 1.0 kg object on Earth is 9.8 N, as we see

4.8

w = mg = (1 . 0 kg) (9.80 {m/s}^{2}) = 9.8 N .

Recall that $g$ can take a positive or negative value, depending on the positive direction in the coordinate system. Be sure to take this into consideration when solving problems with weight.

When the net external force on an object is its weight, we say that it is in free-fall. That is, the only force acting on the object is the force of gravity. In the real world, when objects fall downward toward Earth, they are never truly in free-fall because there is always some upward force from the air acting on the object.

The acceleration due to gravity $g$ varies slightly over the surface of Earth, so that the weight of an object depends on location and is not an intrinsic property of the object. Weight varies dramatically if one leaves Earth’s surface. On the Moon, for example, the acceleration due to gravity is only $1.67 {m/s}^{2}$ . A 1.0-kg mass thus has a weight of 9.8 N on Earth and only about 1.7 N on the Moon.

The broadest definition of weight in this sense is that the weight of an object is the gravitational force on it from the nearest large body, such as Earth, the Moon, the Sun, and so on. This is the most common and useful definition of weight in physics. It differs dramatically, however, from the definition of weight used by NASA and the popular media in relation to space travel and exploration. When they speak of weightlessness and microgravity, they are really referring to the phenomenon we call free-fall in physics. We shall use the above definition of weight, and we will make careful distinctions between free-fall and actual weightlessness.

It is important to be aware that weight and mass are very different physical quantities, although they are closely related. Mass is the quantity of matter or how much stuff and does not vary in classical physics, whereas weight is the gravitational force and does vary depending on gravity. It is tempting to equate the two, since most of our examples take place on Earth, where the weight of an object only varies a little with the location of the object. Furthermore, the terms mass and weight are used interchangeably in everyday language; for example, our medical records often show our weight in kilograms, but never in the correct units of newtons.

Common Misconceptions: Mass vs. Weight

Mass and weight are often used interchangeably in everyday language. However, in science, these terms are distinctly different from one another. Mass is a measure of how much matter is in an object. The typical measure of mass is the kilogram or the slug in English units. Weight, on the other hand, is a measure of the force of gravity acting on an object. Weight is equal to the mass of an object ( $m$ ) multiplied by the acceleration due to gravity ( $g$ ). Like any other force, weight is measured in terms of newtons or pounds in English units.

Assuming the mass of an object is kept intact, it will remain the same, regardless of its location. However, because weight depends on the acceleration due to gravity, the weight of an object can change when the object enters into a region with stronger or weaker gravity. For example, the acceleration due to gravity on the Moon is $1.67 {m/s}^{2},$ which is much less than the acceleration due to gravity on Earth, $9.80 {m/s}^{2}$ . If you measured your weight on Earth and then measured your weight on the Moon, you would find that you weigh much less, even though you do not look any skinnier. This is because the force of gravity is weaker on the Moon. In fact, when people say that they are “losing weight,” they really mean that they are losing mass, which in turn causes them to weigh less.

Take-Home Experiment: Mass and Weight

What do bathroom scales measure? When you stand on a bathroom scale, what happens to the scale? It depresses slightly. The scale contains springs that compress in proportion to your weight—similar to rubber bands expanding when pulled. The springs provide a measure of your weight for an object which is not accelerating. This is a force in newtons or pounds. In most countries, the measurement is divided by 9.80 to give a reading in mass units of kilograms. The scale measures weight but is calibrated to provide information about mass. While standing on a bathroom scale, push down on a table next to you. What happens to the reading? Why? Would your scale measure the same mass on Earth as on the Moon?

Example 4.1 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. The mass of the mower is 24 kg. What is its acceleration?

A man pushing a lawnmower to the right. A red vector above the lawnmower is pointing to the right and labeled F sub net.

Figure 4.7 The net force on a lawn mower is 51 N to the right. At what rate does the lawn mower accelerate to the right?

Strategy

Since $F_{net}$ and $m$ are given, the acceleration can be calculated directly from Newton’s second law as stated in $F_{net} = m a$ .

Solution

The magnitude of the acceleration $a$ is $a = \frac{F_{net}}{m}$ . Entering known values gives

4.9

a = \frac{51 N}{24 kg} .

Substituting the units $kg \cdot {m/s}^{2}$ for N yields

4.10

a = \frac{51 kg \cdot {m/s}^{2}}{24 kg} = 2.1 m {/s}^{2} .

Discussion

The direction of the acceleration is the same direction as that of the net force, which is parallel to the ground. There is no information given in this example about the individual external forces acting on the system, but 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 moves forward, and the vertical forces must cancel if there is to be no acceleration 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.

Example 4.2 What Rocket Thrust Accelerates This Sled?

Prior to 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 4.8. The sled’s initial acceleration is $49 {m/s}^{2},$ the mass of the system is 2,100 kg, and the force of friction opposing the motion is known to be 650 N.

A sled is shown with four rockets, each producing the same thrust, represented by equal length arrows labeled as vector T pushing the sled toward the right. Friction force is represented by an arrow labeled as vector f pointing toward the left on the sled. The weight of the sled is represented by an arrow labeled as vector W, shown pointing downward, and the normal force is represented by an arrow labeled as vector N having the same length as W acting upward on the sled. A free-body diagram is also shown

Figure 4.8 A sled experiences a rocket thrust that accelerates it to the right. Each rocket creates an identical thrust

T

. As in other situations where there is only horizontal acceleration, the vertical forces cancel. The ground exerts an upward force

N

on the system that is equal in magnitude and opposite in direction to its weight,

w

. The system here is the sled, its rockets, and rider, so none of the forces between these objects are considered. The arrow representing friction (

f

) is drawn larger than scale.

Strategy

Although there are forces acting vertically and horizontally, we assume the vertical forces cancel since 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 the figure.

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. Since we have defined the direction of the force and acceleration as acting to the right, we need to consider only the magnitudes of these quantities in the calculations. Hence we begin with

4.11

F_{net} = ma,

where $F_{net}$ is the net force along the horizontal direction. We can see from Figure 4.8 that the engine thrusts add, while friction opposes the thrust. In equation form, the net external force is

4.12

F_{net} = 4 T - f .

Substituting this into Newton’s second law gives

4.13

F_{net} = ma = 4 T - f .

Using a little algebra, we solve for the total thrust 4T

4.14

4 T = ma + f .

Substituting known values yields

4.15

4 T = ma + f = (2,100 kg) ({49 m/s}^{2}) + 650 N .

So the total thrust is

4.16

4 T = 1.0 \times 10^{5} N,

and the individual thrusts are

4.17

T = \frac{1.0 \times 10^{5} N}{4} = 2.6 \times 10^{4} N .

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 designed to protect human subjects in jet fighter emergency ejections. Speeds of 1,000 km/h were obtained, with accelerations of 45 $g$ 's. Recall that $g$ , the acceleration due to gravity, is $9 . {80 m/s}^{2}$ . When we say that an acceleration is 45 $g$ 's, it is $45 \times 9 . {80 m/s}^{2}$ , which is approximately ${440 m/s}^{2}$ . While living subjects are not used any more, land speeds of 10,000 km/h have been obtained with rocket sleds. In this example, as in the preceding one, the system of interest is obvious. We will see in later examples that choosing the system of interest is crucial—and the choice is not always obvious.

Newton’s second law of motion 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. The next section introduces the third and final law of motion.

Applying the Science Practices: Sums of Forces

Recall that forces are vector quantities, and therefore the net force acting on a system should be the vector sum of the forces.

(a) Design an experiment to test this hypothesis. What sort of a system would be appropriate and convenient to have multiple forces applied to it? What features of the system should be held constant? What could be varied? Can forces be arranged in multiple directions so that, while the hypothesis is still tested, the resulting calculations are not too inconvenient?

(b) Another group of students has done such an experiment, using a motion capture system looking down at an air hockey table to measure the motion of the 0.10-kg puck. The table was aligned with the cardinal directions, and a compressed air hose was placed in the center of each side, capable of varying levels of force output and fixed so that it was aimed at the center of the table.


Forces	Measured Acceleration (magnitudes)
3 N north, 4 N west	48 ± 4 m/s²
5 N south, 12 N east	132 ± 6 m/s²
6 N north, 12 N east, 4 N west	99 ± 3 m/s²

Table 4.1

Given the data in the table, is the hypothesis confirmed? What were the directions of the accelerations?

4.3 Newton's Second Law of Motion: Concept of a System

Learning Objectives