# Classical Dynamics

We have discussed how motion is described in terms of velocity and acceleration. Now we deal with the question of why objects move as they do: What makes an object at rest begin to move? What causes an object to accelerate or decelerate? What is involved when an object moves in a curved path? We can answer in each case that a force is required. In this section we will investigate the connection between force and motion, which is the subject called dynamics.

## Force

Intuitively, we experience force as any kind of a push or a pull on an object. When you push a stalled car or a grocery cart, you are exerting a force on it. When a motor lifts an elevator, or a hammer hits a nail, or the wind blows the leaves of a tree, a force is being exerted. We often call these contact forces because the force is exerted when one object comes in contact with another object. On the other hand, we say that an object falls because of the force of gravity.

If an object is at rest, to start it moving requires force-that is, a force is needed to accelerate an object from zero velocity to a nonzero velocity. For an object already moving, if you want to change its velocity-either in direction or in magnitude-a force is required. In other words, to accelerate an object, a force is always required. Later we will discuss the precise relation between acceleration and net force, which is Newton’s second law.

One way to measure the magnitude (or strength) of a force is to use a spring scale (Fig. 1). Normally, such a spring scale is used to find the weight of an object; by weight we mean the force of gravity acting on the object. The spring scale, once calibrated, can be used to measure other kinds of forces as well, such as the pulling force shown in Fig. 1.

A force exerted in a different direction has a different effect. Force has direction as well as magnitude.  We can represent any force on a diagram by an arrow, just as we did with velocity. The direction of the arrow is the direction of the push or pull, and its length is drawn proportional to the magnitude of the force.

Forces can be placed into two broad categories: Contact Forces and Field Forces.

Net Force Changes in motion are produced by a force or combination of forces (in the next section we’ll refer to changes in motion as acceleration). A force, in the simplest sense, is a push or a pull. Its source may be gravitational, electrical, magnetic, or simply muscular effort. When more than a single force acts on an object, we consider the net force. For example, if you and a friend pull in the same direction with equal forces on an object, the forces combine to produce a net force twice as great as your single force. If each of you pull with equal forces in opposite directions, the net force is zero. The equal but oppositely directed forces cancel each other. One of the forces can be considered to be the negative of the other, and they add algebraically to zero, with a resulting net force of zero.

## Newton’s First Law of Motion

What is the relationship between force and motion? Aristotle (384-322 B.c.) believed that a force was required to keep an object moving along a horizontal plane. To Aristotle, the natural state of an object was at rest, and a force was believed necessary to keep an object in motion. Furthermore, Aristotle argued, the greater the force on the object, the greater its speed.

Some 2000 years later, Galileo disagreed: he maintained that it is just as natural for an object to be in motion with a constant velocity as it is for it to be at rest. To understand Galileo’s idea, consider the following observations involving motion along a horizontal plane. To push an object with a rough surface along a tabletop at constant speed requires a certain amount of force. To push an equally heavy object with a very smooth surface across the table at the same speed will require less force. If a layer of oil or other lubricant is placed between the surface of the object and the table, then almost no force is required to keep the object moving. Notice that in each successive step, less force is required. As the next step, we imagine that the object does not rub against the table at all-or there is a perfect lubricant between the object and the table-and theorize that once started, the object would move across the table at constant speed with no force applied. A steel ball bearing rolling on a hard horizontal surface approaches this situation. So does a puck on an air table, in which a thin layer of air reduces friction almost to zero.

It was Galileo’s genius to imagine such an idealized world-in this case, one where there is no friction-and to see that it could lead to a more accurate and richer understanding of the real world. This idealization led him to his remarkable conclusion that if no force is applied to a moving object, it will continue to move with constant speed in a straight line. An object slows down only if a force is exerted on it. Galileo thus interpreted friction as a force akin to ordinary pushes and pulls.

To push an object across a table at constant speed requires a force from your hand that can balance out the force of friction. When the object moves at constant speed, your pushing force is equal in magnitude to the friction force, but these two forces are in opposite directions, so the net force on the object (the vector sum of the two forces) is zero. This is consistent with Galileo’s viewpoint, for the object moves with constant speed when no net force is exerted on it.

Upon this foundation laid by Galileo, Isaac Newton built his great theory of motion. Newton’s analysis of motion is summarized in his famous “three laws of motion.” In his great work, the Principia (published in 1687), Newton readily acknowledged his debt to Galileo. In fact, Newton’s first law of motion is close to Galileo’s conclusions. It states that

Every object continues in its state of rest, or of uniform velocity in a straight
line, as long as no net force acts on it.

The tendency of an object to maintain its state of rest or of uniform velocity in a straight
line is called inertia. As a result, Newton’s first law is often called the law of inertia.

CONCEPTUAL EXAMPLE Newton’s first law. A school bus comes to a sudden stop, and all of the backpacks on the floor start to slide forward. What force causes them to do that?

## Inertial Reference Frames

Newton’s first law does not hold in every reference frame. For example, if your reference frame is fixed in an accelerating car, an object such as a cup resting on the dashboard may begin to move toward you (it stayed at rest as long as the car’s velocity remained constant). The cup accelerated toward you, but neither you nor anything else exerted a force on it in that direction.

Newton’s first law does not hold in every reference frame. For example, if your reference frame is fixed in an accelerating car, an object such as a cup resting on the dashboard may begin to move toward you (it stayed at rest as long as the car’s velocity remained constant). The cup accelerated toward you, but neither you nor anything else exerted a force on it in that direction. Similarly, in the reference frame of the decelerating bus there was no force pushing the backpacks forward. In accelerating reference frames, Newton’s first law does not hold. Reference frames in which Newton’s first law does hold are called inertial reference frames (the law of inertia is valid in them). For most purposes, we usually make the approximation that a reference frame fixed on the Earth is an inertial frame. This is not precisely true, due to the Earth’s rotation, but usually it is close enough.

Any reference frame that moves with constant velocity (say, a car or an airplane) relative to an inertial frame is also an inertial reference frame. Reference frames where the law of inertia does not hold, such as the accelerating reference frames discussed above, are called noninertial reference frames. How can we be sure a reference frame is inertial or not? By checking to see if Newton’s first law holds. Thus Newton’s first law serves as the definition of inertial reference frames.

## Mass

Newton’s second law makes use of the concept of mass. Newton used the term mass as a synonym for quantity of matter. This intuitive notion of the mass of an object is not very precise because the concept “quantity of matter” is not very well defined. More precisely, we can say that mass is a measure of the inertia of an object. The more mass an object has, the greater the force needed to give it a particular acceleration. It is harder to start it moving from rest, or to stop it when it is moving, or to change its velocity sideways out of a straight-line path. A truck has much more inertia than a baseball moving at the same speed, and a much greater force is needed to change the truck’s velocity at the same rate as the ball’s. The truck therefore has much more mass.

To quantify the concept of mass, we must define a standard. In SI units, the unit of mass is the kilogram (kg) as we discussed in our

The terms mass and weight are often confused with one another, but it is important to distinguish between them. Mass is a property of an object itself (a measure of an object’s inertia, or its “quantity of matter”). Weight, on the other hand, is a force, the pull of gravity acting on an object. To see the difference, suppose we take an object to the Moon. The object will weigh only about one-sixth as much as it did on Earth, since the force of gravity is weaker. But its mass will be the same. It will have the same amount of matter as on Earth, and will have just as much inertia-for in the absence of friction, it will be just as hard to start it moving on the Moon as on Earth, or to stop it once it is moving.

## Newton’s Second Law of Motion

Newton’s first law states that if no net force is acting on an object at rest, the object remains at rest; or if the object is moving, it continues moving with constant speed in a straight line. But what happens if a net force is exerted on an object? Newton perceived that the object’s velocity will change. A net force exerted on an object may make its velocity increase. Or, if the net force is in a direction opposite to the motion, the force will reduce the object’s velocity. If the net force acts sideways on a moving object, the direction of the object’s velocity changes (and the magnitude may as well). Since a change in velocity is an acceleration, we can say that a net force causes acceleration.

What precisely is the relationship between acceleration and force? Everyday experience can suggest an answer. Consider the force required to push a cart when friction is small enough to ignore. (If there is friction, consider the net force, which is the force you exert minus the force of friction.) If you push the cart with a gentle but constant force for a certain period of time, you will make the cart accelerate from rest up to some speed, say 3 km/h. If you push with twice the force, the cart will reach 3 km/h in half the time. The acceleration will be twice as great. If you triple the force, the acceleration is tripled, and so on. Thus, the acceleration of an object is directly proportional to the net applied force. But the acceleration depends on the mass of the object as well. If you push an empty grocery cart with the same force as you push one that is filled with groceries, you will find that the full cart accelerates more slowly. The greater the mass, the less the acceleration for the same net force. The mathematical relation, as Newton argued, is that the acceleration of an object is inversely proportional to its mass. These relationships
are found to hold in general and can be summarized as follows:

The acceleration of an object is directly proportional to the net force acting
on it, and is inversely proportional to the object’s mass. The direction of the
acceleration is in the direction of the net force acting on the object.

This is Newton’s second law of motion. It can be written as an equation:

$\ \vec a = \frac{\sum \vec F}{m}$

where $\vec a$ stands for acceleration, m for the mass, and $\sum \vec F$  for the net force on the object. The symbol ∑ (Greek “sigma”) stands for “sum of”; $\vec F$ stands for force, so $\sum \vec F$ means the vector sum of all forces acting on the object, which we define as the net force.

We rearrange this equation to obtain the familiar statement of Newton’s
second law:

$\ \sum \vec F = m \vec a$                                                             (1a)

Newton’s second law relates the description of motion (acceleration) to the cause of motion (force). It is one of the most fundamental relationships in physics. From Newton’s second law we can make a more precise definition of force as an action capable of accelerating an object.

Every force F is a vector, with magnitude and direction. The second law is a vector equation valid in any inertial reference frame. It can be written in component form in rectangular coordinates as

$\sum F_x = ma_x$       $\sum F_y = ma_y$       $\sum F_z = ma_z$         (1b)

where

$\vec F = F_x \vec i + F_y \vec j + F_z \vec k$

The component of acceleration in each direction is affected only by the component
of the net force in that direction.

In SI units, with the mass in kilograms, the unit of force is called the newton (N). One newton, then, is the force required to impart an acceleration of 1 m/s² to a mass of 1 kg. Thus 1 N = 1 kg’m/s².

It is very important that only one set of units be used in a given calculation or problem, with the SI being preferred. If the force is given in, say, newtons, and the mass in grams, then before attempting to solve for the acceleration in SI units, we must change the mass to kilograms. For example, if the force is given as 2.0 N along the x axis and the mass is 500 g, we change the latter to 0.5 kg, and the acceleration will then automatically come out in m/s² when Newton’s second law is used:

$a_x = \frac{\sum F_x}{m} = \frac{2.0 N }{0.5 kg } = 4.0 m/s^{2}$

EXAMPLE 1: ESTIMATE Force to accelerate a fast car.

Estimate the net force needed to accelerate (a) a 1000-kg car at g/2; (b) a 200-g apple at the same rate.

APPROACH We use Newton’s second law to find the net force needed for each object. This is an estimate (the 1/2 is not said to be precise) so we round off to one significant figure.

SOLUTION (a) The car’s acceleration is a = g/2 = (9.8 m/s²)/2 ≈ 5 m/s². We use
Newton’s second law to get the net force needed to achieve this acceleration:

∑ F = ma ≈ (1000 kg)(5 m/s²) = 5000 N.

(b) For the apple, m = 200 g = 0.2 kg, so

∑ F = ma ≈ (0.2 kg)(5 m/s²) = 1 N.

EXAMPLE 2 Force to stop a car. What average net force is required to bring a 1500-kg car to rest from a speed of 100 km/h within a distance of 55 m?

APPROACH We use Newton’s second law, ∑F = ma, to determine the force, but first we need to calculate the acceleration a. We assume the acceleration is constant, so we can use the kinematic equations to calculate it.

SOLUTION We assume the motion is along the + x axis (Fig. above). We are given
the initial velocity $v_0 = 100$ km/h = 27.8 m/s, the final velocity v = 0, and the distance traveled $x - x_0$ = 55 m. From the third kinematic equation we have:

$v^{2} = v_0^{2} +2a(x-x_0)$

$a = \frac{v^{2} - v_0^{2}}{2(x-x_0)} = \frac{0-(27.8m/s)^{2}}{2(55m)} = -7.0 m/s^{2}$

The net force required is then

∑F = ma = (1500 kg)( -7.1 m/s²) = -1.1 x $10^{4}$ N.

The force must be exerted in the direction opposite to the initial velocity, which is
what the negative sign means.

Newton’s second law, like the first law, is valid only in inertial reference frames. In the noninertial reference frame of an accelerating car, for example, a cup on the dashboard starts sliding-it accelerates-even though the net force on it is zero; thus $\sum \vec F = m \vec a$ doesn’t work in such an accelerating reference frame ($\sum \vec F =0$, but $\vec a \neq 0$ in this noninertial frame).

## Newton’s Third Law of Motion

Newton’s second law of motion describes quantitatively how forces affect motion. But where, we may ask, do forces come from? Observations suggest that a force exerted on any object is always exerted by another object. A horse pulls a wagon, a person pushes a grocery cart, a hammer pushes on a nail, a magnet attracts a paper clip. In each of these examples, a force is exerted on one object, and that force is exerted by another object. For example, the force exerted on the nail is exerted by the hammer.

But Newton realized that things are not so one-sided. True, the hammer exerts a force on the nail. But the nail evidently exerts a force back on the hammer as well, for the hammer’s speed is rapidly reduced to zero upon contact. Only a strong force could cause such a rapid deceleration of the hammer. Thus, said Newton, the two objects must be treated on an equal basis. The hammer exerts a force on the nail, and the nail exerts a force back on the hammer. This is the essence of Newton’s third law of motion:

Whenever one object exerts a force on a second object, the second exerts an equal force in the opposite direction on the first.

This law is sometimes paraphrased as “to every action there is an equal and opposite reaction.” This is perfectly valid. But to avoid confusion, it is very important to remember that the “action” force and the “reaction” force are acting on different objects.

As evidence for the validity of Newton’s third law, look at your hand when you push against the edge of a desk ( Fig. below). Your hand’s shape is distorted, clear evidence that a force is being exerted on it. You can see the edge of the desk pressing into your hand. You can even feel the desk exerting a force on your hand; it hurts! The harder you push against the desk, the harder the desk pushes back on your hand. (You only feel forces exerted on you; when you exert a force on another object, what you feel is that object pushing back on you.)

The force the desk exerts on your hand has the same magnitude as the force your hand exerts on the desk. This is true not only if the desk is at rest but is true even if the desk is accelerating due to the force your hand exerts.

As another demonstration of Newton’s third law, consider a person who throws a package out of a small boat (initially at rest), the boat starts moving in the opposite direction. The person exerts a force on the package. The package exerts an equal and opposite force back on the person, and this force propels the person (and the boat) backward slightly.

Rocket propulsion also is explained using Newton’s third law. A common misconception is that rockets accelerate because the gases rushing out the back of the engine push against the ground or the atmosphere. Not true. What happens, instead, is that a rocket exerts a strong force on the gases, expelling them; and the gases exert an equal and opposite force on the rocket. It is this latter force that propels the rocket forward-the force exerted on the rocket by the gases. Thus, a space vehicle is maneuvered in empty space by firing its rockets in the direction opposite to that in which it needs to accelerate. When the rocket pushes on the gases in one direction, the gases push back on the rocket in the opposite direction. Jet aircraft too accelerate because the gases they thrust out backwards exert a forward force on the engines (Newton’s third law).

CHECK YOURSELF : Does a speeding missile possess force?

CHECK YOUR ANSWER: No, a force is not something an object has, like mass, but is part of an interaction between one object and another. A speeding missile may possess the capability of exerting a force on another object when interaction occurs, but it does not possess force as a thing in itself.

CONCEPTUAL EXAMPLE: What exerts the force to move a car? What
makes a car go forward?

RESPONSE A common answer is that the engine makes the car move forward. But it is not so simple. The engine makes the wheels go around. But if the tires are on slick ice or deep mud, they just spin. Friction is needed. On firm ground, the tires push backward against the ground because of friction. By Newton’s third law, the ground pushes on the tires in the opposite direction, accelerating the car forward.

Consider how we walk. A person begins walking by pushing with the foot backward against the ground. The ground then exerts an equal and opposite force forward on the person (Fig. below), and it is this force, on the person, that moves the person forward. (If you doubt this, try walking normally where there is no friction, such as on very smooth slippery ice.) In a similar way, a bird flies forward by exerting a backward force on the air, but it is the air pushing forward (Newton’s third law) on the bird’s wings that propels the bird forward.

We tend to associate forces with active objects such as humans, animals, engines, or a moving object like a hammer. It is often difficult to see how an inanimate object at rest, such as a wall or a desk,  can exert a force. The explanation is that every material, no matter how hard, is elastic (springy) at least to some degree. A stretched rubber band can exert a force on a wad of paper and accelerate it to fly across the room. Other materials may not stretch as readily as rubber, but they do stretch or compress when a force is
applied to them. And just as a stretched rubber band exerts a force, so does a stretched (or compressed) wall, desk, or car fender.

From the examples discussed above, we can see how important it is to remember on what object a given force is exerted and by what object that force is exerted. A force influences the motion of an object only when it is applied on that object. A force exerted by an object does not influence that same object; it only influences the other object on which it is exerted. Thus, to avoid confusion, the two prepositions on and by must always be used-and used with care.

One way to keep clear which force acts on which object is to use double subscripts. For example, the force exerted on the Person by the Ground as the person walks can be labeled $\vec F_{PG}$ . And the force exerted on the ground by the person is $\vec F_{GP}$.  By Newton’s third law:

$\vec F_{PG} = - \vec F_{GP}$                                                                        (2)

$\vec F_{GP}$ and $\vec F_{PG}$ have the same magnitude (Newton’s third law), and the minus sign reminds us that these two forces are in opposite directions.

Note carefully that the two forces shown in the figure above act on different objects-hence we used slightly different colors for the vector arrows representing these forces. These two forces would never appear together in a sum of forces in Newton’s second law,

$\sum \vec F = m \vec a$. Why not? Because they act on different objects: $\vec a$ is the acceleration of one particular object, and $\sum \vec F$ must include only the forces on that one object.

## Weight – the Force of Gravity; and the Normal Force

Galileo claimed that all objects dropped near the surface of the Earth would fall with the same acceleration $\vec g$, if air resistance was negligible. The force that causes this acceleration is called the force of gravity or gravitational force. What exerts the gravitational force on an object? It is the Earth and the force acts vertically downward  toward the center of the Earth. Let us apply Newton’s second law to an object of mass m falling freely due to gravity. For the acceleration $\vec a$, we use the downward acceleration due to gravity $\vec g$. Thus, the gravitational force on an object, $\vec F_G$ can be written as

$\vec F_G = m \vec g$                                                                                                      (3)

The direction of this force is down toward the center of the Earth. The magnitude
of the force of gravity on an object, mg, is commonly called the object’s weight.

In SI units, g = 9.8 m/s² = 9.8 N/kg; so the weight of a 1.0-kg mass on Earth is 1.00 kg X 9.80 m/s² = 9.80 N. We will mainly be concerned with the weight of objects on Earth, but we note that on the Moon, on other planets, or in space, the weight of a given mass will be different than it is on Earth. For example, on the Moon the acceleration due to gravity is about one-sixth what it is on Earth, and a 1.0-kg mass weighs only 1.6 N.

The force of gravity acts on an object when it is falling. When an object is at rest on the Earth, the gravitational force on it does not disappear, as we know if we weigh it on a spring scale. The same force, given by Eq. (3), continues to act. Why, then, doesn’t the object move? From Newton’s second law, the net force on an object that remains at rest is zero. There must be another force on the object to balance the gravitational force. For an object resting on a table, the table exerts this upward force; (see the fig. below).

The table is compressed slightly beneath the object, and due to its elasticity, it pushes up on the object as shown. The force exerted by the table is often called a contact force, since it occurs when two objects are in contact. (The force of your hand pushing on a cart is also a contact force.) When a contact force acts perpendicular to the common surface of contact, it is referred to as the normal force (“normal” means perpendicular); hence it is labeled $\vec F_N$.

The two forces (shown in the fig. above) are both acting on the statue, which remains at rest, so the vector sum of these two forces must be zero (Newton’s second law). Hence $\vec F_G$ and $\vec F_N$ must be of equal magnitude and in opposite directions. But they are not the equal and opposite forces spoken of in Newton’s third
law. The action and reaction forces of Newton’s third law act on different objects, whereas the two forces  $\vec F_G$ and $\vec F_N$ act on the same object. For each of the forces shown in the above figure, we can ask, “What is the reaction force?” The upward force, $\vec F_N$, on the statue is exerted by the table. The reaction to this force is a force exerted by the statue downward on the table. On the figure above it is labeled $\vec F_N'$. This force $\vec F_N'$, exerted on the table by the statue, is the reaction force to $\vec F_N$ in accord with Newton’s third law. What about the other force on the statue, the force of gravity $\vec F_G$ exerted by the Earth? Can you guess what the reaction is to this force? We will see later that the reaction force is also a gravitational force, exerted on the Earth by the statue.

EXAMPLE 3:  Weight, normal force, and a box.

A friend has given you a special gift, a box of mass 10.0 kg with a mystery surprise inside. The box is resting on the smooth (frictionless) horizontal surface of a table.

(a) Determine the weight of the box and the normal force exerted on it by the
table.

(b) Now your friend pushes down on the box with a force of 40.0 N. Again determine the normal force exerted on the box by the table.

(c) If your friend pulls upward on the box with a force of 40.0 N, what now is the normal force exerted on the box by the table?

APPROACH The box is at rest on the table, so the net force on the box in each case is zero (Newton’s second law). The weight of the box has magnitude mg in all three cases.

SOLUTION (a) The weight of the box is mg = (10.0 kg)(9.80 m/s²) = 98.0 N, and this force acts downward. The only other force on the box is the normal force exerted upward on it by the table.

We chose the upward direction as the positive y direction; then the net force $\sum F_y$ on the box is   $\sum F_y = F_N - mg$;  the minus sign means mg acts in the negative y direction (m and g are magnitudes). The box is at rest, so the net force on it must be zero. Thus

$\sum F_y = ma_y = F_N - mg = 0$

so we have

$F_N = mg$

The normal force on the box, exerted by the table, is 98.0 N upward, and has
magnitude equal to the box’s weight.

(b) Your friend is pushing down on the box with a force of 40.0 N. So instead of only two forces acting on the box, now there are three forces acting on the box. The weight of the box is still mg = 98.0 N. The net force is $\sum F_y = F_N - mg - 40 N$, and is equal to zero because the box remains at rest (a = 0). Newton’s second law gives

$\sum F_y = F_N - mg - 40 N = 0$

We solve this equation for the normal force:

$F_N = mg + 40.0 N = 98.0 N + 40.0 N = 138.0 N$

which is greater than in (a). The table pushes back with more force when a person pushes down on the box. The normal force is not always equal to the weight!

(c) The box’s weight is still 98.0 N and acts downward. The force exerted by your
friend and the normal force both act upward (positive direction). The box doesn’t move since your friend’s upward force is less than the weight.

The net force, again set to zero in Newton’s second law because a = 0, is

$\sum F_y = F_N - mg + 40 N = 0$

so

$F_N = mg - 40.0 N = 98.0 N - 40.0 N = 58.0 N$

The table does not push against the full weight of the box because of the upward
NOTE The weight of the box (= mg) does not change as a result of your friend’s
push or pull. Only the normal force is affected.

Recall that the normal force is elastic in origin (the table in the example above sags
slightly under the weight of the box). The normal force in Example above is vertical,
perpendicular to the horizontal table. The normal force is not always vertical, however. When you push against a wall, for example, the normal force with which the wall pushes back on you is horizontal. For an object on a plane inclined at an angle to the horizontal, such as a skier or car on a hill, the normal force acts perpendicular to the plane and so is not vertical.

## Solving Problems with Newton’s laws: Free-Body Diagrams

When solving problems involving Newton’s laws and force, it is very important to draw a diagram showing all the forces acting on each object involved. Such a diagram is called a free-body diagram, or force diagram: choose one object, and draw an arrow to represent each force acting on it. Include every force acting on that object. Do not show forces that the chosen object exerts on other objects. To help you identify each and every force that is exerted on your chosen object, ask yourself what other objects could exert a force on it. If your problem involves more than one object, a separate free-body diagram is needed for each object.

Practice problems: For each of the following situations, specify the system and draw a motion diagram and a free-body diagram. Label all forces with their agents, and indicate the direction of the acceleration and of the net force. Draw vectors of appropriate lengths.

1. A flowerpot falls freely from a windowsill. (Ignore any forces due to air resistance.)
2. A sky diver falls downward through the air at constant velocity. (The air exerts an
upward force on the person.)
3. A cable pulls a crate at a constant speed across a horizontal surface. The surface
provides a force that resists the crate’s motion.
4. A rope lifts a bucket at a constant speed. (Ignore air resistance.)
5. A rope lowers a bucket at a constant speed. (Ignore air resistance.)

Using Newton’s Laws:

1.Draw a sketch of the situation.

2. Consider only one object (at a time), and draw a free-body diagram for that object, showing all the forces acting on that object. Include any unknown forces that you have to solve for. Do not show any forces that the chosen object exerts on other objects.

Draw the arrow for each force vector reasonably accurately for direction and magnitude. Label each force acting on the object, including forces you must solve for,
as to its source (gravity, person, friction, and so on). If several objects are involved, draw a free-body diagram for each object separately, showing all the forces acting on that object (and only forces acting on that object). For each (and every) force, you must be clear about: on what object that force acts, and by what object that force is exerted. Only forces acting on a given object can be included in $\sum \vec F = m \vec a$ for that object.

3. Newton’s second law involves vectors, and it is usually important to resolve vectors into components. Choose x and y axes in a way that simplifies the calculation. For example, it often saves work if you choose one coordinate axis to be in the direction of the acceleration.

4. For each object, apply Newton’s second law to the x and y components separately. That is, the x component of the net force on that object is related to the x component of that object’s acceleration: $\sum F_x = ma_x$, and similarly for the y direction.

5. Solve the equation or equations for the unknown(s).

This Problem Solving Strategy should not be considered a prescription. Rather it is a summary of things to do that will start you thinking and getting involved in the problem at hand.

When we are concerned only about translational motion, all the forces on a given object can be drawn as acting at the center of the object, thus treating the object as a point particle. However, for problems involving rotation or statics, the place where each force acts is also important.

Example Problem 4 Fighting Over a Toy

Ann is holding a stuffed dog, with a mass of 0.30 kg, when Sarah decides that she wants it and tries to pull it away from Ann. If Sarah pulls horizontally on the dog with a force of 10.0 N and Ann pulls with a horizontal force of 11.0 N, what is the horizontal acceleration of the dog?

I.  Analyze and Sketch the Problem

• Sketch the situation.
• Identify the dog as the system and the direction in which Ann pulls as positive.
• Draw the free-body diagram. Label the forces.

Known:

m = 0.30 kg

$F_{Ann On dog} = 11.0 N$

$F_{Sarah On dog} = 10.0 N$

Unknown:

a = ?

II. Solve for the Unknown

$F_{net} = F_{Ann On dog} +(-F_{Sarah On dog})$

Use Newton’s second law.

$a = \frac{F_{net}}{m} = \frac{11.0 N - 10.0 N }{0.3 kg} = 3.3 m/s^{2}$

a = 3.3 m/s² toward Ann.

• Are the units correct? m/s² is the correct unit for acceleration.
• Does the sign make sense? The acceleration is in the positive direction, which is expected, because Ann is pulling in the positive direction with a greater force than Sarah is pulling in the negative direction.
• Is the magnitude realistic? It is a reasonable acceleration for a light, stuffed toy.

Example Problem 5 Two boxes connected by a cord.

Two boxes, A and B, are connected by a lightweight cord and are resting on a smooth (frictionless) table. The boxes have masses of 12.0 kg and 10.0 kg. A horizontal force $F_p$ of 40.0 N is applied to the 10.0-kg box, as shown in the figure below. Find (a) the acceleration of each box, and (b) the tension in the cord connecting the boxes.

APPROACH We streamline our approach by not listing each step. We have two boxes so we draw a free-body diagram for each. To draw them correctly, we must consider the forces on each box by itself, so that Newton’s second law can be applied to each. The person exerts a force $F_p$ on box A. Box A exerts a force $F_T$ on the connecting cord, and the cord exerts an opposite but equal magnitude force $F_T$ back on box A (Newton’s third law). The cord is light, so we neglect its mass. The tension at each end of the cord is thus the same. Hence the cord exerts a force $F_T$ on the second box. There will be only horizontal motion. We take the positive x axis to the right.

SOLUTION (a) We apply $\sum F_x = ma_x$ to box A:

$\sum F_x = F_P-F_T=m_A a_A$

For box B, the only horizontal force is $F_T$ , so

$\sum F_x = F_T = m_B a_B$

The boxes are connected, and if the cord remains taut and doesn’t stretch, then
the two boxes will have the same acceleration a. Thus $a_A=a_B=a$. We are
given $m_A = 10.0 kg$ and  $m_B = 12.0 kg$. We can add the two equations above to eliminate an unknown ($F_T$ ) and obtain

$(m_A +m_B)a = F_P -F_T+F_T=F_P$

or

$a = \frac{F_P}{M_A+m_B}=\frac{40.0 N}{22.0 kg} 1.82 m/s^{2}$

This is what we sought.

Alternate Solution We would have obtained the same result had we considered
a single system, of mass m A + mB, acted on by a net horizontal force equal to $F_p$. (The tension forces $F_T$ would then be considered internal to the system as a whole, and summed together would make zero contribution to the net force on
the whole system.)

(b) From the equation above for box B ($F_T =m_B a_B$) the tension in the cord is

$F_T =m_B a_B= (12.0 kg)(1.82 m/s^{2})=21.8 N$

Thus, $F_T$ is less than $F_p$ (= 40.0 N), as we expect, since $F_T$ acts to accelerate only $m_B$.

NOTE It might be tempting to say that the force the person exerts, $F_p$, acts not only on box A but also on box B. It doesn’t.  $F_p$ acts only on box A. It affects box B via the tension in the cord, $F_T$ , which acts on box B and accelerates it.

Example Problem 2  The advantage of a pulley.

A mover is trying to lift a piano (slowly) up to a second-story apartment. He is using a rope looped over two pulleys as shown. What force must he exert on the rope to slowly
lift the piano’s 2000-N weight?

SOLUTION: The magnitude of the tension force $F_T$ within the rope is the same at
any point along the rope if we assume we can ignore its mass. First notice the forces acting on the lower pulley at the piano. The weight of the piano pulls down on the pulley via a short cable. The tension in the rope, looped through this pulley, pulls up twice, once on each side of the pulley. Let us apply Newton’s second law to the pulley-piano combination (of mass m), choosing the upward direction as positive:

$2F_T - mg = ma$

To move the piano with constant speed (set a = 0 in this equation) thus requires a tension in the rope, and hence a pull on the rope, of $F_T = mg /2$. The mover can exert a force equal to half the piano’s weight. We say the pulley has given a mechanical advantage of 2, since without the pulley the mover would have to exert twice the force.

Example Problem 3 Lifting a Bucket

A 50.0-kg bucket is being lifted by a rope. The rope will not break if the tension is 525 N or less. The bucket started at rest, and after being lifted 3.0 m, it is moving at 3.0 m/s. If the acceleration is constant, is the rope in danger of breaking?

I.  Analyze and Sketch the Problem

• Draw the situation and identify the forces on the system.
• Establish a coordinate system with the positive axis upward.
• Draw a motion diagram including v and a.
• Draw the free-body diagram, labeling the forces.

Known:

m= 50.0 kg

$v_f = 3.0 m/s$

$v_i = 0.0 m/s$

d = 3.0 m

Unknown:

$F_T = ?$

II. Solve for the Unknown

$F_{net}$ is the sum of the positive force of the rope pulling up, $F_T$, and the negative weight force, $-F_g$, pulling down as defined by the coordinate system.

$F_{net} = F_T + (- F_g)$

$F_T = F_{net} + F_g$ = ma + mg = m(a+g)

$v_i$, $v_f$, and d are known. Use this motion equation to solve for a.

$v_f^{2} = v_i^{2} + 2ad$

$a = \frac{v_f^{2} - v_i^{2}}{2d} = \frac{v_f^{2}}{2d}$

$F_T = m( \frac{v_f^{2}}{2d}+g) = 570 N$

The rope is in danger of breaking because the tension exceeds 525 N.

• Are the units correct? Dimensional analysis verifies kg.m/s², which is N.
• Does the sign make sense? The upward force should be positive.
• Is the magnitude realistic? The magnitude is a little larger than 490 N, which is the
weight of the bucket.  $-F_g$ = mg = (50.0 kg)(9.80 m/s²) = 490 N

## Sample test questions/problems

1. When, if ever, can a vector quantity be added to a scalar quantity?

2. Calculate the resultant of a horizontal vector with a magnitude of 4 units and a vertical vector with a magnitude of 3 units.

3. Can the velocity of an object reverse direction while maintaining a constant acceleration? If so, give an example; if not, provide an explanation.

4. Is it possible to move in a curved path in the absence of a force? Defend your answer.

5. A 400-kg bear grasping a vertical tree slides down at constant velocity. What is the friction force that acts on the bear?

6. A crate remains at rest on a factory floor while you push on it with a horizontal force F. How big is the friction force exerted on the crate by the floor? Explain.

7. A race car travels along a raceway at a constant velocity of 200 km/h, What horizontal forces act, and what is the net force acting on the car?

8. Three identical blocks are pulled, as shown, on a horizontal frictionless surface, If tension in the rope held by the hand is 30 N, what is the tension in the other ropes?

9. How much force acts on a tossed coin when it is half way to its maximum height? How much force acts on it when it reaches its peak? (Neglect air resistance.)

10. What is the acceleration of a rock at the top of its trajectory when it has been thrown straight upward?

11. What is the net force acting on a 1-kg ball in free fall?

12. What is the net force acting on a falling 1-kg ball if it encounters 2 N of air resistance?

13. When a parachutist opens her parachute, in what direction does she accelerate?

14. How does the gravitational force on a falling body compare with the air resistance it encounters before it reaches terminal velocity? After reaching terminal velocity?

15. If you drop a pair of tennis balls simultaneously from the top of a building, they will strike the ground at the same time. If you fill one of the balls with lead pellets and then drop them together, which one will hit the ground first? Which one will experience

16. What is the greatest acceleration a runner can muster if the friction between her shoes and the pavement is 90% of her weight?

17. What is the acceleration of a 40-kg block of cement when pulled sideways with a net force of 200 N?

18. What is the acceleration of a 20-kg pail of cement that is pulled upward (not sideways!) with a force of 300 N?

19. How much acceleration does a 747 jumbo jet of mass 30 tons experience in takeoff when the thrust for each of four engines is 30 000 N?

20. Two boxes are seen to accelerate at the same rate when a force F is applied to the first and 4F is applied to the second. What is the mass ratio of the boxes?

21. A firefighter of mass 80 kg slides down a vertical pole with an acceleration of 4 m/s 2 What is the friction force that acts on the firefighter?

22. What will be the acceleration of a skydiver when air resistance builds up to be equal to half her weight?

23. We know that the Earth pulls on the Moon. Does it follow that the Moon also
pulls on the Earth?

24. Two 100-N weights are attached to a spring scale as shown. Does the scale read 0, 100, or 200 N, or does it give some other reading?

25. Consider the two forces acting on the person who stands still- namely, the downward pull of gravity and the upward support of the floor. Are these forces equal and opposite?
Do they form an action-reaction pair? Why or why not?

26. 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?

27. Ken and Joanne are astronauts floating some distance apart in space. They are joined by a safety cord whose ends are tied around their waists. If Ken starts pulling on the cord, will he pull Joanne toward him, or will he pull himself toward Joanne, or will both astronauts move? Explain.

28. Two people of equal mass attempt a tug-of-war with a 12-m rope while standing on frictionless ice. When they pull on the rope, each of them slides toward the other. How do their accelerations compare, and how far does each person slide before they meet?

29. Which of the following are scalar quantities, which are vector quantities, and which are neither? (a) velocity; (b) age; (c) speed; (d) acceleration; (e) temperature.

30. What can you correctly say about two vectors that add together to equal zero?

31. A boxer punches a sheet of paper in midair and brings it from rest up to a speed of 25 m/s in 0.05 s. If the mass of the paper is 3 g, what force does the boxer exert on it?

32. According the graph below, what is the force being exerted on the 16-kg cart?

33. Three blocks that are connected by massless strings are pulled along a frictionless
surface by a horizontal force, as shown in  the Figure below. a. What is the acceleration of each block? b. What are the tension forces in each of the strings? Hint: Draw a separate free-body diagram for each block.

34. Two blocks, one of mass 5.0 kg and the other of mass 3.0 kg, are tied together with a massless rope as in the figure below. This rope is strung over a massless, resistance-free pulley. The blocks are released from rest. Find the following: a. the tension in the rope
b. the acceleration of the blocks Hint: you will need to solve two simultaneous equations.

35. Suppose a 65-kg boy and a 45-kg girl use a massless rope in a tug-of-war on an icy, resistance-free surface as in the figure below. If the acceleration of the girl toward the boy is 3.0 m/s², find the magnitude of the acceleration of the boy toward the girl.

36. What is your weight in newtons?

37. When a softball with a mass of 0.18 kg is dropped, its acceleration toward Earth is equal to g, the acceleration due to gravity. What is the force on Earth due to the ball, and what is Earth’s resulting acceleration? Earth’s mass is 6.0 x $10^{24}$ kg.

38. A window washer pulls herself upward using the bucket-pulley apparatus shown in the figure below. (a) How hard must she pull downward to raise herself slowly at constant speed? (b) If she increases this force by 15%, what will her acceleration be? The mass of the person plus the bucket is 72 kg.

39. A 1.5-kg block rests on top of a 7.5-kg block. The cord and pulley have negligible mass, and there is no significant friction anywhere. (a) What force F must be applied to the bottom block so the top block accelerates to the right at 2.5 m/s²? (b) What is the tension in the connecting cord?