Introduction
From the previous chapters on physical quantities and their corresponding units, you now know that the unit for mass, kilogram, is a fundamental unit. Mass might also be called a fundamental quantity. It relates to motion, forces, and energy, and therefore basically to the whole field of physics. In the following lesson, we will discuss inertia, Newton's laws of classical mechanics that connect motion to forces, and conditions for an equilibrium or non-equilibrium situation when an object is subjected to two or more forces at the same time.
Inertia
Imagine a situation where you have to push some furniture around, for example, if you are moving to a new home. Although you might think that taking clothes out of the dresser drawers before moving the dresser is a waste of time, the reality is that the amount of effort you put into unloading the dresser is much less than the effort you would use to move the full dresser. What opposition do we encounter in the two processes of moving the furniture? It is the mass that makes the difference; it is the mass that causes the resistance. This resistance that an object has to motion is something we call inertia. Mass is the quantity that measures inertia, and the fundamental unit for mass is the kilogram (kg).
Motion on a straight trajectory is called linear motion. Later, we will study objects moving in a circular trajectory, and they too will exhibit inertia; in this case, called rotational inertia.
Newton's First Law of Motion
Newton's first law of motion, or the law of inertia, states that an object continues to stay at rest or move in a uniform linear motion as long as there is no resultant (net) action on it.
Newton's First, Second, and Third Laws of Motion
The idea of resistance to motion was formulated by Newton into what we now call the first law of motion or the law of inertia.
There are two important words in this definition: continues and resultant. To understand this law, you must work with these two words together. Resultant (net) action refers to the idea that there might be an action on the object, or two, or more. The importance is not in individual actions but in the total effect on the object. The other word, continues, refers to one of the two situations: rest or uniform linear motion. That is, the object will maintain, or continue, an identical state of rest or motion at all times if there is no external action.
Most objects around us, when left to act independently of engines, will not continue their uniform linear motion. Also, objects that might be at rest at one time can change their state and accelerate. The motion is a result of interaction, and the objects accelerate (speed up) or decelerate (slow down). The change in the state of motion is due to a resultant force. The relationship among the resultant force, the mass, and the acceleration is the subject of the second law of motion.
As you may remember, physical quantities can be scalars or vectors. In this case, mass is a scalar, but both acceleration and force are vector quantities. Therefore, the second law does not refer only to the proportionality but also to the direction of the acceleration, telling us that the acceleration and the resultant force are in the same direction. The expression of the second law is as follows:
a = 
, or 
where a is the vector acceleration, 
F is the vector representing the resultant force, and m is the mass of the object. This equation defines also the unit for force: a newton, or N. As we replace the unit for acceleration (m/s2) and for mass (kg), we can find the unit for force to be kg . m/s2 = 1 N.
As with all vectors, force will be completely defined by value, unit, and direction. The direction of the net force will also give the type of acceleration; if a > 0 (a positive number is considered to be the direction of motion), then the object is accelerated, and if a < 0 (opposite to motion), then the object is decelerated (slowed down).
Newton's Second Law of Motion
The acceleration of an object is directly proportional to the resultant (net) force and inversely proportional to the mass of the object.
Example 1
A 1,000 kg car starts from a traffic light and moves accelerated in the NE direction with an acceleration of 0.45 m/s2. Find out the resultant force on the car.
Solution 1
in NE direction
F = m · a
1,000 kg · 0.45 m/s2 = 450 N
F = 450 N in NE direction
Example 2
A 2-D force with components Fx = 100 N and Fy = 45 N accelerates an object of mass 5 kg. If this force is the net force on the object, find the value of the acceleration and its direction relative to the x-axis.
Solution 2
According to the second law of motion, we need to divide the force in each direction to the mass of the object to find the two components of the acceleration.
Fx = 100 N
Fy = 45 N
m = 5.0 kg
Because the two components are 90° apart, we can apply Pitagora's theorem to find the acceleration:
To find the direction relative to x-axis, α, we calculate the tangent of the angle and find the angle.
α = 24°
So, acceleration can be written in two ways:
a = (20,9) m/s2 for each component, or
a = 22 m/s2 at 24° north of east
The net force described by Newton's second law or any other force is, by definition, a measure of the interaction. Whenever you interact with an object, you will feel the effect of the interaction yourself. When you kick a ball, the ball will react (due to the interaction) by shooting into the air, but your foot will "feel" the interaction also. This effect is described by the third law of motion.
In this case, one force is called action and the other one reaction. As a very important note: The two forces are acting on different objects and so, although equal and opposite, they still have an effect on the system.
Newton's Third Law of Motion
If you act with a force on an object, the object will react with a force equal and opposite on the first object.
Example 3
Consider a car moving on a road. Determine the action and reaction forces on each tire.
Solution 3
Let's look at the situation for one tire, and the same can be repeated for the remaining tires. The car's left tire acts on the road with a force that we call action, and the road reacts with an equal and opposite force on the tire called reaction. That is how the car can move.
Weight
Let's put to work the knowledge we've accumulated and define some important types of forces in classical mechanics, one of which is weight. Whenever two objects are in proximity, they interact through a force. If you consider Earth and an apple that is about to fall from a tree, the apple is attracted by Earth with the force of gravity, so it will accelerate toward the earth for as long as it falls (if we do not consider other forces present such as friction with air). Because the gravity is the only force on the apple, it will be the only component in the net force. According to the second law, this force is also proportional to the mass of the apple and the acceleration, and we call it weight. The expression of the weight of an object of mass m is:
W = m · g
The symbols in the expression are: W is weight, m is mass, and g is the acceleration. It is very important to note that this expression gives us the argument to answer the question: Are mass and weight the same quantity? The answer is No! Weight is a force, and mass is a scalar quantity that gives a measure of the inertia of the object. In the previous example, the apple falling down due to gravity is in a free fall, and the acceleration of fall, which is the same for all objects at the same place on the earth, is called acceleration due to gravity, g. Although the acceleration due to gravity varies with the mass of the planet and the distance from an object to the center of the planet, for most purposes herein, we will take the acceleration due to gravity on Earth to be g = 9.8 m/s2. Another and last very important note for this section is that weight is a vector always pointing toward the center of Earth.
Example
Consider the apple falling freely toward Earth. Why is Earth not falling toward the apple?
Solution
The third law of motion says that every action has an equal and opposite reaction, so the apple is acted on by gravity, and hence, Earth has to be acted on by an equal and opposite force. The reason the earth does not start moving toward the apple is that the same force acting on a huge object (Earth) has a negligible effect while having a definite effect on a smaller object (the apple).
Normal Force
The fact is that not all objects fall or fly; most of them are at rest or moving in contact with different surfaces. In such a case, we have an action, weight, and a reaction from the surface toward the object, and we call that a normal force, with a symbol N. In the simplest case, an object sits on a horizontal surface and pushes down only with its own weight; the surface responds with an equal and opposite force as shown in Figure 5.4. A drawing that includes all the forces acting on an object is called a free-body diagram.
When the surface is no longer horizontal, part of the weight will not act on the supporting surface, so the normal force of the object is now smaller than the weight of the object as seen in Figure 5.5. In this case, the component of the weight, called Wx , is the one acting on the surface, and normally is:
Wx= – N
The other component of the weight is not equalized by any other force in this case, and therefore, the object will accelerate toward the bottom of the plane. In real-life applications, there will be at least one other force acting. Yes, you guessed it: friction. But we will talk about this later.
In many real-life applications, you will have to take into consideration other forces acting on the object: pushes, pulls, and so on. In these cases, the freebody diagram will be more complex, and the calculation of the normal force can result in amounts smaller or larger than the weight, as we will show in the following example.
Example
Set a book on a table (a mass of 1,000 g) and exert a pushing force on the book of 10.0 N. Now draw the object's free-body diagram and calculate the reaction of the table.
Solution
Start by drawing the free-body diagram that will show the three forces acting in the problem: the weight of the book on the table, the pushing force, and the reaction of the table (normal).
mbook = 1,000 g = 1.00 kg
Fpush = 10.0 N
N = ?
The weight of the book is:
Wbook = m · g = 1 kg · 9.8 m/s2 = 9.8 N
The total force on the table is then:
F = Wbook + Fpush = 19.8 N
And the reaction of the table according to the third law of motion is a force equal and opposite of F:
N = 19.8 N upward
As you can see, this force is larger than the weight of the object.
Tension
In the process of moving small or large objects, we often use tools. One of these tools is a simple string and is the base of more complex machines such as pulleys. A string in a pulley is used to transfer forces from the source of force to the object(s), to reduce the load on the string, and also to change the direction of motion. In the simplest case, it is considered that the string has no mass, hence no energy is needed to accelerate the string (there is no net force on the rope itself).
Example
Consider the following situation for which the freebody diagram needs to be determined.
Solution
The hanging body is acted upon only by its weight and the tension in the rope, which is the reaction of the force on the rope as it can be seen in Figure 5.8.
The object on the surface is acted upon by the tension in the rope and on the vertical by the weight and the normal force.
On the pulley, we have tension from both ends of the rope. (This is similar to a rubber band when you try to pull it down by putting your finger inside it and pulling on it; you can feel each part of the elastic band pulling on your finger.) Also, we have the reaction of the table to which the pulley is attached.
So, the free-body diagram treats the system of objects as independent objects, the connection between them being the tension force (the interaction force).
Kinetic Frictional Force
The magnitude fk of the kinetic frictional force is proportional to the normal force N and to the coefficient of kinetic friction:
fk = μ k · N
where μk is the coefficient of kinetic friction and is a dimensionless quantity.
Friction
So, what happens then in terms of interaction between the object and the surface if we cannot neglect friction? At an atomic level, even the smoothest surface shows some roughness, and so if two such surfaces come into contact, there is a resistance to motion, which we call friction.
The forces of friction are classified in the following ways: (1) static frictional force can have a value of zero up to a maximum frictional force, fsMAX, and has a value dependent on the applied force, and (2) kinetic frictional force, which acts when the object is in motion and is directly opposite to the direction of motion.
Example
Consider a hockey puck of 500 g mass sliding on ice. If the puck's normal force is equal only to the weight of the object, what are the frictional force and the acceleration of the object if the coefficient of kinetic friction is 0.3? In the horizontal plane, consider only the frictional force.
Solution
In order to find the acceleration, construct a free-body diagram and calculate the frictional force. We know:
m = 500 g = 0.5 kg
μk = 0.3
W = N
Then we calculate the normal force (see Figure 5.11):
N = m · g = 0.5 · 9.8 = 4.9 N
fk = μk · N = 0.3 · 4.9 = 1.5 N
fk = 1.5 N
And because there is no other interaction, the second law of mechanics will give us the acceleration of motion. But before calculating acceleration, we determine that forces in the direction of motion will be considered positive and forces opposite to the motion will be negative.
∑F = m · a
∑F = fk = m · a
fk = m · a
Because the motion is in one direction, we can neglect the vector sign and write:
–μk · N = m · a
a = –μk· N/m = –μk · m · g/m =
–μk· g = –3 m/s2
a = –3m/s2
This value of the acceleration tells us that the object is slowing down.
The problem can also be solved numerically using the value of fk = 1.5 N and applying the second law of motion:
a = –fk/m = –1.5/0.5 kg = –3m/s2
Static Equilibrium
An object is in equilibrium when it has zero acceleration. The condition of equilibrium can be expressed as:
∑Fx = 0
∑Fy = 0
These conditions are equivalent to the fact that all forces acting on the object must balance. Then, in the nonequilibrium state, acceleration is nonzero:
∑Fx = m ax
∑Fy = m ay
Example
Consider an object accelerating and decelerating due to gravity on an inclined plane that has a kinetic frictional coefficient μk. Find the expression of the acceleration along the plane in the two cases mentioned.
Solution
Start with the free-body diagram for the first case, applying the conditions of equilibrium appropriate to each direction and then calculate the acceleration (see Figure 5.12).
Then find out the components of the weight along the plane and perpendicular to the plane: Wx and Wy.
Wx = W sin α = m g sin α
Wx = W cos α = m g cos α
Then write Newton's second law for directions along the plane and perpendicular to the plane. Again, we will work with each direction so we can use a scalar equation.
y-axis
ay = 0
m ay = N – Wy
0 = N – Wy
N = Wy = m g cos α
x-axis
m ax = Wx – fk = Wx – μk N
m ax = m g sin α – μk m g cos α
The acceleration is:
axDOWN = g(sin α - μkcos α)
In the second case, when the object is moving up the inclined plane, we have a different free-body diagram. In order for the object to move up, imagine you gave it an intial push. All forces are the same except the fricitonal force, which acts against motion. Then apply Newton's second law along the plane and perpendicular to the plane as follows:
y-axis
ay= 0
m ay = N – Wy
0 = N – Wy
N = Wy = m g cos α
x-axis
–m · ax = Wx + fk = Wx + μkN
–m · ax = m g sin α + μkm g cos α
And the acceleration is:
ax UP = – g(sin α + μk cos α)
Practice problems of this concept can be found at: Newton's Laws of Motion Practice Questions
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