Newton’s (three) Laws of Motion
In the seventeenth century Sir Isaac Newton used the concept of force to describe the motion of a particle by three laws.
Newton’s laws are valid when the velocity of the particle does not approach the speed of light and the size of the particle does not go down to the atomic level.
Newton’s laws lie in the domain of classical mechanics.
So, to talk about Newton’s laws of motion, we have to, first of all, understand what is the meaning of “motion” – an object in motion.
The simplest object would be a dimensionless particle. So, the object we are considering is effectively a “mathematical point”.
In reality, this point can move around all over three-dimensional space.
However, for the sake of simplicity, we restrict its existence only along the x-axis – that is, along a particular line.
At any instant of time t, the position of the object, with respect to a reference point O, can be denoted by x. The relation between x and t can be represented by a function
x = f (t)
Say, x0 is the position of the object at time t0, and then x1, x2, …… at t1, t2, ……..
Let us plot t along horizontal axis, x along vertical axis to obtain the figure on the right.
For a function such as x = 1 + 2 t, we have a simple linear relation between x and t, shown by the blue line.
But the function does not always have to be linear.
It can be x = 1 + 4 t – t2, and we have a curve concaving downwards (red).
Or, x = 1 – 4 t + t2, giving a curve concaving upwards (green).
Let us take the function represented by the green line and examine it further.
Velocity of the object at an instant of time t is given by the rate of change of x with respect to t at that particular instant – represented by the tangent to the curve at (t, x) – and, in terms of calculus, written as
This way, we can calculate velocity at different instants of time.
Similarly, velocity can be obtained for the other functions, and we can plot a v vs t graph (below left) – this time taking t as the horizontal axis and v as the vertical axis.
Then, the rate of change of velocity with respect to time at any instant is the acceleration at that particular instant
Now, we can make the following observations from the graph.
- For the blue function – velocity does not change – remains constant – acceleration a = 0
- For the green function – velocity is increasing – acceleration +ve, and
- For the red function – velocity is decreasing – acceleration –ve
So far, we have been restricted to motion of an object along a straight line (x-axis).
In 3-dimension, location of object is denoted by position vector r.
If the object is at r1 at time t1 and r2 at time t2, then
and the velocity at any instant of time is given by
while the acceleration by
Note that vector quantities have been denoted by bold letters – we can use notations with arrows at the top.
Evidently, it is implied that
- When v = 0, the object is motionless or at rest w.r.t. a particular reference point or frame.
v is nonzero if the magnitude or direction or both of r change with time.
Similarly, if the magnitude or direction or both of v change with time, a is non-zero.
Now, the question
Can and, if so, how the velocity changes with time?
This was answered first by Galileo in his “principle of inertia”
“A body at rest remains at rest and a body in motion continues to move at constant velocity along a straight line unless acted upon by an external force.”
What are these “forces” ?
The interaction of an object with another object is described in terms of a force F (which is a vector).
There are several forces we are familiar with in everyday experience – friction, air drag, fluid resistance, weight, etc.
All forces are expressions of just four distinct classes of fundamental forces, or interactions between particles
- Strong interactions
- Weak interactions
(These are extremely short-ranged – they are responsible for interactions between subatomic particles).
(And there are two other forces we are familiar with in our everyday experience)
3. The electromagnetic or electrical force between two charged particles 1 and 2 with charges q1 and q2 respectively separated by a distance r is given by
where (epsilon)0 is an electrical constant.
4. The gravitational force – between two particles 1 and 2 of masses m1 and m2 respectively the gravitational force is expressed as
G is the gravitational constant.
** Friction is a macroscopic manifestation of electromagnetic force
Newton used the concept of force to describe the motion of an object by three laws – these laws lie in the domain of classical mechanics.
The first law (essentially a restatement of Galileo’ principle of inertia)
“Every body perseveres in its state of rest or of uniform motion in a right line unless it is compelled to change that state by forces impressed upon it.”
Galileo’s principle of inertia or Newton’s laws are valid in any inertial frame of reference, that is, a frame which itself is either “at rest” or moving with a uniform velocity.
The first law provides a qualitative description of a force
– something that can change the motion (velocity) of an object
“Change in motion” can be seen from another perspective – by introducing linear momentum.
Linear momentum is the product of the mass of an object and its velocity : p = m v
The second law (a quantitative description of the changes that a force can produce on the motion of an object).
“The time rate of change of the linear momentum of an object is equal, both in magnitude and direction, to the force impressed upon it”.
If m is constant
If multiple forces act on the object
Now, the force that acts on an object to change its momentum is due to another object in its environment. So,
The third law
“If one object exerts a force on another, the other exerts the same force in opposite direction on the one”.
Consider two particles 1 and 2, let
then, according to the third law.
Application of Newton's laws
The best way to understand Newton’s laws is to solve problems by applying these laws.
To solve problems,
(i) we need units of the quantities involved.
The internationally accepted standards of units – Le Système International d’Unités (abbreviated as SI) units:
Primary SI units
Time : second (s)
Length : meter (m)
Mass : kilogram (kg)
and
Displacement : m
Velocity : m s – 1
Acceleration : m s – 2
In F = m a
Mass : kilogram, acceleration : m s – 2 and Force : Newton
and,
(ii) we have to draw what is known as a free-body diagram
where the object is shown by a dot
and the forces on the object are shown by arrows
Problem # 1
An object is moving in a straight line along the x-axis
The net force exerted on the object as a function of time is represented by the force vs time graph on the left.
Which of the graphs A – E could represent momentum as a function of time ?
Let us examine the given F-t graph segment wise, on the basis of Newton’s 
In segment AB, force is linearly increasing with time.
So, by integration, we calculate the momentum which is increasing parabolically.
In segment BC, force is constant, and in segment CD, force is linearly decreasing, that is, the slope is negative.
The momenta have been calculated in both cases and shown.
a1 to a7, are all constants.
Now, just on the basis of the nature of the first segment AB, we can discard momentum-time graphs A, B and D, since in each case the momentum is seen to be decreasing in the first segment.
So, we are left with graphs E and C.
In both cases, the momentum is increasing in the first segment.
However, in E the momentum is linearly increasing – hence, this is not the correct one.
On the other hand, in C, the momentum is seen to be increasing parabolically, which is consistent with the F-t graph.
Further, in the second segment of C, the momentum is linearly increasing while in the third segment it is parabolically decreasing – both are consistent with the F-t graph.
Hence, C is the correct answer
Problem # 2
We have an object moving along the x-axis – the position is recorded at different time points – and an x(t) curve is plotted.
We need to comment on the velocity and acceleration of the object, and hence the forces acting on it, at different time points.
Answer: Based on 
We find the following:
(i) between time t = 0 sec and t = 3 sec, the x is increasing with t, but the x–t graph is not linear, rather parabolic – this indicates that velocity is increasing – and hence the object is accelerating.
(ii) between t = 3 and t = 7, x is increasing linearly with t, hence velocity is constant and there is no acceleration.
(iii) beyond 7 s, the x–t curve slopes downwards – velocity is decreasing, that is, the object is decelerating (negative acceleration).
(iv) at t = 10.5, the x-t curve levels off and, so, v = 0
Clearly, from the change (or no change) in velocity with time, we can easily say that there is accelerating force between t = 0 and t = 3, no force between t = 3 and t = 7, and a retarding force (in opposite direction) beyond t = 7.
Problem # 3
We have a picture (mass m) hung by two massless strings which make angles 60° and 30° respectively with the horizon.
We have to find the tension in each string.
Free-body diagram
Three forces are acting on the picture – the gravitational force Fg = mg, and the tension forces, T1 and T2.
Since the picture is at rest and does not accelerate, the net force acting on it must be zero. That is,
Decomposing each of the forces into x- and y-directions:
T1x = T1 cos 30° T1y = T1 sin 30°
T2x = T2 cos 60° T2y = T2 sin 60°
Fg = mg is only in y-direction.
Since, 
we can write
T1 cos 30° – T2 cos 60° = 0
T1 sin 30° + T2 sin 60° – mg = 0
So, using the values
we have
If m = 1 kg (g = 10 m s-2) mg = 10 N
T1 = 5 N and T2 = 8.66 N
Problem # 4
Two blocks – mass 5 kg and 3 kg respectively – connected by massless string – pulled on a frictionless surface by force F = 16 N
What is the tension in the string?
This is a three-body problem – we have three objects – two blocks and a (massless) string.
Here are the free-body diagrams of the three:
The 5-kg object is pulled to the right with a force called tension denoted by T
The string is pulled to the left by the 5-kg object with a force T
The string is considered to be massless – so that there is no net force on the string itself – otherwise the acceleration would be infinite – hence it will be pulled to the right by force T
The 3-kg block is pulled to the left by the string with a force T – the block is pulled to the right by a force 16 N.
Applying
We have
(i) for 5-kg block: T = 5 a, and
(ii) for 3-kg block: 16 – T = 3 a
Combining the two equations, which are rather very simple,
a = 2 and T = 10 N