# Newton’s Laws of Motion

Contents

## Summary

• Newton’s First Law of Motion: Every body continues in a state of rest or if uniform motion in a straight line unless it is acted upon by an external force and is compelled to change its state.
• Newton’s Second Law of Motion: The rate of change of momentum of a body is directly proportional to the force acting and takes place in the direction of that force. F = ma
• Newton’s Third Law of Motion: To every action, there is always an equal and opposite reaction.

These Newton’s laws of motion were first discovered by ”Galileo” at the end of the sixteenth century, but Newton stated them explicitly and gave them a proper form in his Mathematical Principles of Natural Philosophy which was first published in 1687.

#### Newton’s First Law of Motion

Every body continues in a state of rest or if uniform motion in a straight line unless it is acted upon by an external force and is compelled to change its state. This law gives us a definition of force. Force is an agent which produces acceleration. If there is no resultant force acting on a body, then the body will continue moving with uniform velocity.

#### Newton’s Second Law of Motion

The rate of change of momentum of a body is directly proportional to the force acting and takes place in the direction of that force. This law gives us the fundamental equation of dynamics. It explicitly defines the unit of force.

Let:

F = Force acting on a body

m = mass of the body

a = acceleration produced by the force, then by the above law

F is directly proportional to the rate of change of momentum

F = K[ Rate of change of Momentum]

$=\quad K\frac { d }{ dt } (mv)$ $=\quad Km\frac { dv }{ dt }$

F = Km a

Taking K = 1, the fundamental equation becomes:

F = ma

When m = 1kg and  $a\quad =\quad 1{ m }/{ { s }^{ 2 } }$,  then F = 1 unit.

Hence we define unit of force Newton as:

A force of 1 Newton produces an acceleration of  $a\quad =\quad 1{ m }/{ { s }^{ 2 } }$  in a body of mass 1 kg.

i) If F = 0, then a = 0 and the body is said to be in equilibrium. This verifies Newton’s first law of motion.

ii) F = ma is a vector equation. Force and acceleration have the same direction.

iii) If F is constant, then the acceleration will also be constant.

iv) If there are many forces acting on a body, then we will first calculate the resultant force F and the body will move in the direction of F.

#### Newton’s Third Law of Motion

To every action, there is always an equal and opposite reaction. This law will be used in detail when we deal with two connected particles. These particles are connected with a light inextensible string. The string passes over a fixed smooth pulley.

We apply Newton’s law of motion or individual particle and solve the two equations simultaneously.

#### Example #1

Q. A particle of mass 6 kg is pulled from rest along a rough horizontal plane by a horizontal force of magnitude 24N. The coefficient of friction between the particle and the plane is  $\frac { 1 }{ 3 }$.

Find:

(i) The normal and frictional contact force acting on the particle.
(ii) Find the acceleration of the particle.
(iii) Find the displacement of the particle after 4 seconds

Solution:

(i) The Normal contact force is R

R = 60

Frictional Force:   $F\quad =\quad \mu R$

$=\quad \frac { 1 }{ 3 } (60)$

$=\quad 20\quad N$

(ii) Newton’s law of motion gives:

$24\quad -\quad \mu R\quad =\quad ma$

$24\quad -\quad 20\quad =\quad 6a$

$6a\quad =\quad 4$

$a\quad =\quad \frac { 2 }{ 3 } \quad{ m }/{ { s }^{ 2 } }$

(iii) u = 0t = 4$a\quad =\quad \frac { 2 }{ 3 }$s = ?

Using:

$s\quad =\quad ut\quad +\quad \frac { 1 }{ 2 } a{ t }^{ 2 }$

$s\quad =\quad 0(4)\quad +\quad \frac { 1 }{ 2 } (\frac { 2 }{ 3 } )\quad (4)^{ 2 }$

$s\quad =\quad \frac { 16 }{ 3 } \quad m$

#### Example #2

Q. A car of mass 500 Kg is brought to rest in 5 seconds from a speed of 15 m/s neglecting resistance. Find the braking force required to achieve this.

Solution:

First we apply equation of motion to find the retardation of the car.

u = 15 m/s, v =0 , t = 5 seconds

Using: v = u + a t

0 = 15 + a(5)

$a\quad =\quad \frac { -15 }{ 5 }$

$a\quad =\quad -3{ \quad m }/{ { s }^{ 2 } }$

The retardation of the car is  $3{ \quad m }/{ { s }^{ 2 } }$.

Now using Newton’s law of motion: F = m a

F = 500 (3)

F = 1500 N