# Newton’s Laws of Motion

### Concept of force :

A force is a push or a pull acting on a body. It is a vector quantity : i.e. it has both magnitude & direction.

In 3 dimensions, using vector notation, we can write ,$\displaystyle \vec{F}= F_x \hat{i} + F_y \hat{j} + F_z \hat{k}$ where Fx , Fy and Fz represent the X , Y , Z components of the force F and  $\hat{i} , \hat{j} , \hat{k}$  represent the (dimensionless) unit vectors along X , Y and Z respectively.

Forces are measured in SI units in Newton .  Newton (N) is that force which when applied to a body of mass 1 kg , causes an acceleration of 1 m/s2.

### System ?

A system consists of bodies, whose motion is to be analysed. In this chapter we will be analysing the motion of rigid bodies.

### Classification of Forces : The first classification is self explanatory.

All forces acting on a body can be classified as:

(a) Contact forces and

(b) Non-contact forces

(a) Contact force: Forces experienced by objects due to contact with each other are contact forces The component of the contact force normal to the surface of contact (or line of contact) is usually known as the normal reaction, and a tangential component of the force, may act along the surface of contact.

(b) Non-contact force: Bodies can exert forces on each other without actual physical contact. This is known as action at a distance. Such forces are known as non-contact forces .e.g. gravitation, coulomb repulsion between like charges, etc.

For the moment, we will deal with actual forces. Suffice it to say that there exist pseudo-forces acting in a non-inertial frame of reference.

Forces may be conservative or non-conservative depending on whether work done against them by an external agent is recoverable or otherwise . Some typical forces we will be dealing with are tension, spring force, normal reaction, etc. Some free body diagrams for these forces are shown below :

(i) Tension in a string : For a block A pulled by a string, (ii) Spring forces : (a) F = kx

Where x = extension in the spring

= present length – normal length (b) F’ = kx’
where x’ = compression in the spring

= normal length − present length

You can use either of the above diagrams: (A) or (B)

(iii) Normal Reaction A block A rests on another block B. The normal reaction N acts between A and B as shown in the diagram.

### Fundamental Forces in nature ?

The following basic forces operate in nature :

(i) Gravitational force: The force of attraction between two objects due to their masses is known as gravitational force All object fall because of the gravitational force of attraction exerted on them by the earth.
The gravitational force has the following properties

(i) It is always attractive in nature.

(ii) It is long range force.

(iii) It is the weakest force operating in nature

(iv) It obeys inverse square law.

(v) It is a central force.

(ii) Electromagnetic force :- The force between two charges is called electrostatic force while the force between two magnetic poles is called magnetic force.
The electromagnetic force has the following properties.

(i) It may be attractive or repulsive in nature.

(ii) It is also a long rang force.

(iii) It is about 1036 times stronger than gravitational force.

(iv) It obeys inverse square law.

(v) It is also a central force

(iii) Strong force :- It is the force of nuclear origin within the atomic nucleus, the stronger nuclear force holds the nucleons- Protons and neutrons together. It has the following properties –

(i) Basically it is an attractive force.

(ii) It is short rang force of about the nuclear dimensions of 10-15m

(iii) It varies inversely with some higher power of distance.

(iv) It is the strongest force operating in nature.

(v) It is a non-central force.

(iv) Weak force :- In the process of radioactive decay. The nucleus emits an electron and neutron . The electron and the neutron interact with each other exclusive through the weak force. The weak forces are about 1025 times as strong as gravitational forces.

## Newton’s Laws of Motion:

### Newton’s first law of motion :

Every body continues in its state of rest or of uniform motion in a straight line unless it is compelled by an externally impressed force to act otherwise.

The tendency of material bodies to maintain their velocity constant is known as inertia Mass is a measure of inertia. When an external force acts on a body, it changes its velocity.

### Newton’s second law of motion :

The rate of change of momentum of a body is directly proportional to the applied force and acts in the direction of the force.

Momentum is the ‘quantity of motion’ present in a body. It is a vector quantity and for small enough velocities (v « c, the velocity of light), it is given by,

$\displaystyle \vec{P} = m\vec{v}$ , where m is the mass of the body and v its velocity.

For a body ,

$\displaystyle \frac{\vec{dp}}{dt} \varpropto \vec{F}$

$\displaystyle \frac{d (m\vec{v})}{dt} \varpropto \vec{F}$

$\displaystyle m\frac{\vec{dv}}{dt} \varpropto \vec{F}$

$\displaystyle \vec{F} = k m \frac{\vec{dv}}{dt}$

where k is a constant. With proper choice of units, k = 1

Thus , $\displaystyle \vec{F} = m \vec{a}$

Solved Example : A body of mass m = 1 kg falls from a height h = 20 m from the ground level
(a) What is the magnitude of the total change in momentum of the body before it strikes the ground ?
(b) What is the corresponding average force experienced by it?
(g = 10m/sec2).

Solution: (a) Since the body falls from rest (u = 0) through a distance h before striking the ground, the speed v of the body is given by

v2 = u2 + 2 as ; putting a = g and s = h

we obtain $\displaystyle v = \sqrt{2 g h}$

The magnitude of total change in momentum of the body

= ΔP = | mv – 0 | = mv

where $\displaystyle v = \sqrt{2 g h}$

⇒ $\displaystyle \Delta P = m \sqrt{2 g h}$

$\displaystyle \Delta P = ( 1 )\sqrt{2 \times 10 \times 20}$

ΔP = 20 kg m/sec

The average force experienced by the body = $\displaystyle F_av = \frac{\Delta P}{\Delta t}$

, where Δt = time of motion of the body = t(say)

We know ΔP = 20 kg m/sec.

Therefore we will have to find t using the given data. We know from kinematics that,

$\displaystyle S = ut + \frac{1}{2}a t^2$

$\displaystyle h = \frac{1}{2}g t^2$ , (Since , u = 0)

$\displaystyle t = \sqrt {\frac{2h}{g} } \, and \, \Delta P = m\sqrt{2gh}$

$\displaystyle F_av = \frac{\Delta P}{\Delta t}$

Putting the general values of ΔP & t we obtain

Fav = mg

where mg is the weight (W) of the body and is directed vertically downwards. Therefore the body experiences a constant vertically downward force of magnitude mg.

Exercise :
A body of mass m = 2 kg changes its velocity v1 = (i + 2j + 3k) m/sec to v2 = (-2i + 2j – k) m/sec during 2 seconds.
(a) Find the magnitude of the change in momentum of the body .
(b) What is the magnitude of the average force experienced by the body ?

### Newton’s third law of motion

Every action has an equal and opposite reaction. This implies that if a force is applied by a body A on a body B, then a reaction force acts on A due to B.

According to the third law,
$\displaystyle \vec{F_{AB}} = -\vec{F_{BA}}$ , is they are equal in magnitude but opposite in direction.

This law is true for all forces – contact forces (such as friction)and non-contact forces (such as gravitation and electrostatics ) but does not apply to pseudo forces.

Solved Example : Two blocks of mass 2 kg and 4 kg are kept in contact with each other on a smooth horizontal surface. A horizontal force of 12 N is applied on the first block due to which the pair moves with a constant acceleration. Calculate the force acting between the blocks. Solution : Let a be the common acceleration of the blocks, R = Force between the two blocks

From FBD of Ist block

F – R = m1 a   . .. . (i)

From FBD of block 2,

R = m2 a     . . . .(ii)

F = (m1 + m2). a

$\displaystyle a = \frac{F}{m_1 + m_2}$

= 12/6 = 2 m/sec2

Force between blocks

= R = m2 a

= (4 ×2) N = 8 N.

### Application of laws of motion : Techniques & Approach

A separate diagram of the body is drawn showing all the different forces exerted by the bodies in the environment. This diagram is known as the free body diagram.

Application of Newton’s Laws to any system (consisting of one or more objects) can be done by following a systematic method.

We Recommend the following steps in the order given below :

(i) Draw the complete free body diagram (FBD), showing all the forces acting on each separate body.

(ii) Select proper coordinates for analysing the motion of each body.
Include any pseudo-forces within the FBD, if required.

(iii) If there are any constraints, write the proper constraint equations.

(iv) Apply Newton’s 2nd law of motion : F-> = ma-> for each body. This leads to a system of equations.

(v) Solve these equations:

(a) Identify the known and unknown quantities. Check that the no. of equations equals the number of unknowns.
(b) Check the equations using dimensional analysis.

(c) After solving, check the final solution using back substitution.

(vi) If the velocity (v ) or, position ( x) is required, proceed from a knowledge of the acceleration ( a->) as found from equations in step (v) and apply kinematics: dv/dt = a (known) and finally integrate.