Unit – I

Basics & Statics of Particles

Introduction – units & Dimensions – Law of Mechanics – Lami’s theorem –

Parallelogram and triangular law of forces – Vectors – Vectorial representation of

forces & moments – Vector operations : additions, subtraction, dot product, cross

product – coplaner forces – Resolution & Composition of forces – Equilibrium of a

particle – Forces in space – Equilibrium of particle in space – Equivalent systems of

forces – Principle of transmissibility – Single equivalent force.

Units of Measurement

A physical quantity can be measured by comparing sample with a known

standard amount.

Types

Basics Units

Derived Units

Basics Units

Basic (or) fundamental quantities likes mass, length, time are measured.

Derived Units

Physical quantities are measured it is also known as the secondary units.

Example : Area, Volume, Speed, Velocity, etc.,

System of Units

Foot pound second system (FPS)

Centimetre, Gram, Second System (CGS)

Metre, Kilogram Second System (MKS)

System of International (SI)

SI Units are on the following ‘6’ fundamental units

(i) Metre – Length

(ii) Kilogram - Mass

(iii) Ampere - Electric current

(iv) Second - Time

(v) Kelvin - Temperature

(vi) Candela - Intensity

Another Addition

Radian - Solid angles.

Law of Mechanics

First Law

A particle remain in its [rest (or) motion] if the resultant force acting on the

particle is zero.

Second Law

Acceleration of the particle will be proportional to the resultant force &

direction if the resultant force is not zero.

Third Law

Action & reaction forces b/w the interacting bodies are in the same line of

action equal in magnitude.

Important Laws

Lame’s theorem

Parallelogram theorem

Triangular Law of Force

Polygon Law of Force

Scalar & Vector Quantities

Scalar Quantities

The quantities which possess magnitude only are called scalar quantities.

Example : Length, Area, Volume, Mass, etc.,

Vector Quantities

The quantities which possess magnitude as well as direction are called vector

quantities.

Example : Force, Velocity, Acceleration, etc.,

Detailed Vector Quantities

A vector is a quantity that has both a magnitude & a direction.

A vector is generally represented by a letter with an arrow written over.

Such as F⃗.

Classifications of Vectors

i) Free Vectors - Move on any where

ii) Sliding Vectors - Any point

iii) Fixed Vectors - Fixed Point (Moment)

iv) Unit Vectors - One unit length

v) Negative Vectors - Opposite Directions

vi) Zero Vectors - Same vectors in opposite

[Null Vectors] ie A⃗ - A⃗= 0

Vectors Operations

Rectangular Unit Vectors i.j.k

Vector addition

Vector subtraction

Multiplication of vectors by scalars

Dot (or) Scalar product of vectors

Cross (or) Vector product of vectors

Position vectors

The moment of a force

Rectangular Unit Vectors i.j.k

The unit vector along OX, OY, OZ are denoted by i.j.k. respectively.

Draw PM ⊥ to the plane XOZ & MZ ⊥ to OZ.

Then,

ON = zk, NM = xi, MP = yj

Find OP.

O⃗P = p⃗ = r⃗ = ON + NM + MP

= zk + xi + yj r⃗ – Position Vectors

r⃗ = xi + yj + zk, Magnitude r⃗ = r = √ x2+ y2+z2

More general any vector A⃗ where components along the axes are respectively

Ax, Ay, Az.

A⃗ = Axi + Ayj + Azk

Magnitude A⃗ = | A⃗| = A = √ Ax2+Ay2+ Az2

Vector addition

Two vectors A⃗ & B⃗ may be added by using (i) Parallelogram law (or) (ii)

Triangle law.

Parallelogram

A⃗

A⃗

R⃗ = A⃗ + B⃗

B⃗ B⃗

Triangle

A⃗A⃗ B⃗

R⃗ = A⃗ + B⃗

B⃗

Vector Subtraction

A⃗ - B⃗ = A⃗ + (-B⃗) = R⃗

A⃗ R⃗ = A⃗ + B⃗ B⃗

A⃗ R⃗ = A⃗ + B⃗ A⃗

B⃗ B⃗ O

Multiplication of vectors by scalars :

1) (m + n) A⃗ = m A⃗ + n A⃗

2) m( A⃗+B⃗) = m A⃗ + mB⃗

3) m (n A⃗) = n(m A⃗) = (mn) A⃗

Dot (or) Scalar product of vectors

1) A⃗.B⃗ = B⃗. A⃗

2) A⃗.(B+C⃗) = a.b + a.c

3) m( A⃗.B⃗) = (m A⃗).B⃗ = A⃗.(mB⃗)

Cross (or) Vector product of vectors

A⃗ x B⃗ = -B⃗ x a

A⃗ = A⃗xi + A⃗yj + A⃗zk, B⃗ = B⃗xi + B⃗yj + B⃗zk

=> Angle

A⃗ x B⃗= | i j kAx A y A z

Bx By B z| A⃗ x B⃗ = AB sin θ

a x b = i(Ay Bz – By Az) – j(AxBz - BxAz) + k(AxBy - BxAy)

Position Vector

Position Vector r⃗ = xi + vj + zk

Magnitude r = √ x2+ y2+z2

r⃗

The Moment of a force :

Moment M⃗ = r⃗ x F⃗

r⃗ = xi + yj + zk

F⃗ = Fxi + Fyj + Fzk

M⃗ = r⃗ x F⃗ = | i j kx y z

Fx F y F z|

Problem Note

Dot Product = Scalar Value

ie. Number values

& Another all product = Vector values

ie. (i.j.k) values

(1) Moment of force

Given by : 1) Co – ordinates (P)

2) Position (r⃗)

(2) A⃗B = O⃗B – O⃗A (Subtraction law)

(3) Right angles do not product Answer = 0

(4) Parallel Cross Product Answer = 0

System of force

Force

A force represents the action of one body on another.

Force is vector Quantity.

Characteristic of a force

Magnitude (importance)

Line of action

Direction

System of Force

Coplanar Non – Coplanar

Collinear Concurrent Non-Concurrent Concurrent Non – Concurrent

Like Unlike Parallel Non – Parallel Parallel Non-Parallel

Like Unlike Like Unlike

Fig : System of Force

Coplanar Force

In Coplanar force system all the forces act in one plane.

Collinear Forces :

The forces which acts on a common line of action are called collinear.

F1 F2 F3 F1 F2 F3

> > > > < >

Concurrent Forces

In concurrent force system forces intersects at a common point.

Parallel Forces

In Parallel force system line of action of forces are parallel to each other.

Same Direction Like Opposite Direction Unlike

Non – Coplanar Forces

In Non – Coplanar forces system, the forces do not act in one plane.

Coplanar Non – Concurrent Force

EF

F1 F1

F2

D C F2

A B

Non – Coplanar Concurrent Force

Forces intersects at one point, but their lines of action do not lie on the same

plane.

F1

F2

Non – Coplanar Non – Concurrent Forces

Forces do not intersects at one point & also their lines of action do not lie on

the same plane.

F1

F2

Resultant Force

Resultant force of all the force system can be determined by two methods.

1. Analytical Method

2. Graphical Method

Analytical Method

Resultant force on coplanar

(i) Collinear forces

(i) Like

(ii) Unlike

Like

P Q S = R = (P+Q+S)

> > > >

Unlike

Horizontal => → (+) ← (-)Vertical => ↑ (+) ↓ (-)

P Q S = R => (- P + Q - S)< > < P

Q = R => (P – Q + S)

S

(2) Concurrent Forces

In order to find the resultant force of concurrent forces we shall take.

(i) Resultant force of two concurrent forces.

(ii) Resultant force of more than two concurrent forces.

Resultant force of the concurrent forces

the analytical method of finding out the resultant of two concurrent forces can

be developed from the parallelogram law of forces.

Parallelogram law of forces

If two forces acting simultaneously at a point be represented in magnitude &

direction by the two adjacent sides of a parallelogram then the resultant of these two

forces is represented in magnitude & direction by the diagonal of that parallelogram

originating from that point.

Proof

θ α θ

(OA + AD)

From the geometry of the parallelogram

OB = AC & OA = BC

Hence the force Q can be represented on the line AC.

In triangle ACD

cosθ = ADAC

= ADQ

AD = Q cosθ --------------------- (1) θ

sinθ = CDAC

= CDQ

CD = Q sinθ --------------------- (2)

And also,

AD2 + CD2 = AC2

AD2 + CD2 = Q2 --------------------- (3)

In triangle OCD

OC2 = OD2 + CD2

OC2 = (OA+AD)2 + CD2

OC2 = OA2 + AD2 + 2.OA.AD + CD2 α

OC2 = OA2 + 2.OA.AD + AC2

We Know, OC = R.OA = P | |

AC = Q.AD = Q cosθ

CD = Q sinθ

R2 = P2 + 2.P.Q cosθ + Q2

R2 = P2 + Q2 + 2.P.Q cos θ

/ R = √ P2+Q2+2PQ cosθ

Inclination of the resultant force :

In triangle OCD

tanα = CDOD

= CD

OA+AD =

Q sinθP+Qcos θ

tanα = Q sinθ

P+Q sinθ

Resultant force of more than two concurrent forces

Horizontal : cosθ

→ Right side (→ +) Positive

→ Left side (←, -) Negative

Vertical : sinθ

→ Upward (↑, +) Positive

→ Downward (↓, -) Negative

Solving Procedure

Step 1 : Find the algebraic sum of horizontal components.

∑H

¿ F1 cos θ + F2 cos θ + ....................

Step 2 : Find the algebric sum of the vertical components :

∑v

¿ F1 sin θ + F2 sin θ – F3 sinθ + ....................

Step 3 : Find the magnitude of Resultant force

R = √¿¿¿

Step 4 : Find the direction of Resultant force

tan α = ∑

V

∑H

α = tan-1 (∑V

∑H

)Graphical Method

→ Resultant force of two concurrent forces

(i) Parallelogram law

(ii) Triangle of forces

Triangle of forces

“If two forces acting at a point are represented by the sides of a triangle taken

in order than resultant force in represented by the 3rd side taken in opposite order”

R2 = P2 + Q2 – 2PQ cosθ θ

sinθ = Qsin(180−α )

R α α

Resultant force of more than two concurrent for

(i) polygon law of forces

Polygon law of forces

“If a number of coplanar concurrent forces are represented in magnitude &

direction by the sides of a polygon taken in an order, then their resultant force is

represented by the closing side of the polygon taken in the opposite order”

α

Lami’s Theorem

It states that “if three coplanar forces acting at a point be in equilibrium then

each force is proportional to the sine of the angle b/w the other two”

Psin α

= Q

sin β =

Ssin γ

Proof

Now we will find the resultant force of the forces P & Q.

This can be determined by parallelogram law of forces.

γ

β α

From the geometry of the figure

We know, sum of interior angles in a triangle in 180o

AOC + ACO + CAO = 180o

CAO = 180 – [ AOC + ACO ]

= 180 – [(180 - β) + (180 - α)]

AOC = (180 - β) ------ (1)

BOC = (180 - α)

ACO = BOC

BOC = (180 - α) ------ (2)

AOC

= 180 – 180 + β – 180 + α

CAO = α + β - 180

We know α + β + γ = 360o

Subtract 180o from either side

(α + β + γ) – 180 = 360 – 180

(α + β - 180) + γ = 180

CAO + r = 180

CAO = (180 - γ)

Apply sine rule AOC

OAsin ACO

= OB

sin AOC =

OCsin OAC

OAsin (180−α) =

ACsin(180−β) =

OCsin(180−γ )

Psinα

= Q

sinβ =

Ssin γ

Note :

1. Lami’s theorem is only applied for three coplanar concurrent forces, which

are in equilibrium.

2. Lami’s theorem cannot be applied directly for three concurrent equilibrium

forces.

Equilibrium of a Particle

Conditions of Equilibrium

For Equilibrium condition of force system the resultant is zero.

ie R = O.

But R = √∑H

2+∑

v

2

Hence both a ∑H

¿∑v

are to be zero for equilibrium condition.

Equations of Equilibrium in two dimensions :

∑H

¿0. HL collinear forces.

∑v

¿0. VL collinear forces.

Both ∑H

¿0, ∑v

¿0 for concurrent forces.

Principle of Equilibrium :

(i) Two force principle

If a body is subjected to two forces, then the body will be in equilibrium if the

two forces are collinear equal & opposite.

(ii) Three force principle

If a body is subjected to three force, then the body will be equilibrium. If the

resultant of any two forces is equal, opposite & collinear with the third force.

Force in space

Force vector interms of co-ordinates

(i) Components of the force

x axis Fx = Fdx

d, y axis Fy =

Fdyd

dx = (x2 – x1)

dy = (y2 – y1)

d = √dx2+dy2

(ii) Angle with co-ordinates

Ox = cos-1 ( FxF ) or

dxd

Oy = cos-1 ( FyF ) or

dyd

(iii) Force vector

F⃗ = Fxi + Fyi

(iv) Resultant force

R = √ Fx2+Fy2

Given by :

Total force F.

Angle = ?

Fx = F cosθx

Fy = F cosθy

Fy = F cosθz

F = Fxi + Fyi + Fzk

Check in direction

cos2θx + cos2θy + cos2θz = 1.

Resultant force of coplanar concurrent force system

Let R⃗ be the resultant force

R⃗ = √ Rx2+Ry2

Direction :

θx = cos-1 ( RxR )

θy = cos-1 ( RyR )

Rx = ∑Fx

¿Ry = ∑Fy

Vector approach for a three – dimensional

F⃗ = Fxi + Fyj + Fzk

(or)

F⃗ = (F cosθx) i + (F cosθy) j + (F cosθz) k

|F⃗| = F = √ Fx2+Fy2+Fz2

Angle :

θx = cos-1 ( FxF )

θy = cos-1 ( FyF )

θz = cos-1 ( FzF )

Equilibrium of particles in space

When a particle is subjected to concurrent forces in space for equilibrium

condition

Resultant force R⃗ = 0

ie. Rxi + Ryj + Rzk = 0

¿i + ¿j + ¿k = 0

∑Fx

¿0 ,∑Fy

¿0 ,∑Fz

¿0

Problems

Unit vector (λ)

“Same for two dimensional only

addition of ‘z’ part”

λ = Fxi+Fyi+Fzk

√Fx2+Fy2+Fz2

Position Vectors

r⃗ = xi + yj + zk

r = √ x2+ y2+z2

Principle of Transmissibility

If a force acts at any point on a rigidbody it may also be considered to act at

any other point on its line of actions.

Step 1 Step 2 Step 3