Engineering Mechanics Notes – Complete Study Material for First Year

Engineering Mechanics is a fundamental subject for all engineering branches. These comprehensive notes cover Statics and Dynamics with solved examples, formulas, and important concepts for university exams.

Course Overview

Topic	Weightage
Statics – Force Systems	20%
Equilibrium	15%
Friction	15%
Centroid & Moment of Inertia	20%
Kinematics	15%
Kinetics	15%

Unit 1: Statics – Force Systems

1.1 Basic Concepts

Force: An action that changes or tends to change the state of rest or motion of a body
Types: Contact forces, Body forces, Point forces, Distributed forces
Characteristics: Magnitude, Direction, Point of application, Line of action

1.2 Resolution of Forces

Rectangular Components:

Fx = F cos θ (Horizontal component)
Fy = F sin θ (Vertical component)
F = √(Fx² + Fy²)
θ = tan⁻¹(Fy/Fx)

1.3 Resultant of Concurrent Forces

Parallelogram Law:

R = √(P² + Q² + 2PQ cos θ)

tan α = Q sin θ / (P + Q cos θ)

1.4 Moment of a Force

Moment = Force × Perpendicular distance

M = F × d (Unit: N-m)

Varignons Theorem: Moment of resultant = Sum of moments of components

1.5 Couple

Two equal, opposite, parallel forces
Moment of couple = F × d (arm of couple)
A couple can only be balanced by another couple

Unit 2: Equilibrium

2.1 Conditions of Equilibrium

For Coplanar Forces:

ΣFx = 0 (Sum of horizontal forces = 0)
ΣFy = 0 (Sum of vertical forces = 0)
ΣM = 0 (Sum of moments about any point = 0)

2.2 Types of Supports

Support	Reactions	Unknowns
Roller	Normal only	1
Hinged/Pinned	Horizontal + Vertical	2
Fixed	H + V + Moment	3

2.3 Free Body Diagram (FBD)

Steps to draw FBD:

Isolate the body from surroundings
Show all external forces
Replace supports with their reactions
Include weight acting at center of gravity

Unit 3: Friction

3.1 Laws of Friction

Friction acts opposite to direction of motion/tendency
Friction is proportional to normal reaction: f = μN
Friction is independent of area of contact
Friction depends on nature of surfaces

3.2 Types of Friction

Static friction: fs ≤ μsN (limiting friction = μsN)
Kinetic friction: fk = μkN (μk < μs)
Angle of friction: tan φ = μ
Angle of repose: α = φ (for inclined plane)

3.3 Applications

Ladder problems
Wedge friction
Belt and pulley friction: T1/T2 = e^(μθ)
Screw jack: P = W tan(α ± φ)

Unit 4: Centroid and Moment of Inertia

4.1 Centroid

For composite areas:

x̄ = ΣAixi / ΣAi

ȳ = ΣAiyi / ΣAi

4.2 Moment of Inertia (Second Moment of Area)

Shape	Ixx (about centroid)	Iyy (about centroid)
Rectangle (b×d)	bd³/12	db³/12
Triangle (base b, height h)	bh³/36	hb³/36
Circle (radius r)	πr⁴/4	πr⁴/4
Semicircle	0.11r⁴	πr⁴/8

4.3 Parallel Axis Theorem

I = Ic + Ad²

Where: Ic = MI about centroidal axis, A = area, d = distance between axes

Unit 5: Kinematics

5.1 Rectilinear Motion

v = u + at
s = ut + ½at²
v² = u² + 2as
s = ½(u + v)t

5.2 Projectile Motion

Range: R = u²sin2θ/g
Max Height: H = u²sin²θ/2g
Time of Flight: T = 2u sinθ/g
Trajectory: y = x tanθ – gx²/2u²cos²θ

5.3 Circular Motion

Angular velocity: ω = dθ/dt (rad/s)
Angular acceleration: α = dω/dt (rad/s²)
v = rω, a_t = rα, a_n = v²/r = rω²

Unit 6: Kinetics

6.1 Newtons Laws of Motion

Second Law: F = ma
DAlemberts Principle: F – ma = 0 (Inertia force = -ma)

6.2 Work, Energy, Power

Work: W = F.s.cosθ (Joules)
Kinetic Energy: KE = ½mv²
Potential Energy: PE = mgh
Power: P = W/t = F.v (Watts)
Work-Energy Theorem: W = ΔKE

6.3 Impulse and Momentum

Momentum: p = mv
Impulse: J = FΔt = Δp
Conservation: m1u1 + m2u2 = m1v1 + m2v2
Coefficient of restitution: e = (v2-v1)/(u1-u2)

Important Solved Problems

Practice these types for exam:

Finding resultant of force system
Equilibrium of rigid bodies
Ladder and friction problems
Centroid of composite sections
Moment of inertia calculations
Projectile motion problems
Collision and momentum problems

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