Engineering Mechanics (Mech) — Overview

Engineering mechanics is the backbone of every mechanical design — before any machine moves, stationary forces must be balanced and moving bodies must obey Newton's laws, making this topic non-negotiable for RRB JE Mechanical candidates.

Statics: forces in balance

Definition: Force — a push or pull that tends to change the state of rest or motion of a body. Measured in newtons (N).

Definition: Equilibrium — a body is in equilibrium when the net force and net moment (torque) acting on it are both zero (ΣF = 0; ΣM = 0).

A resultant force is the single equivalent force that produces the same effect as all the original forces combined. You find it by the parallelogram law (for two forces) or the polygon law (for many forces — draw them head-to-tail; the closing side is the resultant, reversed).

Varignon's Theorem (Principle of Moments): The moment of a resultant force about any point equals the algebraic sum of the moments of the component forces about the same point. This lets you break an awkward force into convenient horizontal and vertical components and sum their moments instead.

Lami's Theorem: When three concurrent, coplanar forces keep a body in equilibrium, each force is proportional to the sine of the angle between the other two:

F₁/sin α = F₂/sin β = F₃/sin γ

where α, β, γ are the angles opposite the respective forces (i.e., the angles between the other two forces). This is the fastest tool for three-force equilibrium problems on exams.

Free Body Diagram (FBD): Isolate the body; replace every contact and external agent with its force arrow. The FBD is the single most important tool in statics — draw it before writing any equation.

Centre of gravity and centroid

Definition: Centre of gravity (CG) — the point where the total weight of the body effectively acts. For uniform gravity, CG coincides with the centroid (geometric centre).

Key centroids to remember:

• Triangle: at 1/3 height from base (h/3 from base).
• Semi-circle: 4r/3π from the diameter.
• Rectangle/Square: at the intersection of diagonals.

Moment of inertia (I): A measure of a cross-section's resistance to bending. For a rectangle (width b, depth h): I = bh³/12 about the centroidal axis. For a circle (diameter d): I = πd⁴/64.

Truss analysis

A truss is a structure of two-force members connected by pin joints. All loads apply at joints; no moments at joints.

• Method of joints: Resolve forces at each joint (ΣFx = 0, ΣFy = 0). Start at a joint with only two unknowns.
• Method of sections: Cut through the truss (through no more than 3 unknown members), isolate one side, and apply equilibrium. Faster when only one member's force is needed.

Dynamics: bodies in motion

Kinematics deals with motion without asking why (no forces). The three SUVAT equations for uniform acceleration:

• v = u + at
• s = ut + ½at²
• v² = u² + 2as

where u = initial velocity, v = final velocity, a = acceleration, t = time, s = displacement.

Newton's three laws:

1. A body stays at rest (or uniform motion) unless a net force acts on it.
2. Net force = mass × acceleration → F = ma.
3. Every action has an equal and opposite reaction.

Work-Energy Theorem

The net work done on a body equals its change in kinetic energy:

W_net = ΔKE = ½mv² − ½mu²

This is enormously useful: when friction is present, set the work done against friction equal to the loss of KE.

Momentum and Impulse

Definition: Momentum (p) = mass × velocity = mv (kg·m/s).

Impulse = F × t = Δp (change in momentum). A large force applied briefly, or a small force over a long time, can produce the same change in momentum — the principle behind airbags (increasing collision time reduces peak force on the occupant).

Conservation of momentum: In the absence of external forces, total momentum before collision = total momentum after. Applies to every collision type.

Friction

Definition: Friction — a contact force that opposes relative motion (or tendency of motion) between surfaces.

• Static friction (f_s): Acts when the body is stationary. Can vary from zero up to a maximum: f_s(max) = μ_s × N, where μ_s is the coefficient of static friction and N is the normal reaction.
• Kinetic (sliding) friction (f_k): Acts once the body slides: f_k = μ_k × N. Always f_k < f_s(max) for the same surfaces.

Angle of friction (φ): The angle the total reaction (N + friction vector resultant) makes with the normal. tan φ = μ.

Angle of repose (θ): The maximum slope angle at which a block rests without sliding. tan θ = μ_s. Numerically, angle of repose = angle of friction.

Inclined plane problems

For a block of mass m on a plane inclined at angle θ (coefficient of friction μ):

• Component along plane (gravity): mg sin θ (downward along slope)
• Normal reaction: N = mg cos θ
• Friction force: f = μ mg cos θ (opposing motion)
• Condition to slide: mg sin θ > μ mg cos θ → tan θ > μ

Screw jack: A screw jack converts torque to large lifting force. Efficiency depends on lead angle and friction angle. The self-locking condition (jack stays put when torque removed): lead angle < friction angle.

Belt friction (Euler's equation): T₁/T₂ = e^(μθ), where T₁ and T₂ are tensions on tight and slack sides, θ is the angle of wrap in radians. Exam favourite for belt-drive problems.

Simple Machines

A simple machine multiplies force (mechanical advantage, MA = Load/Effort) or changes its direction.

Velocity ratio (VR) = distance moved by effort / distance moved by load.
Efficiency = MA/VR × 100%.

Worked Example

Question: A 10 kg block on a horizontal surface (μ = 0.3) is pushed by a horizontal force of 40 N. Find the acceleration. (g = 10 m/s²)

Solution:

Step 1: Normal reaction N = mg = 10 × 10 = 100 N.

Step 2: Kinetic friction f_k = μ_k × N = 0.3 × 100 = 30 N.

Step 3: Net force = Applied force − Friction = 40 − 30 = 10 N.

Step 4: F = ma → a = F/m = 10/10 = 1 m/s².

Conclusion: The block accelerates at 1 m/s² in the direction of the applied force.

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Question: Three coplanar concurrent forces of 10 N, 20 N, and F are in equilibrium. The angle between the 10 N and 20 N forces is 150°, and between 20 N and F is 120°. Find F using Lami's Theorem.

Solution:

Step 1: Angle opposite to F (between the other two forces, 10 N and 20 N) = 150°.

Step 2: 10/sin 120° = 20/sin 90° = F/sin 150°.

Step 3: 20/sin 90° = 20/1 = 20. F = 20 × sin 150° = 20 × 0.5 = 10 N.

Conclusion: The third force F = 10 N.

Real-world example

Indian railways uses truss bridges (e.g., Howrah Bridge uses a cantilever truss) to span wide rivers with minimum material. Method of sections is the exact tool designers use to find the most stressed members. When a heavily loaded goods train crosses, every member must carry its design load — the equilibrium equations keep it standing.

Common misconception

Many students think friction always opposes motion. More precisely, static friction opposes the tendency of relative motion — it can point in any direction needed to keep a body stationary, including uphill along a slope. Only kinetic friction always acts exactly opposite to the direction of sliding.

Engineering Mechanics is the single highest-yield technical chapter in RRB JE CBT-2 Mechanical — expect 6–9 questions on force systems, equilibrium, friction, centroid/moment of inertia, trusses and particle dynamics. It is scoring because most questions are direct formula-substitution, not derivation.

Why it matters

Every later subject (Strength of Materials, Theory of Machines, Fluid Mechanics) rests on free-body diagrams and equilibrium. Master ΣF=0 / ΣM=0 and you unlock half the paper.

Must-know concepts

• Resultant of two concurrent forces: R = √(P² + Q² + 2PQ·cosθ); direction tanα = Q·sinθ/(P + Q·cosθ).
• Equilibrium of coplanar forces: ΣFx = 0, ΣFy = 0, ΣM = 0.
• Lami's theorem (three concurrent forces): P/sinα = Q/sinβ = R/sinγ.
• Friction: limiting friction F = μN; angle of friction φ, tanφ = μ; angle of repose = φ.
• Moment = Force × perpendicular distance; a couple = F×d (free vector).

Key data table

Quantity	Formula
Centroid of triangle (from base)	h/3
MOI of rectangle (centroidal)	bd³/12
MOI of circle (dia d)	πd⁴/64
Parallel-axis theorem	I = I_G + A·h²
Projectile range	R = u²·sin2θ/g
Max height	H = u²·sin²θ/(2g)
Time of flight	T = 2u·sinθ/g
Plane-truss determinacy	m + r = 2j

Newton's 2nd law F = m·a; work–energy: Work done = change in KE; momentum p = m·v; impulse = F·t = Δp.

Exam Tricks & Tips

• 🎯 Range is maximum at θ = 45°; complementary angles (30° & 60°) give the same range.
• 🎯 On a ladder/block problem, always draw the free-body diagram first — friction acts opposite to impending motion.
• 🎯 Angle of repose = angle of friction (tanφ = μ) — a favourite one-liner.
• 🎯 For a truss, use method of joints when all member forces are wanted, method of sections for one or two members.
• 🎯 A couple has zero resultant force but non-zero moment — it cannot be balanced by a single force.
• ❌ Common mistake: forgetting that moment of inertia of a circle is πd⁴/64 (diameter), but πr⁴/4 (radius) — mixing d and r doubles the error.

Expected exam pattern

Numericals on resultant forces, Lami's theorem, friction on inclined planes, centroid/MOI of composite sections, projectile motion, and 1–2 truss (zero-force member) questions.

Quick recap

Equilibrium = ΣF=0 & ΣM=0. tanφ = μ. MOI rectangle bd³/12, circle πd⁴/64. Range max at 45°. Truss check: m + r = 2j.

Cover the answer, recall the value/formula, then check. 12 cards on RRB JE Engineering Mechanics essentials.

Q1. Resultant of two concurrent forces P and Q with angle θ between them?
A1. R = √(P² + Q² + 2PQ·cosθ). Maximum when θ=0° (R=P+Q), minimum when θ=180° (R=|P−Q|).

Q2. State Lami's theorem.
A2. For three concurrent coplanar forces in equilibrium: P/sinα = Q/sinβ = R/sinγ, where each angle is opposite its force.

Q3. Relation between coefficient of friction and angle of friction?
A3. μ = tanφ. The angle of repose equals the angle of friction.

Q4. Moment of inertia of a rectangle (b×d) about its centroidal axis parallel to the base?
A4. I = bd³/12. About the base it is bd³/3.

Q5. Moment of inertia of a solid circular section of diameter d?
A5. I = πd⁴/64 (about a centroidal diameter). Polar MOI J = πd⁴/32.

Q6. State the parallel-axis theorem.
A6. I = I_G + A·h², where I_G is centroidal MOI, A the area, h the distance between the two parallel axes.

Q7. Where is the centroid of a triangle located?
A7. At h/3 from the base (2h/3 from the apex), along the median, at intersection of medians.

Q8. Projectile: range, max height and time of flight for launch speed u at angle θ?
A8. R = u²·sin2θ/g; H = u²·sin²θ/(2g); T = 2u·sinθ/g. Range is maximum at θ=45°.

Q9. Condition for equilibrium of a coplanar force system?
A9. ΣFx = 0, ΣFy = 0 and ΣM = 0 (three independent equations).

Q10. Static determinacy condition for a plane (pin-jointed) truss?
A10. m + r = 2j (m = members, r = reactions, j = joints). If m+r > 2j it is indeterminate.

Q11. What is a couple?
A11. Two equal, opposite, parallel non-collinear forces. Net force = 0, moment = F×d (arm), same about every point (free vector).

Q12. Work–energy principle for a particle?
A12. Work done by the net force = change in kinetic energy: W = ½m(v² − u²).

Worked Example

Problem: Two forces of 3 N and 4 N act at a point at right angles to each other. Find the magnitude and direction of their resultant.

Solution:
For two forces at 90°, the resultant magnitude follows the Pythagoras theorem:
R = √(F₁² + F₂²) = √(3² + 4²) = √(9 + 16) = √25 = 5 N.
Direction (angle θ from the 4 N force):
tan θ = opposite/adjacent = 3/4 = 0.75 → θ = tan⁻¹(0.75) ≈ 36.87°.

Answer: Resultant = 5 N, acting at ≈ 36.87° from the 4 N force.