Free-Body Diagrams & Equilibrium of Forces

Engineering Mechanics contributes ~5–7 marks to GATE ME, and every problem in it starts with a correct free-body diagram (FBD). Master equilibrium here and the whole subject becomes bookkeeping.

Core principles

A rigid body is in static equilibrium when the net force and net moment are zero:

• ΣFx = 0, ΣFy = 0 (2D coplanar), plus ΣM = 0 about any point.
• In 3D: ΣFx = ΣFy = ΣFz = 0 and ΣMx = ΣMy = ΣMz = 0 (six equations).

Types of forces on an FBD: applied loads, self-weight (at centroid), normal reactions (⊥ to surface), friction (along surface), tension (along a cable, pulling away), and support reactions.

Support reactions: roller → 1 reaction (⊥ to surface); pin/hinge → 2 reactions (Rx, Ry); fixed → 3 reactions (Rx, Ry, M). A two-force member carries force only along the line joining its two pins.

Moment M = F·d (perpendicular distance) = r × F. A couple is two equal-opposite forces; its moment is the same about every point.

Exam Tricks & Tips

• 🎯 Take moments about a point where an unknown acts — that force drops out, giving one equation in one unknown.
• 🎯 Identify two-force members first: their force acts purely along the member's axis, eliminating direction unknowns instantly.
• 🎯 Count reactions vs equilibrium equations to test determinacy: 2D determinate if reactions = 3 (for a single body).
• 🎯 A frictionless pulley just changes cable direction — tension is the same on both sides, a common quick simplification.
• 🎯 Assume a sense for each reaction; a negative answer simply means the true sense is opposite — no need to redo the FBD.
• ❌ Common mistake: omitting a reaction (e.g. the horizontal pin reaction) or drawing an internal force on the FBD of the whole body — only external forces belong on a single-body FBD.

Expected exam pattern

A 1-mark reaction or moment calculation on a simple beam/lever, and a 2-mark equilibrium problem needing a well-chosen moment centre. Determinacy (statically determinate vs indeterminate) is a favourite conceptual MCQ.

Quick recap

Equilibrium: ΣF = 0 and ΣM = 0 (three equations in 2D, six in 3D). Draw every external force; roller=1, pin=2, fixed=3 reactions. Take moments about an unknown's line of action to isolate it. Two-force members carry axial force only.

Cover the answer, recall, then check. 11 cards on statics equilibrium for GATE ME.

Q1. State the equilibrium conditions for a 2D rigid body.
A1. ΣFx = 0, ΣFy = 0, and ΣM = 0 about any point — three independent equations.

Q2. How many equilibrium equations exist in 3D?
A2. Six: ΣFx = ΣFy = ΣFz = 0 and ΣMx = ΣMy = ΣMz = 0.

Q3. How many reactions do roller, pin, and fixed supports provide (2D)?
A3. Roller → 1 (perpendicular to surface); pin/hinge → 2 (Rx, Ry); fixed → 3 (Rx, Ry, moment M).

Q4. What is a two-force member?
A4. A member loaded only at two pins with no other load; the force in it acts along the line joining the two pins (pure tension or compression).

Q5. Define the moment of a force.
A5. M = F × d, where d is the perpendicular distance from the point to the force's line of action (M = r × F vectorially).

Q6. What is a couple, and its key property?
A6. Two equal, opposite, parallel forces. Its moment is constant (= F·separation) about every point — a pure turning effect.

Q7. Strategy: which point should you take moments about?
A7. A point where one or more unknown forces act, so those unknowns vanish, isolating the one you want.

Q8. What does a negative reaction value mean?
A8. The assumed direction was wrong; the true reaction acts in the opposite sense. Magnitude is still correct.

Q9. How do you test static determinacy of a single 2D body?
A9. Determinate if number of unknown reactions = 3 (matches the 3 equilibrium equations); more → indeterminate.

Q10. What forces belong on the FBD of a whole structure?
A10. Only external forces: applied loads, self-weight, and support reactions. Internal member forces cancel and are excluded.

Q11. For a frictionless pulley, how do the two cable tensions compare?
A11. They are equal in magnitude; the pulley only redirects the cable, not its tension.

The free-body diagram (FBD) is the single most important habit in all of mechanical engineering — get it right and statics, dynamics, SOM and machine design all fall into place; get it wrong and every later number is wrong. GATE rewards clean FBDs directly and indirectly across half the syllabus.

Core concept: isolate a body, draw every external force and moment acting on it (and nothing internal), then impose that a body in equilibrium has zero net force and zero net moment.

Deep explanation

Beginner — the equilibrium equations

A rigid body in 2D equilibrium satisfies three scalar equations:
ΣFx = 0, ΣFy = 0, ΣM = 0.
In 3D there are six (three force, three moment). "Equilibrium" means at rest or moving at constant velocity — acceleration is zero, so inertia forces vanish.

Intermediate — reactions and supports

Each support removes degrees of freedom and supplies reactions:

• Roller / smooth surface: one reaction, normal to the surface.
• Pin / hinge: two reactions (Rx, Ry), no moment.
• Fixed support: two forces + one moment.
  Count unknown reactions r vs. equilibrium equations. If r = 3 (2D) and the body is properly constrained, it is statically determinate; if r > 3, it is indeterminate and needs compatibility (SOM), not statics alone.

Advanced — moment tricks and two/three-force members

• Moment = force × perpendicular distance. Taking moments about the point where the most unknowns act eliminates them from that equation — the fastest route to a single unknown.
• A two-force member (loaded only at two points) carries force along the line joining them — a huge simplification for frames.
• A three-force member in equilibrium has its three forces concurrent (meeting at one point); this converts a hard problem into simple geometry.

Worked example

A uniform 100 N horizontal beam, 4 m long, is pinned at A (left) and rests on a roller at B (right). A 300 N load hangs 1 m from A. Find the reactions.
Take moments about A (eliminates Ax, Ay): R_B(4) − 300(1) − 100(2) = 0 → 4R_B = 300 + 200 = 500 → R_B = 125 N (up).
ΣFy = 0: R_A + R_B − 300 − 100 = 0 → R_A = 400 − 125 = 275 N (up). (Self-weight acts at the 2 m centroid.)

GATE relevance

Directly tested as reaction/equilibrium problems, and it is the prerequisite step in trusses, friction, beam SFD/BMD, and machine-element loading. A wrong FBD is the most common root cause of a wrong final answer across the whole ME paper.

Exam tricks & shortcuts

• Take moments about the point with the most unknown forces — they drop out, leaving one equation, one unknown.
• Exploit two-force and three-force member rules to avoid simultaneous equations.
• Mnemonic: "Isolate, Replace supports with reactions, Sum to zero."

Key formulas

• Equilibrium: ΣFₓ = 0, ΣF_y = 0, ΣM = 0.
• Three concurrent forces (Lami): F₁/sinα = F₂/sinβ = F₃/sinγ.
• Resultant: R = √(ΣFₓ² + ΣF_y²); direction tanθ = ΣF_y/ΣFₓ.
• Moment: M = F × perpendicular distance.
• Couple: two equal opposite forces; moment = F × arm.

Draw the free-body diagram, resolve forces along two axes, and set force and moment sums to zero.

Worked Example

Problem: A weight W = 200 N is suspended from a ceiling by two strings. One string makes 30° with the horizontal and the other makes 60° with the horizontal. Find the tension in each string.

Solution:
Draw a free-body diagram of the knot where the two strings meet the weight. Three forces act: T₁ (30° string), T₂ (60° string), and the weight 200 N straight down. For equilibrium ΣF_x = 0 and ΣF_y = 0.

Horizontal equilibrium (the two strings pull in opposite horizontal directions):
T₁·cos30° = T₂·cos60°
T₁(0.866) = T₂(0.5) ⇒ T₁ = 0.5774·T₂.

Vertical equilibrium (both vertical components support the weight):
T₁·sin30° + T₂·sin60° = 200
T₁(0.5) + T₂(0.866) = 200.

Substitute T₁ = 0.5774·T₂:
0.5774·T₂(0.5) + 0.866·T₂ = 200
0.2887·T₂ + 0.866·T₂ = 1.1547·T₂ = 200
T₂ = 173.2 N, and T₁ = 0.5774 × 173.2 = 100 N.

Check: vertical → 100(0.5) + 173.2(0.866) = 50 + 150 = 200 ✓.

Answer: T₁ = 100 N (30° string) and T₂ = 173.2 N (60° string).