Equilibrium of a Particle — Concurrent Forces, Lami's Theorem (NCERT 4.8)
1. What Is Equilibrium?
A particle is in equilibrium when the vector sum of all forces acting on it is zero — ΣF = 0, meaning ΣFx = 0 and ΣFy = 0 independently. Equilibrium means zero acceleration, not zero force: the body may sit at rest (static equilibrium) or coast at constant velocity (dynamic equilibrium — the case the first law protects). This page is the statics workshop of the chapter: hanging lamps, signboards, blocks held on inclines — any scene where nothing accelerates and the unknowns are the forces themselves. NTA's version of these questions is extremely consistent: two unknown tensions or a tension plus a normal, resolved along two axes, always solvable in two lines once the free-body diagram is honest.
The particle idealisation matters: we treat the body as a point so every force's line of action passes through it — concurrent forces. Then no rotational bookkeeping is needed (that belongs to rigid bodies in System of Particles), and the single vector condition ΣF = 0 is the complete story. Most exam systems are two-string lamps or block-on-incline setups where the concurrency is automatic.
2. Complete Theory: Conditions, Lami, and the FBD Habit
The two-force and three-force rules. Two concurrent forces can balance only if they are equal, opposite and collinear — otherwise a resultant survives. Three concurrent forces balance when, taken in order (head-to-tail), they close a triangle: the triangle rule of equilibrium. That closed triangle is exactly Lami's theorem in geometry clothing: applying the sine rule to the force triangle gives each force proportional to the sine of the angle between the other two. Four or more forces: drop to components — ΣFx = 0, ΣFy = 0 — and solve the simultaneous equations.
Lami's theorem, stated for exams. If three concurrent (coplanar) forces keep a body in equilibrium, then F₁/sin α = F₂/sin β = F₃/sin γ, where α is the angle between F₂ and F₃ (opposite F₁), and similarly for β, γ. The angle assignment is the theorem's only trap: α is not the angle adjacent to F₁ in your sketch — it is the angle facing it. With a 37°/53° lamp (Example 1) the angles between the cords and the vertical force are 143° and 127°, whose sines equal sin 37° and sin 53° — the supplementary-angle identity sin(180° − θ) = sin θ is why students who grab the wrong angle still stumble to the right answer half the time, and lose the derivation mark the other half.
The FBD habit. Every statics answer begins with a free-body diagram: isolate the body, draw only the forces on it (weight down from its centre; normal perpendicular to each contact surface; tension along each string, away from the body; friction along the surface, opposing impending sliding). Then choose axes — for inclines, along and perpendicular to the surface — and write one zero-sum equation per axis. The problem-solving page turns this habit into a five-step routine; here we apply it twice and let the numbers speak.
Block held on a frictionless incline — the template system. A block of mass m on a smooth incline of angle θ, held by a string parallel to the slope: resolving along the incline, T = mg sin θ; perpendicular to it, N = mg cos θ. Note the beautiful consistency check — N < mg on an incline because the surface only needs to balance the perpendicular component of weight. The same two numbers (mg sin θ, mg cos θ) reappear in the friction page and in the banking derivation of the circular dynamics page; memorise the resolution, not the answers.
3. Visualising the Lamp on Two Cords
Figure 4.5 — Three concurrent forces, one closed triangle
The two cord tensions pull along the cords toward the ceiling; the weight pulls straight down. All three lines meet at the lamp — concurrency in action.
Exam read-out: horizontal bookkeeping: T₁ cos 37° = T₂ cos 53°. Vertical bookkeeping: T₁ sin 37° + T₂ sin 53° = mg. Two equations, two unknowns — and with the 3-4-5 pair from Chapter 3, the answers come out as clean integers: 120 N and 160 N for a 20 kg lamp.
4. Solved Examples
Example 1 — The 37°/53° lamp (g = 10 m s⁻²)
A 20 kg lamp hangs from two cords making 37° and 53° with the horizontal ceiling. Find both tensions.
Solution (step by step): FBD of the lamp: T₁ along the 37° cord, T₂ along the 53° cord, mg = 200 N down. Horizontal: T₁ cos 37° = T₂ cos 53° → 0.8T₁ = 0.6T₂ → T₂ = (4/3)T₁. Vertical: T₁ sin 37° + T₂ sin 53° = 200 → 0.6T₁ + 0.8 × (4/3)T₁ = (5/3)T₁ = 200 → T₁ = 120 N, T₂ = 160 N. Check: the steeper cord (53°) carries the bigger share — geometry decides the load split, not symmetry. Lami cross-check: the three force directions are 143°, 53° and −90°, giving pairwise angles of 90° (between the tensions), 143° (between T₂ and mg) and 127° (between T₁ and mg); then T₁/sin 143° = 120/sin 37° = 200, T₂/sin 127° = 160/sin 53° = 200, and mg/sin 90° = 200 — one common ratio, ledger closed.
Example 2 — Block held on a frictionless 37° incline (g = 10 m s⁻²)
A 10 kg block rests on a smooth incline of 37°, held by a light string parallel to the slope. Find the tension and the normal reaction.
Solution: Axes along/perpendicular to the incline (the choice that makes both equations one-force-per-unknown). Along: T = mg sin 37° = 10 × 10 × 0.6 = 60 N. Perpendicular: N = mg cos 37° = 10 × 10 × 0.8 = 80 N. Note N < mg = 100 N — the incline carries only the perpendicular component. When the string is cut, the 60 N along-incline resultant accelerates the block at g sin 37° = 6 m s⁻², and the incline friction (if any) would enter through the 80 N — exactly where the next-but-one page picks up the story.
5. Practice Questions
Q1. A 5 kg picture frame hangs at rest from two wires at 45° to the horizontal on either side. Find the tension in each wire (g = 10 m s⁻²).
ANSWER: Symmetry halves the job: vertical bookkeeping gives 2T sin 45° = 50 → T = 25/sin 45° = 25√2 ≈ 35.4 N per wire. Horizontal components cancel by mirror symmetry. Note T > mg/2 — the flatter the wires, the crueler the tension: at 30° it would be 50 N per wire.
Q2 (MCQ). A block is at rest on a rough incline of angle θ (below the repose angle). The friction force on it equals: (a) μs mg cos θ (b) mg sin θ (c) μk mg sin θ (d) zero
ANSWER: (b) — equilibrium along the incline demands friction = mg sin θ, whatever μ happens to be; static friction is self-adjusting and only caps at μsmg cos θ. Option (a) is the "impending motion" value, correct only at the repose angle.
Q3. A 2 kg ball hangs from a string; it is pushed sideways by a horizontal 15 N force and holds equilibrium off-vertical. Find the string's tension and its angle from the vertical (g = 10 m s⁻²).
ANSWER: Horizontal: T sin θ = 15; vertical: T cos θ = 20. Divide: tan θ = 15/20 = 0.75 → θ = 37°; T = 20/cos 37° = 25 N. The closed force triangle (20, 15, 25) is the 3-4-5 triple — NTA builds these to stay integral.
6. Key Formulas & Takeaways
| Relation | Condition / remark |
|---|---|
| ΣF = 0 ⇔ ΣFx = 0 and ΣFy = 0 | Complete equilibrium condition for a particle (a = 0) |
| F₁/sin α = F₂/sin β = F₃/sin γ | Lami: exactly 3 concurrent coplanar forces; α opposite F₁ |
| Two forces balance: equal, opposite, collinear | Otherwise a resultant survives |
| Three forces balance: closed force triangle | Head-to-tail closure ⇔ Lami ⇔ components agree |
| Incline held by string: T = mg sin θ, N = mg cos θ | Smooth incline; axes along/perpendicular to slope |
| Hanging lamp: Tx balance + Ty sum = mg | Steeper cord carries more; flatter wires, higher tension |
Equilibrium is a = 0; give the resultant a nonzero value and the same resolution machinery produces accelerations — the solving-problems page swaps the zero on the right-hand side for ma.
7. Frequently Asked Questions
What is meant by equilibrium of a particle?
A particle is in equilibrium when the vector sum of all forces acting on it is zero: ΣF = 0, equivalently ΣFx = 0 and ΣFy = 0 separately. Equilibrium means zero acceleration — the particle is either at rest (static equilibrium) or moving with constant velocity (dynamic equilibrium). It does not mean no forces act; it means the forces balance. The condition follows directly from Newton's first law and is the a = 0 special case of the second law.
State Lami's theorem and its conditions.
If three concurrent forces acting on a body keep it in equilibrium, then each force is proportional to the sine of the angle between the other two: F1/sin α = F2/sin β = F3/sin γ. The conditions are: exactly three forces, their lines of action passing through one point (concurrent), the forces coplanar, and the body in equilibrium. Crucially, each angle in the formula is the angle opposite that force — the angle between the other two forces — not the angle adjacent to it. Misreading this angle assignment is the standard error.
What are concurrent forces?
Concurrent forces are forces whose lines of action pass through a common point. For such a system the rotational effect vanishes automatically and only the translational condition ΣF = 0 matters — which is why the particle idealisation works. Two concurrent forces can balance only if they are equal and opposite along the same line; three concurrent forces balance when they can form a closed triangle taken in order (the triangle rule), which is also the geometric content of Lami's theorem.
What is a free-body diagram and why is it drawn?
A free-body diagram isolates one body and shows every force acting on it — and only on it — as arrows from a point or a simple outline. Drawing it is the first step of every mechanics problem because it converts a messy physical scene into a clean force inventory: weights down, normals perpendicular to contact surfaces, tensions along strings, friction along surfaces. Once the diagram is drawn, equilibrium questions reduce to resolving along chosen axes and setting each axis sum to zero. Forces the body exerts on other bodies must never appear on its diagram.
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