1. Equilibrium defined
A particle is in equilibrium when the net external force on it is zero. By Newton's first law, this produces two equivalent states: the particle is at rest, or it moves with constant velocity in a straight line. The vector condition is
\[ \sum \vec{F} = 0 \]
The sum runs over every external force on the particle — gravity, tension, normal, friction, applied push, spring force.
NCERT footnote: equilibrium of a particle requires only translational equilibrium, \(\sum \vec{F} = 0\). An extended body also needs rotational equilibrium, \(\sum \vec{\tau} = 0\) (deferred to the rotational motion chapter). This article is about the particle only.
The equilibrium condition
- \(\sum \vec{F} = 0\) on the particle — vector sum of all external forces.
- Equivalent in 2-D: \(\sum F_x = 0\) and \(\sum F_y = 0\) (two scalar equations).
- Equivalent in 3-D: \(\sum F_x = \sum F_y = \sum F_z = 0\) (three scalar equations).
- Implies acceleration \(a = 0\); velocity is constant (possibly zero).
2. Two-force equilibrium
With two forces \(\vec{F}_1, \vec{F}_2\), equilibrium requires \(\vec{F}_1 = -\vec{F}_2\) — equal magnitude, opposite direction, same line of action. A book on a table is the classic case: weight \(mg\) down balanced by normal \(N = mg\) up. These are not a third-law pair (they act on the same body); they are equal because equilibrium forces them to be.
Equilibrium does not mean rest
A body moving with uniform velocity is also in equilibrium — Newton's first law lumps "at rest" and "uniform straight-line motion" into the same category because both have zero acceleration. NEET distractors often add "the velocity must be zero" or "the body must be stationary" to an otherwise correct equilibrium definition.
Rule: a parachutist at terminal velocity, a car cruising at constant 60 km/h, and a block sliding at a steady speed on a rough floor are all in equilibrium.
3. Triangle of forces — three concurrent forces
For three concurrent forces — three forces all meeting at the same point — equilibrium reads
\[ \vec{F}_1 + \vec{F}_2 + \vec{F}_3 = 0 \]
The parallelogram law gives a strong geometric reading: if three concurrent forces are in equilibrium, they can be represented in magnitude and direction by the three sides of a closed triangle taken head-to-tail. This is the triangle of forces.
Lay \(\vec{F}_1\) as one side; from its tip draw \(\vec{F}_2\); the closing side back to the tail of \(\vec{F}_1\) must equal \(\vec{F}_3\). The triangle closes only when equilibrium holds. Sanity check: three magnitudes 1 N, 2 N, 10 N cannot form a triangle, so they can never be in equilibrium (triangle inequality).
4. Polygon law — n forces in equilibrium
The triangle rule generalises: if n forces acting at a point keep a particle in equilibrium, they can be represented by the n sides of a closed polygon taken head-to-tail (the polygon law). For computation with five or six forces, the component method is fastest; the polygon is best as a conceptual check.
5. Component-form solving
The vector equation \(\sum \vec{F} = 0\) gives three scalar equations:
\[ \sum F_x = 0, \quad \sum F_y = 0, \quad \sum F_z = 0 \]
NEET almost always tests 2-D, so two equations suffice. Pick axes that align with as many forces as possible — along the incline for wedge problems, horizontal/vertical for hanging masses.
| Equation | What it expresses | NEET use |
|---|---|---|
\(\sum F_x = 0\) | Horizontal forces cancel | Pulls, tensions, friction |
\(\sum F_y = 0\) | Vertical forces cancel | Weight, normal, vertical tensions |
\(\sum F_z = 0\) | 3-D balance (rare) | Tripod, table-leg problems |
Two unknowns need exactly two equations. Three unknown tensions in 2-D need extra information (a string angle, symmetry) to close the system.
6. Lami's theorem
Lami's theorem is the sine-rule version of the triangle of forces — a NEET favourite because it bypasses components for three concurrent coplanar forces.
Statement. If three concurrent coplanar forces \(\vec{F}_1, \vec{F}_2, \vec{F}_3\) keep a particle in equilibrium, then
\[ \frac{F_1}{\sin\alpha} = \frac{F_2}{\sin\beta} = \frac{F_3}{\sin\gamma} \]
where \(\alpha\) is the angle opposite \(\vec{F}_1\) (between \(\vec{F}_2\) and \(\vec{F}_3\)), and similarly for \(\beta, \gamma\). The three angles sum to \(360°\).
Why. The closed force-triangle has interior angles \(180°-\alpha, 180°-\beta, 180°-\gamma\); the sine rule on that triangle gives Lami's relation since \(\sin(180°-\theta)=\sin\theta\).
Want 20 Lami's-theorem MCQs at NEET difficulty with timed feedback? Start the Laws of Motion chapter test.
Lami's theorem only for three concurrent coplanar forces
If four forces meet at a point, Lami's theorem does not apply — fall back to components. NEET sometimes plants a 4-force figure with answer options shaped like Lami's ratios; the figure has one extra force you must spot. Always confirm: three forces, all concurrent, all in one plane.
Rule: count the forces in the FBD. If \(n=3\) and they all meet at one point in one plane, Lami's is fastest. Otherwise components.
7. Worked: NCERT Example 4.6 — hanging mass with a horizontal pull
A 6 kg mass is suspended by a 2 m rope from the ceiling. A horizontal 50 N force is applied at the midpoint P of the rope. Find the angle the upper rope makes with the vertical. (\(g = 10\,\text{m/s}^2\); massless rope.)
Step 1. Equilibrium of the 6 kg mass: the lower rope is vertical, so \(T_2 = mg = 60\,\text{N}\).
Step 2. At P three forces meet — tension \(T_1\) along the upper rope at angle \(\theta\) from vertical, \(T_2 = 60\) N straight down, and the 50 N horizontal pull. Resolve:
\[ \text{Horizontal:}\ T_1 \sin\theta = 50; \quad \text{Vertical:}\ T_1 \cos\theta = 60 \]
Step 3. Divide: \(\tan\theta = 50/60 = 5/6\), so \(\theta = \tan^{-1}(5/6) \approx 39.8°\). Square-and-add gives \(T_1 = \sqrt{50^2 + 60^2} \approx 78.1\,\text{N}\).
Note. The answer does not depend on the rope length or on where the horizontal force is applied (rope is massless). The 2 m is a red herring.
8. NEET-favourite equilibrium scenarios
Hanging from two strings
A mass \(m\) hangs from two ceiling strings at angles \(\theta_1, \theta_2\). At the knot, three forces meet — the two tensions and \(mg\). Solve by Lami's or by components. Symmetric case (\(\theta_1=\theta_2=\theta\) from horizontal): \(T = mg/(2\sin\theta)\).
Block on an inclined plane
On a rough incline, equilibrium holds while \(mg\sin\theta \le \mu_s mg\cos\theta\), i.e. \(\tan\theta \le \mu_s\). On a smooth incline the block cannot stay at rest for any \(\theta > 0\) — it slides. An external agent (string, push, wedge acceleration as in NEET 2018) is required to hold it.
Bird on a sagging wire
For a weight \(W\) hung from the middle of a light wire with each half at angle \(\theta\) from horizontal, vertical balance gives \(2T\sin\theta = W\), so \(T = W/(2\sin\theta)\). As \(\theta \to 0\), \(T \to \infty\). No finite tension makes the wire perfectly horizontal — the geometry forbids it.
Action-reaction is NOT an equilibrium pair
When you push a wall with 50 N, the wall pushes you back with 50 N (third law). But these two forces act on different bodies. Equilibrium of you is established by balancing the forces on you (the wall's push, friction at your feet, gravity, normal from the floor), not by the third-law pair. NEET 2019 caught many candidates here.
Rule: when testing equilibrium of body A, only include forces acting on A. Forces A exerts on other bodies are irrelevant.
Tension explodes as a string becomes horizontal
For a weight hung from the middle of a string between two supports, tension is \(T = W/(2\sin\theta)\) where \(\theta\) is the angle from horizontal. As \(\theta \to 0\), \(T \to \infty\). NEET likes to ask "what is the tension when the string is horizontal" — and the correct answer is "no finite tension can keep it horizontal".
Rule: a fully horizontal string carrying a transverse weight is impossible for any massive load — the geometry forces some sag.
9. Worked examples
A 10 kg mass hangs from two ceiling strings making \(30°\) and \(60°\) with the ceiling. Find the tensions. (\(g=10\,\text{m/s}^2\).)
\(W = 100\,\text{N}\) downward. The angle between the two strings at the knot is \(180°-30°-60°=90°\). Lami's angles (opposite each force) are: opposite \(W\) is \(90°\); opposite \(T_1\) (along the \(30°\)-string) is \(120°\); opposite \(T_2\) (along the \(60°\)-string) is \(150°\).
\[ \frac{T_1}{\sin 120°} = \frac{T_2}{\sin 150°} = \frac{100}{\sin 90°} \]
\(T_1 = 100\sin 120° \approx 86.6\,\text{N};\ T_2 = 100\sin 150° = 50\,\text{N}\). The string closer to vertical carries more — as expected.
Three forces 6 N, 8 N, 10 N keep a particle in equilibrium. Show the 6 N and 8 N forces are perpendicular.
By the triangle of forces, the three magnitudes form a closed triangle. Check: \(6^2 + 8^2 = 100 = 10^2\) — a right-angled triangle by the converse of Pythagoras, hypotenuse 10 N. In the original FBD, the angle between the 6 N and 8 N forces (drawn from the particle outwards) is \(90°\). Component check: place 6 N along \(+x\) and 8 N along \(+y\); their resultant has magnitude 10 N, exactly opposed by the third force.
Two bodies, 10 kg and 20 kg, on a smooth surface are linked by a light spring (\(k = 10^4\,\text{N/m}\)). A horizontal \(F = 600\,\text{N}\) acts on the 10 kg. Find the steady-state compression.
Common acceleration: \(a = 600/30 = 20\,\text{m/s}^2\). For the 20 kg body, the only horizontal force is the spring push: \(F_s = 20 \times 20 = 400\,\text{N}\). Hooke's law: \(x = F_s/k = 400/10^4 = 0.04\,\text{m} = 4\,\text{cm}\). The massless spring's two ends exert equal force — its "dynamic equilibrium".