Equilibrium of a particle

A tightrope walker stands dead-still thirty feet above the ground, the rope sagging just enough on either side to hold her weight. A traffic light hangs at a crossroads, swinging gently but never falling. A block rests on a rough incline, refusing to slide even though gravity is clearly pulling it downhill. In every one of these scenes several forces are pulling and pushing at once, yet nothing accelerates. That stillness is not an absence of forces — it is a balance of forces, and learning to read that balance is one of the most useful skills Class 11 mechanics hands you. This chapter's idea sounds almost too simple to be a whole topic: when the forces acting on a particle add up to zero, the particle is in equilibrium — it stays at rest, or it keeps moving in a straight line at constant speed, exactly as Newton's first law demands. The power is in the method, not the sentence. Once you can draw a clean free-body diagram, resolve every force into horizontal and vertical components, and set each sum to zero, you can solve for unknown tensions, unknown angles, and unknown normal reactions in almost any static picture a question can draw — a lamp on cords, a block on a rope-and-pulley, a knot pulled three ways. For exactly three concurrent forces there is an even faster shortcut, Lami's theorem, built from nothing more than the sine rule. This section builds the equilibrium condition from Newton's first and second laws, introduces the everyday contact and non-contact forces you will resolve (weight, normal reaction, tension, spring force, friction), derives Lami's theorem from the closed triangle of forces, and works through the classic hanging-lamp and three-string problems end to end — including where students reliably lose marks.

Key takeaways

A particle is in equilibrium when the net external force on it is zero: $\vec{F}_{\text{net}} = \sum_i \vec{F}_i = 0$ , which is Newton's first law restated as a working condition.
Equilibrium covers both rest and uniform (constant-velocity) motion — zero net force does not mean zero velocity.
In a plane, the single vector equation splits into two independent scalar equations, $\sum F_x = 0$ and $\sum F_y = 0$ , found by resolving every force along chosen axes on a free-body diagram.
For exactly three concurrent forces, Lami's theorem gives $\dfrac{F_1}{\sin\alpha} = \dfrac{F_2}{\sin\beta} = \dfrac{F_3}{\sin\gamma}$ , where each angle is the one opposite (between the other two) forces.
The forces you actually resolve are the everyday ones: weight $W = mg$ , normal reaction $N$ , tension $T$ , spring force $F = -kx$ , and friction, which is special because static friction is an inequality, $f_s \le \mu_s N$ , not a fixed value.
Drawing the force vectors head-to-tail for a body in equilibrium always produces a closed polygon (a closed triangle for three forces) — a fast visual check that nothing has been left out.
Key idea: a particle is in equilibrium exactly when its concurrent forces sum to zero vectorially, which splits into two component equations or, for three forces, into Lami's theorem drawn from a closed force triangle.

The core insight

Equilibrium of a particle means the vector sum of all concurrent forces acting on it is zero, $\sum \vec{F}_i = 0$ , so it either stays at rest or moves at constant velocity, and for three such forces this reduces to Lami's theorem, the sine rule applied to the closed triangle the force vectors trace out.

Where it fits in the syllabus

Equilibrium of a particle sits inside the CBSE Class 11 Physics chapter Laws of Motion, directly downstream of Newton's first law (the law of inertia) and the idea of force. The syllabus expects you to handle a particle acted on by several coplanar, concurrent forces — forces that all pass through one point — and to find unknown forces or angles when their resultant is zero, using either component resolution or Lami's theorem for the three-force case. Because the body is idealised as a point particle, only translational equilibrium is in scope here; turning effects, torque, and the rotational equilibrium of extended rigid bodies belong to the later System of Particles and Rotational Motion chapter, and pushing this topic into rotation is outside this section's brief. Everyday contact forces — normal reaction, tension, spring force, and friction — are also part of this unit because they are exactly what you resolve when you apply the equilibrium condition to a real object.

The physical picture

Newton's first law says a body's velocity does not change unless a net external force acts on it. Turn that statement around and you get the working definition of equilibrium: whenever a particle's velocity is not changing — whether it is sitting still or gliding along at constant speed in a straight line — the vector sum of every force on it must be zero.

\vec{F}_{\text{net}} = \sum_i \vec{F}_i = 0.

where: $\vec{F}_i$ are the individual forces acting on the particle (weight, tensions, normal reactions, friction, applied pushes or pulls, and so on) and $\vec{F}_{\text{net}}$ is their vector sum. The physical reading is that this is one vector equation, and a vector is zero only if every one of its components is zero — so in a plane it is really two independent conditions in disguise, one for each direction you choose to resolve along.

\sum F_x = 0, \qquad \sum F_y = 0.

where: $x$ and $y$ are any pair of convenient perpendicular axes you draw on the free-body diagram — they need not be horizontal and vertical, though that is the usual choice. Two scalar equations mean you can solve for at most two unknowns from a single equilibrium picture: two tensions, or a tension and an angle, or a normal reaction and a friction force. The method is always the same: isolate the particle, draw every force acting on it (and only the forces acting on it, never the forces it exerts on other things), pick axes, resolve, and set each sum to zero. A special case appears constantly in problems: exactly three concurrent forces holding a particle in equilibrium. Because three vectors that sum to zero must, when drawn head-to-tail, form a closed triangle, there is a neat shortcut called Lami's theorem. If the three forces are $F_1$ , $F_2$ , $F_3$ , and $\alpha$ , $\beta$ , $\gamma$ are the angles between the other two forces (that is, the angle facing, or opposite, each force), then

\frac{F_1}{\sin\alpha} = \frac{F_2}{\sin\beta} = \frac{F_3}{\sin\gamma}.

where: each force is paired with the sine of the angle that lies between the two other forces — not the angle the force itself makes with some axis. The three angles $\alpha$ , $\beta$ , $\gamma$ always add up to $360^\circ$ when you go all the way around the point. Lami's theorem is nothing more than the ordinary sine rule applied to the closed force triangle, and for three-force problems it is usually faster than setting up $\sum F_x = 0$ and $\sum F_y = 0$ separately, because it avoids choosing axes altogether. The forces you resolve are drawn from a short, familiar list. Weight $W = mg$ acts vertically downward, a non-contact force of gravity. Normal reaction $N$ is the contact push a surface exerts perpendicular to itself, preventing the body from sinking into it. Tension $T$ is the pull a stretched string or cord exerts along its own length, directed away from the body at both ends. Spring force follows Hooke's law, $F = -kx$ , proportional to the stretch or compression $x$ , with $k$ the spring constant; the minus sign says the force opposes the displacement. Friction resists relative sliding between two surfaces in contact, and is treated below. Weight is non-contact; the rest arise only where surfaces touch, and all are, at bottom, large-scale electromagnetic effects between atoms — one of the four fundamental forces named qualitatively in this unit. There is a beautifully visual way to check equilibrium without any algebra at all. If you draw the force vectors one after another, head-to-tail, a particle in equilibrium always produces a closed polygon — the tip of the last arrow lands exactly back on the tail of the first, because the vector sum is zero. For three forces this closed shape is a triangle (the triangle behind Lami's theorem); for four or more forces it is a closed polygon. If your arrows do not close up when you draw them this way, either a force is missing from your diagram or the particle is not actually in equilibrium.

Figure

Three concurrent forces meeting at a knot, two tensions pulling upward along cords and a weight pulling straight down, alongside the same three vectors redrawn head to tail forming a closed triangle

Common forces and friction in equilibrium

Weight and normal reaction usually appear together: a block resting on a table has $W$ pulling it down and $N$ pushing it up, and on a horizontal surface with nothing else acting, $N = W$ . Tension appears wherever a string, cord, or rope is taut, and it always pulls the body it acts on, never pushes. Spring force is the only one on this list that changes with the body's own position, since $x$ is the spring's stretch or compression measured from its natural length. Friction deserves special care because, unlike the other forces here, it is not a fixed value you can simply write down — it is self-adjusting. As long as a body stays at rest under an applied push, static friction $f_s$ grows or shrinks automatically to exactly cancel the other forces along the surface, up to a maximum value $f_s^{\max} = \mu_s N$ . The moment the applied force would need more than that maximum to hold the body still, it slips, and kinetic friction takes over at $f_k = \mu_k N$ , with $\mu_k$ slightly smaller than $\mu_s$ . Rolling friction is smaller still, which is why wheels beat sliding. The ratio of friction to normal force at the point of slipping defines the angle of friction, $\tan\lambda = \mu_s$ ; on an incline, the steepest angle at which a body still sits without sliding is the angle of repose, numerically equal to the angle of friction. Friction is both foe and friend — it wastes energy and wears surfaces down, yet it is also what lets you walk and lets tyres grip the road; engineers tame the unwanted part with lubrication rather than trying to eliminate it.

Derivation ALami's theorem from the closed triangle of forces

The goal here is to show that when three concurrent forces $F_1$ , $F_2$ , $F_3$ hold a particle in equilibrium, they must obey $\dfrac{F_1}{\sin\alpha} = \dfrac{F_2}{\sin\beta} = \dfrac{F_3}{\sin\gamma}$ , where each angle is the one opposite that force (between the other two). The argument uses nothing beyond the equilibrium condition itself and the ordinary sine rule for a triangle, so it is well within Class 11 tools, and it explains why the shortcut works rather than asking you to memorise it.

Step 1. Start from equilibrium. Since $\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = 0$ , place $\vec{F}_1$ , then $\vec{F}_2$ head-to-tail, then $\vec{F}_3$ : because they sum to zero, the third vector must bring you exactly back to the starting point. The three vectors, drawn head-to-tail in order, close into a triangle.

Step 2. Relate the triangle's angles to the point-angles. At the point where the forces act, the angle between $\vec{F}_2$ and $\vec{F}_3$ is $\alpha$ , between $\vec{F}_3$ and $\vec{F}_1$ is $\beta$ , between $\vec{F}_1$ and $\vec{F}_2$ is $\gamma$ , with $\alpha+\beta+\gamma=360^\circ$ . Translating (not rotating) a vector into the head-to-tail triangle turns each point-angle into the supplement of the corresponding interior angle of the triangle, whose three interior angles add to $180^\circ$ .

Step 3. Apply the sine rule to the force triangle: for sides $a,b,c$ opposite angles $A,B,C$ , $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$ . Here the sides are the magnitudes $F_1,F_2,F_3$ , and after the supplement relation of Step 2 the angle opposite each side is exactly the point-angle facing that force: $\alpha$ opposite $F_1$ , $\beta$ opposite $F_2$ , $\gamma$ opposite $F_3$ .

Step 4. Read off the result. Substituting into the sine rule gives exactly

\frac{F_1}{\sin\alpha} = \frac{F_2}{\sin\beta} = \frac{F_3}{\sin\gamma},

Lami's theorem. It is worth noticing what it buys you: instead of choosing axes and resolving twice, you read three angles off the diagram and one ratio equation gives you every force magnitude directly, provided you already know one of them (usually the weight).

Worked examples

Board level - the hanging lamp on two symmetric cords

A lamp of mass $5.0\ \text{kg}$ hangs from a hook by two cords, each making $30^\circ$ with the horizontal ceiling, one on each side. Find the tension in each cord. Take $g = 9.8\ \text{m/s}^2$ .

Step 1. Identify the forces at the point where the cords meet the lamp: the weight $W = mg$ acting straight down, and the two tensions $T$ , equal by the left-right symmetry of the picture, each pulling up and outward along its cord at $30^\circ$ above the horizontal. The weight is $W = 5.0 \times 9.8 = 49\ \text{N}$ .

Step 2. Resolve horizontally, $\sum F_x = 0$ . The two tensions' horizontal components point opposite ways and, by symmetry, cancel automatically — this axis gives only a consistency check, not new information.

Step 3. Resolve vertically, $\sum F_y = 0$ . Each cord contributes a vertical component $T\sin 30^\circ$ , and together they support the full weight: $2T\sin 30^\circ = W$ .

Step 4. Solve for $T$ : $2T(0.5) = 49$ , so $T = 49\ \text{N}$ . Each cord carries a tension of $49\ \text{N}$ . Note the sensitivity to angle: a nearly horizontal cord would make $\sin\theta$ tiny and $T$ blow up — why cables always keep some sag.

Competitive level - three-string equilibrium by Lami's theorem

A street sign of weight $W = 40\ \text{N}$ hangs from a knot held by two strings. String 1 pulls up and to the left, making an angle of $120^\circ$ with the downward weight vector; string 2 pulls up and to the right, making an angle of $150^\circ$ with the weight vector. Use Lami's theorem to find the tensions $T_1$ and $T_2$ .

Step 1. List the three concurrent forces at the knot: the weight $W$ acting straight down, and the two tensions $T_1$ and $T_2$ . Lami's theorem needs, for each force, the angle between the other two. The angle between $T_1$ and $T_2$ (the one opposite $W$ ) is what remains once you subtract the two given angles from a full turn: $360^\circ - 120^\circ - 150^\circ = 90^\circ$ .

Step 2. Write Lami's theorem with $W$ opposite the $90^\circ$ angle, $T_1$ opposite the $150^\circ$ angle (the angle between $W$ and $T_2$ ), and $T_2$ opposite the $120^\circ$ angle (the angle between $W$ and $T_1$ ):

\frac{W}{\sin 90^\circ} = \frac{T_1}{\sin 150^\circ} = \frac{T_2}{\sin 120^\circ}.

Step 3. Solve for each tension using $W = 40\ \text{N}$ , $\sin 90^\circ = 1$ , $\sin 150^\circ = 0.5$ , $\sin 120^\circ \approx 0.866$ :

T_1 = \frac{W\sin 150^\circ}{\sin 90^\circ} = \frac{40\times 0.5}{1} = 20\ \text{N}, \qquad T_2 = \frac{W\sin 120^\circ}{\sin 90^\circ} = \frac{40\times 0.866}{1} \approx 34.6\ \text{N}.

Step 4. Sanity-check the answer. Here the angle between $T_1$ and $T_2$ is exactly $90^\circ$ , so the two tensions are perpendicular and Pythagoras must also give back the weight: $\sqrt{T_1^2+T_2^2} = \sqrt{20^2+34.6^2} = \sqrt{400+1197} \approx \sqrt{1597} \approx 40.0\ \text{N} = W$ . It checks out. $T_1 = 20\ \text{N}$ and $T_2 \approx 34.6\ \text{N}$ . Timing note: solvable in under two minutes once you trust Lami's theorem — the only real work is the subtraction $360^\circ-120^\circ-150^\circ=90^\circ$ for the third angle. When a problem hands you three concurrent forces with their mutual angles, reach for Lami's theorem first; reserve component resolution for awkward angles or when a friction inequality is also in play.

Common traps

Missing a force on the free-body diagram. A forgotten normal force, weight, or friction term throws off every equation that follows; list every force touching the particle first.
Treating friction as a fixed number instead of an inequality. Static friction is not automatically $\mu_s N$ ; it only reaches that maximum at the point of slipping. Plugging in $\mu_s N$ when the body is not on the verge of sliding overstates the friction force.
Using the wrong angle in Lami's theorem. Each force pairs with the sine of the angle between the other two forces, never the angle it makes with a fixed axis.
Confusing action-reaction pairs with equilibrium forces. Third-law pairs act on two different bodies and never cancel in one free-body diagram; equilibrium sums different forces on the same body.
Forgetting equilibrium includes uniform motion. Zero net force does not mean at rest — constant-velocity motion is equally equilibrium.
Inventing a centrifugal force in an inertial frame. In the ground frame there is no outward pseudo-force to add; the net force on a body in circular motion is the real, inward centripetal force, not zero.

Common trapEquilibrium is not "nothing is happening" — it is forces cancelling, and a body moving at constant velocity is just as much in equilibrium as one sitting still

The single costliest misreading of this topic is assuming equilibrium always means "at rest." It does not. Newton's first law, and the equilibrium condition $\sum \vec{F} = 0$ built on it, treats rest and uniform straight-line motion identically — both are states of zero acceleration, and zero acceleration is exactly what zero net force produces. A ball rolling at a steady speed across frictionless ice, a lift moving up at constant velocity, an aircraft cruising level at fixed speed: all are in equilibrium in exactly the same sense as a book lying still on a table. The moment any of these speeds up, slows down, or changes direction, the net force is no longer zero and the situation is no longer equilibrium.

You should be able to…

State the equilibrium condition $\vec{F}_{\text{net}} = \sum \vec{F}_i = 0$ and explain why it splits into $\sum F_x = 0$ and $\sum F_y = 0$ in a plane.
Draw a correct free-body diagram for a particle, showing weight, normal reaction, tension, spring force, and friction only where each genuinely acts.
Apply Lami's theorem to a particle held by exactly three concurrent forces, correctly matching each force to the angle opposite it.
Distinguish static friction ( $f_s \le \mu_s N$ , self-adjusting) from kinetic friction ( $f_k = \mu_k N$ , fixed once sliding), and relate both to the angle of friction and angle of repose.
Recognise that equilibrium includes constant-velocity motion, not just rest, and that Newton's third-law force pairs never appear together in one body's equilibrium equation.
Use the closed-polygon (triangle, for three forces) picture as a fast visual check that a set of forces genuinely sums to zero.

Ask the tutor: stuck on setting up a free-body diagram or picking the right Lami angle — open the AI tutor on this page and walk through your own equilibrium problem step by step.

If two forces are equal and opposite, do they always cancel out to give equilibrium?

Only if they act on the same body and are the only forces present, or combine with other forces to sum to zero. This is the trap with Newton's third-law pairs: when you push on a wall, the wall pushes back on you with an equal and opposite force, but those two forces act on two different bodies — you and the wall — so they never appear together in one object's equilibrium equation. Equilibrium only cares about the forces acting on one chosen particle; a third-law partner belongs to a different free-body diagram entirely.

Why does static friction only "go up to" mu_s times N instead of always being mu_s times N?

Because static friction's job is to prevent sliding, not to have a fixed size. Imagine gently pushing a heavy crate: for a light push, friction only needs to supply a small force to keep the crate still, so $f_s$ is small. Push harder, and $f_s$ grows to match, right up until the push exceeds $\mu_s N$ — the maximum grip the surfaces can supply — at which point the crate starts to slide and kinetic friction, $f_k = \mu_k N$ , takes over. So $f_s \le \mu_s N$ is an inequality, and its actual value in any equilibrium problem is whatever is needed to balance the other forces, found from $\sum F = 0$ , not read off a formula.

Why do the two cords in the hanging-lamp problem need such large tension for a small weight?

It comes down to the angle. Each cord only contributes its vertical component, $T\sin\theta$ , to support the weight, and the rest of the tension is "wasted" pulling sideways (which cancels against the other cord). When $\theta$ is small — the cords are nearly horizontal — $\sin\theta$ is small, so $T$ has to be huge just to supply the same small vertical component. This is exactly why a clothesline sags: pulling it perfectly taut and horizontal would in principle require infinite tension, since $\sin 0^\circ = 0$ . Real cables and ropes always keep some sag or a decent angle from the horizontal to keep the tension manageable.

Is a car moving at constant speed on a straight road really in equilibrium, even though the engine is running and pushing it forward?

Yes, if the speed and direction really are constant. The engine does supply a forward force through the wheels, but at constant velocity that forward push is exactly cancelled by air resistance and rolling friction acting backward, plus the vertical forces (weight and the road's normal reaction) also balance. The net force on the car is zero even though several individual forces, including a running engine, are very much "doing something." Equilibrium is about the sum of forces, not about whether any force happens to be zero individually.