The two conditions for equilibrium
NCERT defines mechanical equilibrium precisely: a rigid body is in equilibrium if both its linear momentum and angular momentum are not changing with time — equivalently, the body has neither linear acceleration nor angular acceleration. For a deformable system this would be subtle, but for a rigid body it reduces to two clean vector statements.
Condition 1 — Translational equilibrium. The vector sum of all external forces on the body is zero:
$$\sum_{i} \vec{F}_i = 0$$
If the total force is zero, the total linear momentum does not change with time. This is the condition a single particle must also satisfy.
Condition 2 — Rotational equilibrium. The vector sum of all external torques on the body is zero:
$$\sum_{i} \vec{\tau}_i = 0$$
If the total torque is zero, the total angular momentum does not change with time. A particle has no rotational equilibrium condition because rotation is not defined for a point — only a rigid body needs both.
Each vector equation stands for three scalar equations. For the common case of coplanar forces, NCERT notes that only three independent conditions survive: the sums of force components along two perpendicular axes in the plane must each vanish (translational), and the sum of torque components about any axis perpendicular to that plane must vanish (rotational). Almost every NEET problem in this topic is coplanar, so you work with exactly these three scalar equations.
One practical consequence is worth memorising. NCERT proves that if the translational equilibrium condition holds, the rotational equilibrium condition is independent of the origin about which torques are taken. So once you know the net force is zero, you may take moments about any convenient point — typically a point where an unwanted unknown acts, so that unknown drops out.
Partial equilibrium
A body may satisfy one condition and not the other; NCERT calls this partial equilibrium. The two reference cases come from a light rod $AB$ of negligible mass with $C$ the midpoint, $CA = CB = a$.
- Rotational but not translational. Two equal, parallel forces in the same direction, applied perpendicular to the rod at $A$ and $B$. The two moments $aF$ are equal in magnitude but opposite in sense, so the net moment is zero — yet $\sum\vec F = 2F \ne 0$. The rod is in rotational equilibrium only.
- Translational but not rotational. Reverse the force at $B$ so the two equal forces are now opposite in direction with different lines of action. The total force is zero, so the body is in translational equilibrium; but the two moments now act in the same sense and cause anticlockwise rotation. The rod undergoes pure rotation without translation.
The second case is exactly a couple, which the next section develops.
Couple and its torque
NCERT defines a couple as a pair of forces of equal magnitude but acting in opposite directions with different lines of action. Because the two forces cancel, a couple produces rotation without translation. Turning a bottle lid with your fingers and a compass needle in the Earth's magnetic field — equal, opposite forces on the two poles — are the textbook examples.
The key property is that the moment of a couple does not depend on the point about which moments are taken. NCERT proves it: with forces $\vec F$ at $B$ (position $\vec r_1$) and $-\vec F$ at $A$ (position $\vec r_2$), the total moment about the origin $O$ is
$$\vec r_1 \times (-\vec F) + \vec r_2 \times \vec F = (\vec r_2 - \vec r_1) \times \vec F = \vec{AB} \times \vec F$$
which contains only the relative vector $\vec{AB}$, not the origin. The magnitude of the couple's moment is therefore $F\,d$, where $d$ is the perpendicular distance between the two lines of action.
Principle of moments
An ideal lever is a light (negligible-mass) rod pivoted at a point called the fulcrum. A see-saw is the standard example. Two parallel forces $F_1$ and $F_2$, usually perpendicular to the lever, act at distances $d_1$ and $d_2$ from the fulcrum, with the support providing a reaction $R$.
The lever is in mechanical equilibrium, so both conditions apply. Translational equilibrium gives
$$R - F_1 - F_2 = 0$$
For rotational equilibrium, take moments about the fulcrum. The reaction $R$ acts at the fulcrum and so has zero moment arm; it drops out. With anticlockwise moments positive,
$$d_1 F_1 - d_2 F_2 = 0 \quad\Rightarrow\quad d_1 F_1 = d_2 F_2$$
In words: load arm × load = effort arm × effort. This is the principle of moments. Here $F_1$ is usually the weight to be lifted — the load, at the load arm $d_1$ — and $F_2$ is the effort, at the effort arm $d_2$. NCERT notes the principle still holds even when the parallel forces act at some angle to the lever rather than perpendicular to it.
Mechanical advantage
The ratio of load to effort is the mechanical advantage. Using the principle of moments,
$$\text{M.A.} = \frac{F_1}{F_2} = \frac{d_2}{d_1}$$
When the effort arm $d_2$ is larger than the load arm $d_1$, the mechanical advantage exceeds one — meaning a small effort can lift a large load. The beam of a balance is itself a lever, as is a see-saw; recognising the fulcrum, load arm and effort arm in any device lets you apply the principle directly.
The moment of a force is torque. For the full vector treatment $\vec\tau = \vec r \times \vec F$, see torque and angular momentum.
Centre of gravity vs centre of mass
Balance an irregular cardboard of mass $M$ on the tip of a pencil and you locate a single point $G$ where it stays horizontal. The pencil supplies an upward reaction equal and opposite to $Mg$, so the cardboard is in translational equilibrium; it is also in rotational equilibrium, otherwise the unbalanced torque would tilt it. This balance point is the centre of gravity (CG).
NCERT defines the CG as the point about which the total gravitational torque on the body is zero. If $\vec r_i$ is the position of the $i$-th particle measured from the CG, then
$$\vec\tau_g = \sum_i \vec r_i \times m_i \vec g = 0$$
Because gravity is uniform over a small body, $\vec g$ is the same for every particle and comes out of the sum, leaving $\sum_i m_i \vec r_i = 0$ — which is precisely the defining equation of the centre of mass (CM). Hence the CG coincides with the CM in uniform gravity.
One practical method follows from the definition: suspend the body from a point $A$, and the vertical line through $A$ passes through the CG. Suspending from another point $B$ gives a second vertical; their intersection is the CG. Since a laboratory body is small enough for uniform gravity, this also fixes its centre of mass.
Types of equilibrium
For a body resting under gravity, the response to a small displacement classifies the equilibrium. The discriminant is what happens to the height of the centre of gravity.
CG rises on disturbance
A small displacement raises the centre of gravity. Gravity supplies a restoring torque that returns the body to its original position. Example: a cone resting on its base.
CG falls on disturbance
A small displacement lowers the centre of gravity. The torque drives the body further from its original position, and it topples. Example: a cone balanced on its apex.
CG height unchanged
The displacement leaves the centre of gravity at the same height. There is no restoring or toppling torque, so the body simply rests in its new position. Example: a cone or sphere lying on its side.
The same logic explains why broad-based, low-CG objects are hard to topple: tilting them lifts the CG sharply, so the restoring torque is strong, and the body remains in stable equilibrium over a wide range of displacements.
Worked example — metal bar on knife-edges
A metal bar 70 cm long and 4.00 kg in mass is supported on two knife-edges placed 10 cm from each end. A 6.00 kg load is suspended at 30 cm from one end. Find the reactions at the knife-edges. Take $g = 9.8~\text{m s}^{-2}$.
Geometry. Bar $AB = 70$ cm. The CG $G$ is at the centre, so $AG = 35$ cm. The load hangs at $P$ with $AP = 30$ cm, giving $PG = 5$ cm. Knife-edges $K_1$ and $K_2$ sit 10 cm from each end, so $AK_1 = BK_2 = 10$ cm and $K_1G = K_2G = 25$ cm. The bar's weight $W = 4.00g$ acts down at $G$; the load $W_1 = 6.00g$ acts down at $P$; reactions $R_1, R_2$ act up at the knife-edges.
Translational equilibrium. $R_1 + R_2 - W_1 - W = 0$, so $R_1 + R_2 = 10.00g = 98.00~\text{N}$.
Rotational equilibrium about $G$. Taking moments about $G$ (anticlockwise positive): $-R_1(K_1G) + W_1(PG) + R_2(K_2G) = 0$, i.e. $-0.25 R_1 + 0.05 W_1 + 0.25 R_2 = 0$. This gives $R_1 - R_2 = 1.2g = 11.76~\text{N}$.
Solve. Adding and subtracting the two relations: $R_1 = 54.88~\text{N} \approx 55~\text{N}$ at $K_1$, and $R_2 = 43.12~\text{N} \approx 43~\text{N}$ at $K_2$. The knife-edge nearer the load carries the larger reaction, as expected.
Worked example — ladder against a wall
A 3 m long ladder weighing 20 kg leans on a frictionless wall, its feet resting on the floor 1 m from the wall. Find the reaction forces of the wall and the floor. Take $g = 9.8~\text{m s}^{-2}$.
Geometry and forces. Foot $A$ is 1 m from the wall, so by Pythagoras the top reaches $BC = 2\sqrt{2}$ m up the wall. Forces on the ladder: weight $W$ at its centre of gravity $D$ (midway); wall reaction $F_1$, horizontal, since the frictionless wall can only push perpendicular to itself; and the floor reaction $F_2$, resolved into a vertical normal $N$ and a horizontal friction $F$ directed toward the wall (it must prevent the foot from sliding out).
Translational equilibrium. Vertical: $N - W = 0$. Horizontal: $F - F_1 = 0$. With $W = 20g = 196.0~\text{N}$, immediately $N = 196.0~\text{N}$.
Rotational equilibrium about $A$. Taking moments about the foot $A$ removes both floor components $N$ and $F$ (they act at $A$): $2\sqrt 2\, F_1 - \tfrac12 W = 0$, so $F_1 = \dfrac{W}{4\sqrt 2} = \dfrac{196.0}{4\sqrt 2} = 34.6~\text{N}$.
Assemble. From the horizontal equation $F = F_1 = 34.6~\text{N}$. The floor reaction magnitude is $F_2 = \sqrt{N^2 + F^2} = \sqrt{196.0^2 + 34.6^2} \approx 199.0~\text{N}$, making an angle $\alpha$ with the horizontal where $\tan\alpha = N/F = 196.0/34.6 = 4\sqrt 2$.
Worked example — balanced rod (see-saw type)
A uniform rod of length 200 cm and mass 500 g is balanced on a wedge at the 40 cm mark. A 2 kg mass hangs at the 20 cm mark and an unknown mass $m$ hangs at the 160 cm mark. Find $m$ for equilibrium. Take $g = 10~\text{m s}^{-2}$.
Set the pivot. The wedge at the 40 cm mark $O$ is the fulcrum; take moments about it. Distances from $O$: the 2 kg mass at 20 cm is $0.20$ m on one side; the rod's CG at the centre (100 cm) is $0.60$ m on the other side; the mass $m$ at 160 cm is $1.20$ m on the same side as the CG.
Moments about $O$. Anticlockwise from the 2 kg mass: $\tau_1 = 2 \times 10 \times 0.20 = 4~\text{N m}$. Clockwise from the rod's weight ($0.5$ kg at its CG): $\tau_2 = 0.5 \times 10 \times 0.60 = 3~\text{N m}$. Clockwise from $m$: $\tau_3 = m \times 10 \times 1.20 = 12m~\text{N m}$.
Balance. Net torque zero: $4 - 3 - 12m = 0 \Rightarrow 12m = 1 \Rightarrow m = \dfrac{1}{12}~\text{kg}$. This is just the principle of moments with the rod's own weight included as a third moment about the fulcrum.
Equilibrium of a rigid body in one breath
- Mechanical equilibrium needs both $\sum\vec F = 0$ (translational) and $\sum\vec\tau = 0$ (rotational). One alone gives only partial equilibrium.
- A couple is two equal, opposite forces on different lines of action; net force zero, so pure rotation, moment $= F\,d$, independent of origin.
- Principle of moments for a lever: load arm × load = effort arm × effort, $d_1 F_1 = d_2 F_2$.
- Mechanical advantage $= F_1/F_2 = d_2/d_1$; greater than one when the effort arm is longer, so a small effort lifts a large load.
- CG is the point of zero gravitational torque; it coincides with the CM only in uniform gravity. They are different concepts.
- Stable / unstable / neutral equilibrium $\Leftrightarrow$ a small displacement raises / lowers / leaves unchanged the height of the CG.
- Solving strategy: take moments about a point where an unknown reaction acts so it drops out, then use $\sum\vec F = 0$ for the rest.