Centre of gravity

Balance a pencil on your fingertip and it wobbles until, at one exact spot, it sits still — that spot is the pencil's centre of gravity, the single point through which its entire weight can be imagined to act, no matter how the pencil is tilted. Every one of the countless tiny particles making up a body is pulled down by its own share of gravity, $m_ig$ ; the centre of gravity is where all those pulls can be replaced, without changing any turning effect, by one force $Mg$ . For everyday objects on Earth this point is identical to the centre of mass, so the two ideas can be used interchangeably here. Centre of gravity is the key that unlocks rigid-body equilibrium: a beam, ladder, or see-saw balances only when the moments of every force about a pivot cancel exactly, a rule called the principle of moments. This article builds that rule from the idea of centre of gravity, applies it to a metre rule and a supported beam, and flags the traps ISC examiners set around equilibrium and moments.

The core insight

The centre of gravity is the single point where a body's total weight $Mg$ effectively acts, and — coinciding with the centre of mass under uniform $g$ — it lets rigid-body equilibrium be checked with the principle of moments, $\sum\tau_{clockwise}=\sum\tau_{anticlockwise}$ , alongside zero net force.

The core idea

Weight of a body, particle by particle. Every particle in a rigid body has mass $m_i$ and feels a tiny weight $m_ig$ pulling it straight down. Added over the whole body these weights are equivalent, for every purpose — net force and net turning effect — to a single force $Mg$ acting at one point: the centre of gravity (CG). This is not a physical dent in the object; it is the mathematical balance point of all those individual pulls.

CG and centre of mass are the same point here. The centre of mass (COM) is defined purely from how mass is distributed, with no reference to gravity at all. The CG is defined from where weight acts. The two only have to coincide if $g$ is the same strength and direction at every particle of the body — which is true to extremely high accuracy for any object you can pick up, since $g$ changes negligibly over a height of a few metres. Only for a body many kilometres tall, or in a strongly non-uniform field, would CG and COM separate. For this course: treat CG and COM as one and the same point.

Torque, or moment, of a force. A force can turn a body about a pivot even if it does not move it. The turning effect is the moment (or torque) of the force, defined as

\tau = F \times d

where: $F$ is the magnitude of the force and $d$ is the perpendicular distance from the pivot to the line of action of that force. A larger force, or the same force applied farther from the pivot, produces a bigger turning effect — exactly why a long spanner loosens a tight nut more easily than a short one.

Two conditions for rigid-body equilibrium. A rigid body stays still — no sliding and no spinning — only if both of the following hold at once: The vector sum of every external force on it is zero, $\sum \vec F = 0$ , so there is no net push in any direction. The vector sum of every external torque about any chosen point is zero, $\sum \vec\tau = 0$ , so there is no net turning effect. A pair of equal, opposite, non-collinear forces can satisfy the first condition and still spin the body — this is called a force couple, and it is exactly why the second condition is needed separately. For a body resting under gravity plus supports — a ladder leaning against a wall, held by floor friction and a wall's normal reaction, is the classic three-coplanar-force case in this syllabus — both conditions together pin down every unknown force. Practically, stability also depends on where the CG sits: if the vertical line through the CG falls inside the body's base of support, the body stays up; if it falls outside, the body topples.

Figure

A seesaw balanced on a central pivot with a heavier block closer to the pivot on the left and a lighter block farther out on the right, with dashed lines showing each perpendicular distance to the pivot

Derivation AReaction forces at two supports of a loaded beam

The goal of this derivation is to show, from the two equilibrium conditions alone, exactly how the two support forces holding up a horizontal beam under its own weight can be found without any calculus — purely by writing one force equation and one moment equation and solving them together, which is precisely the ISC syllabus expectation for beam and lever problems. Take a uniform rigid beam of weight $W$ and length $L$ , resting horizontally on two supports at its two ends, with the beam's own weight acting at its centre of gravity, the midpoint at $L/2$ from either end.

Step 1. Write the translational equilibrium condition. Let $R_1$ and $R_2$ be the (unknown) upward reactions at the two supports. Since nothing else pushes the beam vertically besides its weight and the two reactions, $\sum F = 0$ gives

R_1 + R_2 = W

This single equation has two unknowns, so it alone cannot fix both reactions — a second, independent equation is needed.

Step 2. Write the rotational equilibrium condition about one support. Choose the support carrying $R_1$ as the pivot; then $R_1$ itself has zero moment about that point (its distance from the pivot is zero). The weight $W$ acts at the midpoint, a perpendicular distance $L/2$ from this pivot, producing a clockwise moment $W \times (L/2)$ . The reaction $R_2$ acts upward at distance $L$ from this pivot, producing an anticlockwise moment $R_2 \times L$ . Setting clockwise equal to anticlockwise (the principle of moments),

R_2 \times L = W \times \frac{L}{2}

Step 3. Solve for $R_2$ , then back-substitute. Dividing both sides by $L$ gives $R_2 = W/2$ . Substituting into the Step 1 equation, $R_1 = W - W/2 = W/2$ . By the symmetry of a centrally loaded, symmetrically supported beam this equal split is exactly what intuition demands — but the derivation shows why, and the same two-equation method works unchanged even when the supports are not symmetric or the load is not central, simply by keeping track of each moment's perpendicular distance correctly.

Step 4. Check consistency by choosing the other support as pivot. Taking moments about the support carrying $R_2$ instead, $R_1$ now acts at distance $L$ (anticlockwise) and $W$ acts at distance $L/2$ (clockwise), giving $R_1 \times L = W \times L/2$ , i.e. $R_1 = W/2$ again — the same answer, confirming that the choice of pivot never changes the physics, only the bookkeeping.

Worked example

A uniform beam AB of length $4\ \text{m}$ and weight $200\ \text{N}$ rests horizontally on two supports, one at end A and one at $3\ \text{m}$ from A. Find the reaction force at each support.

Step 1. Locate the centre of gravity. The beam is uniform, so its weight $200\ \text{N}$ acts at its midpoint, which is $2\ \text{m}$ from A.

Step 2. Apply translational equilibrium. Let $R_A$ be the reaction at A and $R_B$ the reaction at the support $3\ \text{m}$ from A. Since only these two reactions and the weight act vertically, $R_A + R_B = 200\ \text{N}$ .

Step 3. Apply rotational equilibrium about A. The weight, at $2\ \text{m}$ from A, gives a clockwise moment $200 \times 2 = 400\ \text{N}\,\cdot\,\text{m}$ . The reaction $R_B$ , at $3\ \text{m}$ from A, gives an anticlockwise moment $R_B \times 3$ . Setting these equal,

R_B \times 3 = 400 \implies R_B = 133.3\ \text{N}

Step 4. Solve for $R_A$ and check. From Step 2, $R_A = 200 - 133.3 = 66.7\ \text{N}$ . Checking by taking moments about the $3\ \text{m}$ support instead: $R_A \times 3 = 200 \times (3-2) = 200\ \text{N}\,\cdot\,\text{m}$ , giving $R_A = 66.7\ \text{N}$ again — consistent. Answer: $R_A = 66.7\ \text{N}$ and $R_B = 133.3\ \text{N}$ .

Competitive stretch

The same beam is reconsidered as a metre-rule problem: a uniform metre rule of mass $100\ \text{g}$ balances on a knife-edge at its $50\ \text{cm}$ mark, confirming uniformity. A $60\ \text{g}$ mass is hung at the $20\ \text{cm}$ mark. Where must $40\ \text{g}$ be hung to restore balance? Since the rule's own weight already acts at the $50\ \text{cm}$ pivot, only the two hung masses need balancing: the $60\ \text{g}$ mass sits $30\ \text{cm}$ to the left, giving moment $60 \times 30 = 1800\ \text{g}\,\cdot\,\text{cm}$ ; setting the unknown distance $d$ on the right so that $40 \times d = 1800$ gives $d = 45\ \text{cm}$ , so the $40\ \text{g}$ mass hangs at the $95\ \text{cm}$ mark.

Common traps

Treating centre of gravity and centre of mass as always different quantities — under the uniform- $g$ assumption used throughout this course, they are numerically identical points; the distinction is conceptual, never a source of two different numbers in a calculation.
Forgetting to include a beam or rod's own weight (acting at its midpoint) as one of the forces when applying the principle of moments — the single most common slip in beam and lever problems.
Taking moments about the wrong pivot, or mixing clockwise and anticlockwise sign conventions inconsistently within the same equation, which silently flips an answer's sign.
Assuming $\sum \vec F = 0$ alone guarantees equilibrium — a rigid body also needs $\sum \vec\tau = 0$ ; a force couple has zero net force yet spins the body.

Ask the tutor: Want to work through the ladder-against-a-wall three-force equilibrium problem step by step, or see how CG connects to stability and toppling?

Why do CG and COM ever separate at all, if they are the same for every problem we solve

They separate only when the gravitational field is not uniform across the body's size — for instance, a body many kilometres tall, where g at the top is measurably weaker than at the bottom, or a body placed in an artificially non-uniform field. For every object in this course, from a pencil to a beam to a metre rule, g varies by an utterly negligible amount over the object's dimensions, so CG and COM sit at exactly the same point. The distinction matters conceptually, in definitions, but never changes a number you calculate here.

Why does it matter which point I choose as the pivot when taking moments

It does not change the final answer, but it changes how much algebra you do. Any point at all can serve as the pivot for the moment equation — equilibrium means the net torque is zero about every point, not just one special one. Choosing a pivot that passes through an unknown force (as in the beam example, pivoting at a support) makes that force's moment vanish automatically, leaving one equation in one unknown instead of two. Choosing badly just means more algebra, never a wrong answer, provided the moments are added up correctly and consistently.

Is the ladder-against-a-wall problem solved the same way as the beam problems here

Yes, in method, though it needs care with directions. A ladder leaning on a smooth wall and rough floor experiences three coplanar forces: its weight at the CG, the wall's horizontal normal reaction, and the floor's reaction (a combination of vertical normal force and horizontal friction). You still apply $\sum \vec F = 0$ in both the horizontal and vertical directions, plus $\sum \vec\tau = 0$ about a convenient pivot such as the foot of the ladder, exactly as in the beam examples. The only extra step is resolving forces into horizontal and vertical components before applying the conditions, since the forces here are not all vertical as they were for the horizontal beam.

Centre of gravity

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads