Center of Gravity Formula: x = ΣWx/ΣW with Examples

The center of gravity formula is $x_{cg} = \frac{\sum W_i x_i}{\sum W_i}$ : multiply each weight by its position, add the products, and divide by the total weight. The answer is the single point at which the whole weight of a body can be taken to act. In Indian books you will see it written as the centre of gravity formula.

Since each weight is $W_i = m_i g$ , and g is the same everywhere in a small object near Earth, g cancels top and bottom. The result is then identical to the center of mass formula. The two points separate only when g varies across the body, which needs something as tall as a mountain or a very long structure in orbit.

Quick facts

Formula
$x_{cg} = \frac{\sum W_i x_i}{\sum W_i}$
Uniform g
Center of gravity = center of mass
SI unit
metre (m), a position
Balance condition
Net torque of weights about the CG is zero
Stability
Lower CG and wider base mean more stable
Finding it in a lab
Hang the body from two points; the plumb lines cross at the CG

Why the center of gravity formula works

Take moments of all the separate weights about any pivot. Their total turning effect is $\sum W_i x_i$ . A single force equal to the total weight, $\sum W_i$ , placed at $x_{cg}$ , must give the same moment: $(\sum W_i)\,x_{cg} = \sum W_i x_i$ . Divide and you have the formula. It is really the principle of moments, the same idea behind the formula of torque.

Support a body exactly below or above this point and it balances, because the weights produce no net torque about it. Support it anywhere else and it rotates until the CG hangs directly below the support.

Worked examples

Example 1: a plank with a paint tin

A uniform 2 m plank weighs 100 N, so its own CG sits at its midpoint, 1.0 m along. A 50 N tin sits at 1.8 m. Find the CG of plank plus tin.

$x_{cg} = \frac{100(1.0) + 50(1.8)}{100 + 50} = \frac{100 + 90}{150} \approx 1.27\ \text{m}$

Example 2: masses given instead of weights

A light rod 1.2 m long carries 5 kg at the left end and 15 kg at the right end. Where should you lift it so it stays level?

Weights are 49 N and 147 N. From the left end: $x = \frac{49(0) + 147(1.2)}{49 + 147} = \frac{176.4}{196} = 0.9\ \text{m}$ .

Notice that 9.8 cancels; using the masses directly gives the same 0.9 m.

Example 3: an L shaped bracket

An L shape is made of two thin uniform strips. Strip A weighs 6 N and lies along the x axis from 0 to 0.4 m. Strip B weighs 3 N and lies along the y axis from 0 to 0.2 m. Find the CG.

CG of A: (0.2, 0). CG of B: (0, 0.1). $x = \frac{6(0.2) + 3(0)}{9} \approx 0.133\ \text{m}$ , $y = \frac{6(0) + 3(0.1)}{9} \approx 0.033\ \text{m}$ .

The point lies off the metal, in the corner space between the strips.

Center of gravity and stability

A body resting on a base is stable as long as the vertical line through its CG falls inside the base. Tilt it, and the CG rises; let go, and the weight's torque brings it back. Tilt it far enough that the line passes outside the base, and the same torque tips it over.

Stable equilibrium: the CG rises when the body is disturbed, as with a cone on its base.
Unstable equilibrium: the CG falls when disturbed, as with a cone balanced on its tip.
Neutral equilibrium: the CG stays at the same height, as with a ball on a level floor.

That is why racing cars sit low, why loaded buses keep luggage underneath, and why you widen your feet on a moving train.

When CG and centre of mass differ

For any object of everyday size near Earth, treat them as the same point. They differ when g changes across the body. For a tall tower, the lower parts sit slightly nearer Earth's centre and feel slightly larger g, so the CG lies a tiny distance below the center of mass. The variation of g with height is set out on the acceleration due to gravity page. In free space, with no gravity, a body has a center of mass but no center of gravity at all.

Exam answer: in a uniform gravitational field the center of gravity coincides with the centre of mass.

Avoid these slips

Mixing masses and weights in the same sum. Use one or the other throughout.
Measuring each xᵢ from a different reference point.
Placing a uniform rod's own weight at one end. It acts at the rod's midpoint.
Assuming the CG must be inside the material.

For moment of inertia about an axis through this point, and the parallel axis theorem, see the moment of inertia formula. The rotation section groups every related page.

Frequently asked questions

What is the centre of gravity formula?

x = ΣWᵢxᵢ / ΣWᵢ along each axis. It locates the single point where the total weight of the body acts.

Is center of gravity the same as center of mass?

In a uniform gravitational field, yes. They differ only when g changes noticeably from one part of the body to another.

How do you find the center of gravity of an irregular sheet?

Hang it from one point and draw the vertical line using a plumb line. Hang it from a second point and draw again. The lines cross at the center of gravity.

Why does a lower center of gravity make a body more stable?

A low CG must rise more before the vertical line through it passes outside the base, so a larger tilt is needed to topple the body.

What is the unit of the center of gravity?

It is a position, so it is given in metres from a chosen origin.