8.1 Distinguish centroid, centre of mass, and centre of gravity | University of Alberta ENGG 130: Engineering Mechanics: Statics

University of Alberta ENGG 130: Engineering Mechanics: Statics

Centroids and Centres of Gravity

How geometry, mass, and weight define different locations

These three points may occupy the same location, but they answer different questions. The centroid describes geometry, the centre of mass describes how mass is distributed, and the centre of gravity describes where the resultant gravitational force acts. A body in this lesson means the object, or collection of parts, being considered. We will use coordinates measured from a stated origin. The central calculation idea is a weighted average: a part with more area, mass, or weight has more influence, depending on which point is being found.

What you will learn

Define centroid, centre of mass, and centre of gravity and identify what each describes.
Use area-weighted and mass-weighted averages to locate simple centroids and centres of mass.
Explain when these points coincide, including the assumptions of uniform density and a uniform gravitational field.
Locate the resultant of parallel weights by matching its force and moment to those of the individual weights.

1. Three meanings, three kinds of weighting

The centroid is the geometric average location of a line, area, or volume. For a flat area, imagine dividing the shape into many small area elements. Each element contributes according to its area and position. The centroid depends on the shape, not on the material density or the gravitational field. For a thin plate with uniform thickness and density, the area centroid is also its centre of mass.

The centre of mass is the mass-weighted average location of a body or system. If more mass is located on one side, the centre of mass is drawn toward that side. For separate components, multiply each component's mass by its position, add the products, and divide by the total mass. For a continuous body, the same idea uses small mass elements.

The centre of gravity is the point through which the resultant gravitational force, or total weight, acts. It depends on both the mass distribution and the gravitational field. In a uniform gravitational field, the gravitational acceleration has the same magnitude and direction throughout the body. Weight is then proportional to mass everywhere, so the centre of gravity coincides with the centre of mass.

Keep the questions distinct: “Where is the geometric average of this shape?” asks for a centroid. “Where is the mass distributed on average?” asks for a centre of mass. “Where does the resultant weight act?” asks for a centre of gravity. A weighted average is not necessarily the midpoint of the object's outer dimensions.

\bar{x}=\frac{\int x\,dA}{\int dA}

Centroid: position weighted by geometry, such as area.
Centre of mass: position weighted by mass.
Centre of gravity: location of the resultant gravitational force.

2. Equations and conditions for coincidence

For a plane area, choose axes and measure every small area element from the same origin. Its horizontal centroid coordinate is the sum of each element's horizontal position times its area, divided by the total area. The vertical coordinate is found in the same way. For a shape made from simple pieces, use each piece's area and centroid location. A removed region can be included as a negative area.

For separate components at positions along a line, the centre-of-mass coordinate is the sum of mass times position divided by total mass. In a plane, apply the calculation separately to the horizontal and vertical coordinates. The result has units of length because the mass units cancel.

To find a centre of gravity using a position-weighted average, a key condition is that the individual gravitational forces are parallel and point in the same direction. Under that condition, the resultant weight magnitude is the sum of the individual weight magnitudes, and its position is found by matching moments. For point masses with weight magnitudes

W_i=m_i g_i

, the horizontal coordinate is the weight-weighted average shown below. If the gravitational force directions differ, the resultant is a vector sum; its magnitude need not equal the sum of the individual weight magnitudes, and this scalar average does not in general locate its line of action.

If the gravitational field is uniform, all masses have the same gravitational acceleration vector. Their weights are then proportional to their masses, and the centre of gravity and centre of mass coincide. If a body's density is also uniform, its centre of mass coincides with its geometric centroid. Therefore, all three points coincide for a uniform-density body in a uniform gravitational field.

x_G=\frac{\sum_i W_i x_i}{\sum_i W_i},\qquad W_i=m_i g_i

Use area weights for an area centroid and mass weights for a centre of mass.
Use the scalar weight-weighted position rule only when the individual weights are parallel and point in the same direction.
Uniform density makes centroid and centre of mass coincide; a uniform gravitational field makes centre of mass and centre of gravity coincide.

3. Choose the model and check the location

Start by naming the point requested and the information supplied. A flat outline with dimensions usually calls for a centroid. Component masses or a stated mass distribution call for a centre of mass. A question about weights or gravitational forces calls for a centre of gravity. Do not assume uniform density just because a shape has a simple outline.

For a one-dimensional calculation, state the origin and positive direction. For a two-dimensional location, state both axes. Keep each position measured from the same origin, and keep units visible. For example, area may be measured in square metres, mass in kilograms, weight in newtons, and a final coordinate in metres.

For positive areas, masses, or parallel weights, the weighted average lies between the smallest and largest input coordinates. It is pulled toward the larger contributors. A centroid can lie outside the material for some shapes; it still represents the geometric average location. Similarly, the centre of gravity is a point used to represent the resultant force and need not be a visible feature of the body.

When using the centre of gravity to represent a set of parallel weights, check both the total force and the moment about the chosen origin. The resultant must have the same total weight as the separate weights, and its moment must equal the sum of their moments. These checks confirm that the replacement force represents the same loading.

Identify whether the weighting quantity is area, mass, or weight.
Use one origin, one positive direction, and consistent units.
For a resultant of parallel weights, check total force and moment.

Worked example

1. Centroid of a triangular area

A uniform triangular plate has vertices at

(0,0)

,

(6,0)

, and

(0,3)

, with coordinates in metres. Find its area centroid. Only the geometry is needed.

Identify the requested point
This is a centroid problem, so area—not mass or weight—is the relevant weighting. The right-angle vertex is at the origin, and the legs lie along the positive axes. The base is $6\ \mathrm{m}$ and the height is $3\ \mathrm{m}$ .
Apply the triangle centroid rule
For a triangle, the centroid is one-third of the way along each median from the base toward the opposite vertex. For this right triangle, that places it one-third of the way along each leg from the right-angle vertex.
$\bar{x}=\frac{b}{3},\qquad \bar{y}=\frac{h}{3}$
Substitute the dimensions
Substitute the base and height, keeping both coordinates measured from the origin at the right-angle vertex.
$\bar{x}=\frac{6\ \mathrm{m}}{3}=2.00\ \mathrm{m},\qquad \bar{y}=\frac{3\ \mathrm{m}}{3}=1.00\ \mathrm{m}$

Answer: The area centroid is at

(2.00\ \mathrm{m},\ 1.00\ \mathrm{m})

.

Check: The point is inside the triangle. It is also closer to the right-angle vertex than the far ends of either leg, consistent with the triangle's area distribution.

Worked example

2. Centre of mass of two components

Two small components are fixed along a horizontal support. Component A has mass

2.0\ \mathrm{kg}

at

x=0.0\ \mathrm{m}

, and component B has mass

6.0\ \mathrm{kg}

at

x=4.0\ \mathrm{m}

. Find the centre of mass of the two-component system.

Define the system and coordinate
The system consists of both components. Take the origin at A and positive $x$ to the right. Treat each component's mass as located at its stated position for this calculation.
Set up the mass-weighted average
Each mass-position product contributes to the average. Dividing by total mass gives a coordinate with units of length.
$x_{\mathrm{cm}}=\frac{m_Ax_A+m_Bx_B}{m_A+m_B}$
Substitute and calculate
The numerator has units of kilogram-metres and the denominator has units of kilograms.
$x_{\mathrm{cm}}=\frac{(2.0\ \mathrm{kg})(0.0\ \mathrm{m})+(6.0\ \mathrm{kg})(4.0\ \mathrm{m})}{2.0\ \mathrm{kg}+6.0\ \mathrm{kg}}=3.0\ \mathrm{m}$

Answer: The centre of mass is

3.0\ \mathrm{m}

to the right of the origin.

Check: The coordinate lies between the component positions and is closer to the heavier component B. The mass units cancel, leaving metres.

Worked example

3. Centre of gravity for parallel weights

Two masses lie on a horizontal line:

m_1=2.0\ \mathrm{kg}

at

x_1=0.0\ \mathrm{m}

and

m_2=3.0\ \mathrm{kg}

at

x_2=4.0\ \mathrm{m}

. Their local gravitational acceleration magnitudes are

g_1=9.0\ \mathrm{m/s^2}

and

g_2=10.0\ \mathrm{m/s^2}

. Assume both gravitational forces point vertically downward, so they are parallel and in the same direction. Find the resultant weight and its line of action.

Find each weight
Calculate the magnitude of each gravitational force from its mass and local gravitational acceleration. Both forces act downward under the stated assumption.
$W_1=m_1g_1=18\ \mathrm{N},\qquad W_2=m_2g_2=30\ \mathrm{N}$
Set up force and moment equivalence
Because the forces are parallel and point in the same direction, their resultant magnitude is their sum. Match moments about the origin to locate its line of action. Take downward force magnitudes as positive for this calculation.
$W_R=W_1+W_2,\qquad W_Rx_G=W_1x_1+W_2x_2$
Calculate the location
The moment numerator is in newton-metres and the resultant weight is in newtons, so the coordinate is in metres.
$W_R=48\ \mathrm{N},\qquad x_G=\frac{(18\ \mathrm{N})(0.0\ \mathrm{m})+(30\ \mathrm{N})(4.0\ \mathrm{m})}{48\ \mathrm{N}}=2.50\ \mathrm{m}$
Verify the equivalent force and moment
The separate weights total $48\ \mathrm{N}$ downward, matching the resultant. About the origin, the resultant moment magnitude is $48\ \mathrm{N}$ times $2.50\ \mathrm{m}$ , which matches the sum of the two separate moment magnitudes.
$48\ \mathrm{N}=18\ \mathrm{N}+30\ \mathrm{N},\qquad (48\ \mathrm{N})(2.50\ \mathrm{m})=(18\ \mathrm{N})(0.0\ \mathrm{m})+(30\ \mathrm{N})(4.0\ \mathrm{m})=120\ \mathrm{N\cdot m}$

Answer: The resultant weight is

48\ \mathrm{N}

downward, acting at

x_G=2.50\ \mathrm{m}

from the origin.

Check: Both the total force and the moment about the origin agree with the two original weights. The position is between the masses and closer to the larger weight.

Common mistakes and how to avoid them

Using centroid, centre of mass, and centre of gravity as interchangeable names.

Correction: Ask what is being weighted: geometry, mass, or gravitational force. The points coincide only when the relevant distributions make them coincide.

Assuming an object's centroid or centre of mass is always at the midpoint of its width.

Correction: A midpoint applies only to suitable symmetric or equally weighted arrangements. Use the shape or mass distribution in the calculation.

Assuming a simple outline means the material has uniform density.

Correction: Shape and density are separate properties. A centroid calculation uses geometry; a centre-of-mass calculation uses mass distribution.

Using a scalar weight-weighted average when gravitational force directions differ.

Correction: The scalar rule applies when individual weights are parallel and point in the same direction. With differing directions, the resultant is a vector sum, and its magnitude need not be the sum of the weight magnitudes.

Forgetting to check the force and moment represented by a resultant weight.

Correction: Confirm that the resultant has the same total force and the same moment about the chosen origin as the individual parallel weights.

Lesson summary

The centroid is the geometric average location of a line, area, or volume.
The centre of mass is the mass-weighted average location of a body or system.
The centre of gravity is the location of the resultant gravitational force.
Uniform density makes the centre of mass coincide with the centroid; a uniform gravitational field makes the centre of gravity coincide with the centre of mass.
For parallel weights pointing in the same direction, add their magnitudes and match moments to find the resultant's line of action.

Check your understanding

Question 1

A plate has uniform density and lies in a uniform gravitational field. Which statement is correct?

Its centroid, centre of mass, and centre of gravity coincide.
Its centroid must differ from its centre of mass.
Its centre of gravity depends only on its outline, not its mass distribution.
The three points coincide only if the plate is rectangular.

Show answer and explanation

Its centroid, centre of mass, and centre of gravity coincide.

Uniform density makes mass distributed in proportion to the plate's geometry, and a uniform field makes gravitational force proportional to mass throughout the plate. Thus the three locations coincide.

Question 2

Two masses of

1\ \mathrm{kg}

and

3\ \mathrm{kg}

are at

x=0\ \mathrm{m}

and

x=4\ \mathrm{m}

. Where is their centre of mass?

$1\ \mathrm{m}$
$2\ \mathrm{m}$
$3\ \mathrm{m}$
$4\ \mathrm{m}$

Show answer and explanation

3\ \mathrm{m}

The mass-weighted coordinate is

\frac{(1\ \mathrm{kg})(0\ \mathrm{m})+(3\ \mathrm{kg})(4\ \mathrm{m})}{1\ \mathrm{kg}+3\ \mathrm{kg}}=3\ \mathrm{m}

. It lies closer to the larger mass.

Question 3

Two downward weights act at

x=0\ \mathrm{m}

and

x=4\ \mathrm{m}

. Their magnitudes are

10\ \mathrm{N}

and

30\ \mathrm{N}

, respectively. Where does their resultant act?

$1\ \mathrm{m}$
$2\ \mathrm{m}$
$3\ \mathrm{m}$
$4\ \mathrm{m}$

Show answer and explanation

3\ \mathrm{m}

The weights are parallel and point in the same direction, so

x_G=\frac{(10\ \mathrm{N})(0\ \mathrm{m})+(30\ \mathrm{N})(4\ \mathrm{m})}{10\ \mathrm{N}+30\ \mathrm{N}}=3\ \mathrm{m}

. The result is closer to the larger weight.

Key terms

Centroid: The geometric average location of a line, area, or volume.
Centre of mass: The mass-weighted average location of a body or system.
Centre of gravity: The point through which the resultant gravitational force on a body acts.
Uniform density: Density is the same throughout the body.
Uniform gravitational field: The gravitational acceleration has the same magnitude and direction throughout the region considered.
Resultant: A single force that represents the combined effect of a set of forces.

Continue through ENGG 130

View the complete ENGG 130 University of Alberta ENGG 130: Engineering Mechanics: Statics curriculum and lessons

About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows University of Alberta ENGG 130: Engineering Mechanics: Statics, study topic 8.1. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

Official curriculum reference

Report a correction or ask a question

8.1 · Distinguish centroid, centre of mass, and centre of gravity

University of Alberta ENGG 130: Engineering Mechanics: Statics

How geometry, mass, and weight define different locations

What you will learn

1. Three meanings, three kinds of weighting

2. Equations and conditions for coincidence

3. Choose the model and check the location

1. Centroid of a triangular area

2. Centre of mass of two components

3. Centre of gravity for parallel weights

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through ENGG 130

About this lesson and its review