8.4 Find centres of gravity for composite bodies | University of Alberta ENGG 130: Engineering Mechanics: Statics

University of Alberta ENGG 130: Engineering Mechanics: Statics

Centroids and Centres of Gravity

ENGG 130 study topic 8.4

A composite body is made from simpler parts whose sizes, positions, and weights can be handled separately. Its centre of gravity,

G

, is the point where the combined weight can be represented as acting. The key idea is a weighted average of positions: a larger or heavier part has more influence on the result. For a thin plate with uniform material and thickness in a uniform gravitational field, the weights of its parts are proportional to their areas, so the area centroid is also the centre of gravity. If parts have different densities or thicknesses, use their weights instead of their areas. In this lesson, each example is a plate, and the calculations use a single coordinate system for the whole plate.

What you will learn

Explain how component weights determine the centre of gravity of a composite body.
Find the centre of gravity by dividing a body into simple shapes and combining their first moments.
Use signed areas to account for holes in uniform plates.
Check a result by verifying the total weight and its moments about the coordinate axes.

1. The weighted-position idea

Imagine each component's weight acting downward at that component's own centre of gravity. The total weight is the sum of the component weights. To replace these separate weights by one total weight at

G

, the moments about the coordinate axes must be preserved. For components with weights

W_i

and centres

(x_i,y_i)

, this gives the weighted-position equations.

When every part of a plate has the same material and thickness, its weight per unit area is the same everywhere. Each part's weight is then proportional to its area, and the proportionality factor cancels from the equations. Use areas in place of weights only under this assumption. A hole is not material belonging to the body, so represent it as a negative-area component.

Choose an origin and positive directions before locating component centres. Use the same origin for every component. Coordinates can be negative if a centre lies left of or below the origin. With only positive weights or areas, each combined coordinate lies between the smallest and largest component-centre coordinates along that axis.

x_G=\frac{\sum W_i x_i}{\sum W_i},\qquad y_G=\frac{\sum W_i y_i}{\sum W_i}

Combine coordinates by weighting them with area or weight; do not average them equally unless the contributions are equal.
Use area weighting for a uniform, equal-thickness plate; otherwise use component weights.
A removed region has negative area and negative first-moment contributions.

2. Divide the body and locate each centre

Define the complete body, choose convenient axes, and divide it into simple non-overlapping regions. Rectangles, triangles, and circles are often convenient. For each region, record its signed area or weight and the coordinates of its own centre in the shared coordinate system. Do not count any material twice.

A rectangle's centroid is at the midpoint of its width and height. A triangle's centroid is the average of its three vertex coordinates: for vertices

(x_1,y_1)

,

(x_2,y_2)

, and

(x_3,y_3)

, its coordinates are

((x_1+x_2+x_3)/3,(y_1+y_2+y_3)/3)

. A circle's centroid is at its centre. These are component-centre locations, not the final centre of the composite body.

For area calculations, add the signed areas and the signed first moments. The first moment about the vertical axis uses each area's horizontal coordinate; the first moment about the horizontal axis uses its vertical coordinate. Divide each total first moment by the total signed area. Area has units of length squared, and a first moment has units of length cubed, so the resulting coordinate has units of length.

x_G=\frac{\sum A_i x_i}{\sum A_i},\qquad y_G=\frac{\sum A_i y_i}{\sum A_i}

Locate each component centre before combining contributions.
Use one origin and one coordinate convention throughout.
Apply the negative sign for a hole to its area and both first moments.

3. Moment balance and reasonableness checks

The weighted equations express moment balance for the component weights. If the total weight is

W_G

, the single weight acting at

G

must equal the sum of component weights and produce the same moments about both axes. This is a statics interpretation used to locate the centre; it does not require solving for supports or reactions.

After calculating a centre, verify both first-moment sums. For an area model, check that total area multiplied by each calculated coordinate equals the corresponding sum of signed area-coordinate products. For a weight model, make the same check with weights. The downward force balance is that the resultant weight equals the sum of the component weights. A mismatch points to an arithmetic, sign, or coordinate error.

Also check whether the result makes geometric sense. Symmetry can determine a coordinate directly. A hole tends to shift the centre away from the hole, while adding a region tends to shift it toward that region. With only positive component areas, compare the result with the range of component-centre coordinates. Such checks do not replace the calculation, but they can reveal mistakes.

W_G=\sum W_i,\qquad W_Gx_G=\sum W_ix_i,\qquad W_Gy_G=\sum W_iy_i

The resultant weight equals the sum of the component weights.
The resultant at the calculated centre must preserve moments about both axes.
Use symmetry, coordinate ranges, and the direction of a hole's effect as reasonableness checks.

4. A consistent calculation

A short component list helps prevent sign and unit errors. For each region, record its signed area or weight, its centre coordinates, and the products used in the two moment sums. Keep the sign attached to the region in every column. In particular, do not subtract a hole's area but then add its first moments.

Set the units before calculating. If dimensions are in millimetres, areas are in square millimetres and first moments are in cubic millimetres; final centre coordinates are in millimetres. If using weights, use a consistent force unit. Keep exact values such as

\pi

as long as practical, and round coordinates only at the end.

The worked examples use thin plates in a uniform gravitational field. For these plates, gravity acts downward on each part, and uniform material and thickness allow areas to represent relative weights. The diagrams identify the body and coordinate axes; the geometry and component locations are specified in the problem text.

Keep areas or weights and their moment contributions in consistent units.
State when area represents weight; use weights if thicknesses or densities differ.
Round after evaluating the sums, then verify the result with unrounded values.

Worked example

Two joined rectangular plates

A uniform plate consists of a horizontal rectangle 120 mm wide and 40 mm high, with a second rectangle 40 mm wide and 80 mm high attached above its right-hand end. Let the origin be the lower-left corner of the first rectangle, with

x

rightward and

y

upward. Find the plate's centre of gravity.

Set the model
The plate has uniform material and thickness, so use area as the weight contribution. Divide the plate into the base and upper rectangles. Both regions belong to the plate, so their areas are positive.
Locate component centres
The base centre is halfway across its width and height. The upper rectangle spans from $x=80$ mm to $x=120$ mm and from $y=40$ mm to $y=120$ mm; its centre is at the midpoint of each interval.
$A_1=120(40)=4800\ \mathrm{mm^2},\quad (x_1,y_1)=(60,20)\ \mathrm{mm};\quad A_2=40(80)=3200\ \mathrm{mm^2},\quad (x_2,y_2)=(100,80)\ \mathrm{mm}$
Combine first moments
Add the area-coordinate products and divide by total area. This gives the point at which the plate's total weight can act while preserving the moments of the two component weights.
$x_G=\frac{4800(60)+3200(100)}{8000}=76\ \mathrm{mm},\quad y_G=\frac{4800(20)+3200(80)}{8000}=44\ \mathrm{mm}$
Verify
The total area is $8000\ \mathrm{mm^2}$ . The two first-moment sums are $608000\ \mathrm{mm^3}$ and $352000\ \mathrm{mm^3}$ . The total weight is the sum of the two component weights, and acting at the calculated centre it preserves both first moments.
$8000(76)=4800(60)+3200(100),\quad 8000(44)=4800(20)+3200(80)$

Answer: The centre of gravity is at

(76\ \mathrm{mm},44\ \mathrm{mm})

from the stated origin.

Check: The result lies between the component-centre coordinates along both axes. It is shifted toward the upper rectangle relative to the base centre.

Worked example

A rectangular plate with a circular hole

A uniform plate is a 160 mm by 100 mm rectangle. A circular hole of radius 20 mm is centred at

(110,55)

mm, measured from the rectangle's lower-left corner. Find the centre of gravity of the remaining plate. Keep

\pi

exact in intermediate expressions and round the coordinates to 0.1 mm.

Assign signed areas
Use positive area for the rectangle and negative area for the removed circle. The rectangle centre is $(80,50)$ mm and the hole centre is given. Apply the negative sign to the hole in both moment sums.
$A_R=160(100)=16000\ \mathrm{mm^2},\quad A_C=-\pi(20)^2=-400\pi\ \mathrm{mm^2}$
Calculate the centroid
Divide each signed first-moment sum by the remaining signed area. The denominator is positive because the rectangle area exceeds the area of the hole.
$x_G=\frac{16000(80)-400\pi(110)}{16000-400\pi}\approx77.4\ \mathrm{mm},\quad y_G=\frac{16000(50)-400\pi(55)}{16000-400\pi}\approx49.6\ \mathrm{mm}$
Verify the moment sums
The remaining area is approximately $14743.4\ \mathrm{mm^2}$ . Before rounding the coordinates, multiplying that area by each coordinate reproduces the rectangle's first moment minus the hole's first moment. The resultant plate weight also equals the remaining area multiplied by the common weight per unit area.
$A_G=16000-400\pi,\quad A_Gx_G=16000(80)-400\pi(110),\quad A_Gy_G=16000(50)-400\pi(55)$

Answer: The centre of gravity of the remaining plate is approximately

(77.4\ \mathrm{mm},49.6\ \mathrm{mm})

.

Check: The hole is right of and above the rectangle centre, so removing it shifts the centre of the remaining plate left and downward.

Worked example

A rectangle combined with a triangular region

A uniform plate consists of a rectangle 100 mm wide and 60 mm high, and a right triangle attached along the rectangle's right edge. The triangle's vertices are

(100,0)

mm,

(160,0)

mm, and

(100,60)

mm. With the rectangle's lower-left corner as origin, find the centre of gravity.

Find areas and centres
The rectangle centre is at its midpoint. For the triangle, average the three vertex coordinates to locate its centroid. Both regions belong to the plate, so both areas are positive.
$A_R=100(60)=6000\ \mathrm{mm^2},\quad (x_R,y_R)=(50,30)\ \mathrm{mm};\quad A_T=\frac{1}{2}(60)(60)=1800\ \mathrm{mm^2},\quad (x_T,y_T)=(120,20)\ \mathrm{mm}$
Combine contributions
Add the area-coordinate products for both regions and divide by the total area. This is an area-weighted average because the plate has uniform material and thickness.
$x_G=\frac{6000(50)+1800(120)}{7800}=66.2\ \mathrm{mm},\quad y_G=\frac{6000(30)+1800(20)}{7800}=27.7\ \mathrm{mm}$
Verify moments
The total area is $7800\ \mathrm{mm^2}$ , and the first-moment totals are $516000\ \mathrm{mm^3}$ and $216000\ \mathrm{mm^3}$ . The total weight is the sum of the component weights, and its moments at the calculated centre match these area-based first moments multiplied by the common weight per unit area.
$7800x_G=6000(50)+1800(120),\quad 7800y_G=6000(30)+1800(20)$

Answer: The centre of gravity is approximately

(66.2\ \mathrm{mm},27.7\ \mathrm{mm})

from the rectangle's lower-left corner.

Check: The triangle adds area to the right and below the rectangle centre. The combined centre is therefore right of and below the rectangle centre.

Common mistakes and how to avoid them

Taking the unweighted average of component-centre coordinates.

Correction: Weight each coordinate by its area or component weight before adding. A larger or heavier component contributes more.

Giving a hole positive area.

Correction: Treat removed material as negative area in the total and in both first-moment sums.

Measuring component centres from different origins.

Correction: Choose one origin for the whole body and express every component centre in that coordinate system.

Using area weighting when components have different densities or thicknesses.

Correction: Use component weights (or masses) when area does not represent relative contribution to the total weight.

Reporting a coordinate in square units.

Correction: A first moment has units of length cubed and area has units of length squared, so their ratio has units of length.

Lesson summary

Divide the body into simple regions and locate each region's centre.
Use area as the weight only for a uniform, equal-thickness plate; otherwise use component weights.
Add signed areas or weights and their first moments, then divide each total moment by the total area or weight.
Verify both moment balances, check units and signs, and judge whether the location is geometrically reasonable.

Check your understanding

Question 1

Two equal-area uniform plate pieces have centres at

x=20

mm and

x=80

mm. What is their combined

x

coordinate?

$20$ mm
$50$ mm
$80$ mm
$60$ mm

Show answer and explanation

50

mm

Equal areas contribute equally, so the combined coordinate is the midpoint:

x_G=(20+80)/2=50

mm.

Question 2

A hole is to the right of a rectangular plate's centre. What sign is assigned to the hole's area in the signed-area method?

Negative
Positive
Zero
The sign depends on its radius

Show answer and explanation

Negative

A hole is removed material, so its area and first-moment contributions are subtracted.

Question 3

A plate has two regions of different density but equal area. Which quantity should be used to combine their centre coordinates?

Their areas alone
Their weights
The distance between their centres
The perimeter of each region

Show answer and explanation

Their weights

Different densities mean equal areas need not have equal weights, so weight is the appropriate contribution.

Key terms

Centre of gravity: The point at which the total weight of a body can be represented as acting.
Centroid: The geometric centre of an area; for a uniform plate of constant thickness and material, it is also the centre of gravity.
First moment of area: An area multiplied by a coordinate distance, used in a centroid calculation.
Composite body: A body that can be divided into simpler parts for calculation.

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8.4 · Find centres of gravity for composite bodies

University of Alberta ENGG 130: Engineering Mechanics: Statics

ENGG 130 study topic 8.4

What you will learn

1. The weighted-position idea

2. Divide the body and locate each centre

3. Moment balance and reasonableness checks

4. A consistent calculation

Two joined rectangular plates

A rectangular plate with a circular hole

A rectangle combined with a triangular region

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