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• Centre of gravityCentre of gravity

• • Centre of massCentre of mass

1.6 Centre of gravity1.6 Centre of gravity

• • Determine the centre of Determine the centre of gravity of a laminagravity of a lamina


Centre of gravity

1.6 Centre of gravity (SB p. 68)

The centre of gravity (CG) of an object is the point where the line of action of the weight of the object passes.

move fingers towards each other

centre of gravity = fingers meet (in equilibrium)


Centre of gravity


Bend over with your legs against wall without toppling?

You cannot

Line of action of your weight acts outside the base of your feet and you topple over.


Centre of gravity


You must not lean against the wall.

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Example 12Example 12


Centre of mass


The centre of mass of an object is the point where the whole mass of the object is assumed to be concentrated.

Centre of mass = Centre of gravity

if the object is small or it is in a uniform gravitational field


Determine the centre of gravity of a lamina


1. Experimental method

hung freely at A

at rest, draw vertical line

hung freely at B

at rest, draw vertical line

G is centre of gravity

Line of action of its weight must pass through the centre of gravity.




2. Mathematical method

Assume: Lamina is made up of n particles

coordinate of each particle

coordinate of CG





Taking moments about the y-axis,

Mg( ) = (m1g1)x1 + (m2g2)x2 + (m3g3)x3 + . . . + (mngn)xn

Mg

xgmx

xgm

n

iiii

n

iiii

1

1

)(

)(

x





If g = constant (in a uniform gravitational field), g1 = g2 = g3 = ... = gn = g, then

Similarly, by taking moments about the x-axis,

M

xmx

n

iii

1)(

M

ymy

n

iii

1)(





coordinates of centre of gravity:

M

xmx

n

iii

1)(

M

ymy

n

iii

1)(

Note: These are in fact the coordinates of the centre of mass.

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Q:Q: Fig. (a) shows a person bending over with his back horizontal

to lift a load of 60 N. The spine is considered as a rod pivoted at its base. The weight of the upper part of the body is 250 N acting at the centre of gravity G as shown in Fig. (b). Find(a) the tension T in his back muscle,(b) the horizontal compressional force H acted on the spine, and(c) the vertical reaction V at the pivot.

Solution


Fig. (a) Fig. (b)


Solution:Solution:(a) Since the spine is in equilibrium, sum of the moments about the pivot = 0.

(b) Resultant of the forces along horizontal = 0.

T cos12° − H = 0

H = 1 335 cos12° = 1 306 N

(c) Resultant of forces vertically = 0.

T sin12° + V – 60 – 250 = 0

V = 310 – 1 335 sin12° = 32.4 N


N 335 112sin

32

185

012sin3

22

25060

T

LTLL

It is noted that to lift heavy objects by the above way can strain the muscles in the back. Thus, we should lift objects with our “legs” instead, i.e. bend our keens while keeping the back straight and then pick up the object slowly.

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Q:Q: An object consists of two masses m1 and m2

attached by a light rod of length d as shown in the figure. Find the centre of gravity of the object in

terms of m1, m2 and d.

Solution



Solution:Solution:

Let G be the centre of gravity of the object,


21

2

21

21

0

mmdm

mmdmm

x

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Q:Q: Find the CG of the system of particles as shown in the

figure.

Solution



Solution:Solution:


.,

M

ymy

M

xmx

iii

iii

41

41 isCG theof sCoordinate

41

121110201

)(

41

121010211

)(

3

1

3

1

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Q:Q: The figure shows a metal sheet in the shape of a

square ABCE and an isosceles triangle DEC. If the length of thesquare is 12 cm and the height of the triangle is 9 cm, find the centre of mass of the metal sheet.

Note: Centre of mass of a triangle = 2/3length of median from the apex Solution



Solution:Solution:Since the metal sheet is symmetric about OD, the centre of mass should lie on OD, i.e. x = 6 cm from AE. Centre of mass of the square is at G1, 6 cm from the point O. Centre of mass of the triangle is at G2, (2/3) × 9 = 6 cm from D.

∴ The distance OG2= 12 + 3 = 15 cm

The mass of each portion is directly proportional to its area.

∴ Mass of the square ABCE = 144 units

Mass of the triangle DEC = 54 units

Total mass of the metal sheet = 198 units

If the distance of the centre of mass from O along OD is y, then

198 y = 144 × 6 + 54 × 15

∴ y = 8.45 cm from O


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