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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
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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)
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Centre of gravity
1.6 Centre of gravity (SB p. 68)
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.
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Centre of gravity
1.6 Centre of gravity (SB p. 68)
You must not lean against the wall.
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Example 12Example 12
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Centre of mass
1.6 Centre of gravity (SB p. 69)
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
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Determine the centre of gravity of a lamina
1.6 Centre of gravity (SB p. 70)
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.
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Determine the centre of gravity of a lamina
1.6 Centre of gravity (SB p. 70)
2. Mathematical method
Assume: Lamina is made up of n particles
coordinate of each particle
coordinate of CG
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Determine the centre of gravity of a lamina
1.6 Centre of gravity (SB p. 71)
2. Mathematical method
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
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Determine the centre of gravity of a lamina
1.6 Centre of gravity (SB p. 71)
2. Mathematical method
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)(
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Determine the centre of gravity of a lamina
1.6 Centre of gravity (SB p. 71)
2. Mathematical method
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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Example 13Example 13
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Example 14Example 14
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Example 15Example 15
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End
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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
1.6 Centre of gravity (SB p. 68)
Fig. (a) Fig. (b)
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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
1.6 Centre of gravity (SB p. 69)
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
1.6 Centre of gravity (SB p. 71)
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Solution:Solution:
Let G be the centre of gravity of the object,
1.6 Centre of gravity (SB p. 71)
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
1.6 Centre of gravity (SB p. 72)
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Solution:Solution:
1.6 Centre of gravity (SB p. 72)
.,
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
1.6 Centre of gravity (SB p. 73)
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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
1.6 Centre of gravity (SB p. 73)
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