HKASL Physics
Examination format and allocation of marks
Paper
Format
Weighting
1 (1 hour 10 min.)
Structured-type
questions (4)
33%
MC (25)
30%
Essays (2 out of 4)
22%
TAS
15%
2 (1 hour 50 min.)
3 (2 years)
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1.1 Statics 靜力學
and
Dynamics 動力學
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What is statics?
• Statics is the branch of
physics concerned with the
analysis of loads (force,
torque/moment) on physical
systems in equilibrium.
•When in equilibrium, the
forces in the system will
have zero resultant and zero
turning effect which will not
cause any change in the
motion of the object.
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Zero resultant
F1
q
F2
F3
4
•If the block is in
equilibrium, the
resultant force acting
on it is zero.
•Resolve horizontally,
F1 cos q = F2
•Resolve vertically,
F1 sin q = F3
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Vector approach
F1
q
F2
F1  F2  F3F3 0
•If an equilibrium system consists of n forces,
F1  F2    Fn  0 i.e.
5
F
i
0
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Vector Addition
Triangular law
Parallelogram law
C
A
B
AB  BC  AC
6
B
A
D
C
AB  AC  AD
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•Polygon law
C
B
D
A
F
E
AB  BC  CD  DE  EF  AF
7
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Zero resultant forces
F4
F3
F5
F6
F2
F1
F1  F2    Fn  0 i.e.
8
F
i
0
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12.1 Turning effect of a force (SB p.170)
Turning effect of a force ─ rotate about axes
Pivot (or fulcrum) ─ position of axes
axis
pivot
axis
pivot
pivot
pivot
axis
9
axis
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12.1 Turning effect of a force (SB p.170)
Moment
Moment ─ the turning effect of a force
Moment arm ─ perpendicular distance between
the force and the pivot
moment arm
door
hinge
(pivot)
door
hinge
10
Moment
＝Force  Moment arm
＝F  d
Unit of moment: N m
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12.1 Turning effect of a force (SB p.171)
Moment
Moment
Moment of F1 = F1  d1
Moment of F2 = F2  d2
door
hinge
Same turning effect
F1  d1 = F2  d2
As d1 > d2,
F1 < F2
Note: the longer d, the smaller will be the force required.
11
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12.1 Turning effect of a force (SB p.172)
Moment
Which requires a smaller force ? (d2 > d1)
force
force
pivot
case 1
12
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case 2

12.1 Turning effect of a force (SB p.172)
Moment
Moment ─ clockwise or anticlockwise
clockwise
pivot
pivot
anticlockwise
an anticlockwise moment
13
a clockwise moment
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12.2 Principle of moments (SB p.174)
Principle of moments
pivot
Moment of F1
= 10  0.4
= 4 N m (anticlockwise)
Moment of F2
= 5  0.8 = 4 N m (clockwise)
Two moments：
same in magnitude, but in
opposite direction
cannot turn
14
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12.2 Principle of moments (SB p.175)
Take the mid-point of the ruler as the pivot
Total clockwise moment
= 1.6  10  0.1 + 1  10  0.4
= 5.6 N m
pivot
15
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12.2 Principle of moments (SB p.175)
Take the mid-point of the ruler as the pivot
Total anticlockwise moment
= 0.4  10  0.2 + 1.2  10  0.4
= 5.6 N m
pivot
16
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12.2 Principle of moments (SB p.175)
Principle of moments
anticlockwise
moment
clockwise
moment
When a body is in balance,
Total clockwise moment
＝Total anticlockwise moment
17
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12.2 Principle of moments (SB p.176)
Class Practice 1：Edmond, Jessie and Tracy are sitting
on a seesaw at the positions shown in the figures. Given that
their masses are 65 kg, 40 kg and 50 kg respectively, and the
mass of the seesaw is negligible. Find the distance of Jessie
from the pivot (d) when the seesaw is balanced.
1.7 m
d
1m
pivot
400 N 500 N
650 N
Apply the principle of moment,
Clockwise moment = Anticlockwise moment
650 x 1.7 = 500 x 1 + 400 x d
d = 1.51 m
18
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Answer
•Find the reaction from the pivot.
R
1.7 m
d
1m
pivot
400 N 500 N
650 N
Since the system is in equilibrium, the resultant is zero.
R = 400 + 500 + 650
= 1550 N
19
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Take moment about other points
R = 60 N
0.2m 0.2m
0.5m
Y
X
20 N
10 N
Pivot O
30 N
•Take moment about X
Clockwise moment
= 30 x 0.7 + 10 x 0.9
= 30 Nm
20
Anti-clockwise moment
= 60 x 0.5
= 30 Nm
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Take moment about other points
R = 60 N
0.2m 0.2m
0.5m
0.1m
Y
X
20 N
10 N
Pivot O
30 N
•Take moment about Y
Clockwise moment
= 60 x 0.5
= 30 Nm
21
Anti-clockwise moment
= 10 x 0.1 + 30 x 0.3 + 20 x 1
= 1 + 9 + 20 = 30 Nm
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Conclusion
•When an object is in equilibrium, the
sum of clockwise moments about any
point is equal to the sum of
anticlockwise moments about that point.
22
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•Two blocks are placed on a light metre
ruler as shown. Find the reactions at X
3 kg
and Y.
2 kg
X
Y
0.25 m
0.75 m
23
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•Step 1: Draw all forces acting on the metre ruler
(in equilibrium).
RX
3 kg
2 kg
RY
X
Y
0.25 m 20 N
0.75 m
24
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30 N
•Step 2: Apply the principle of moment for objects in
equilibrium.
3 kg
RX
2 kg
RY
X
Y
0.25 m 20 N
30 N
0.75 m
Take moment about X (any point), Resolve vertically,
20 x 0.25 + 30 x 0.75 = RY x 1
RX + RY = 20 + 30
RY = 27.5 N
RX = 50 – 27.5 = 22.5 N
25
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More on moment of force
Find the moment of F about O.
d sin q
F
q
Pivot O
q
d
F
Moment of F about O
= Force x perpendicular distance from O
= F x d sin q = Fd sin q
26
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More on moment of force
Find the moment of F about O. (Alternative method)
F cos q
Pivot O
q
d
F sin q
F
Moment of F about O
= Force x perpendicular distance from O
= F sin q x d = Fd sin q
27
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Example 2
A sign of mass 5 kg is hung from the end B of a
uniform bar AB of mass 2 kg. The bar is hinged to a wall at A
and held horizontal by a wire joining B to a point C which is
on the wall vertically above A. If angle ABC = 30o, find the
tension in the wire and the reaction exerted from the hinge.
C
Ry
A
hinge
T sin 30o
T
R
q
Rx
30o
T cos 30o
20 N
50 N
d
d
Take moment about A,
(Tsin 30o)(2d) = (20)(d) + (50)(2d)
T = 120 N
Resolve vertically,
Ry + Tsin 30o = 20 + 50
Ry = 10 N
Resolve horizontally,
Rx = T cos 30o
Rx = 103.9 N
R = sqrt(Rx2 + Ry2) = 104.4 N
q = tan -1(Ry/Rx) = 5.50o
Tension is 120 N and the reaction
from the hinge is 104.4 N at an angle
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of 5.50o to
28 the horizontal.
© 2001
Moment and couple
•Couple ─ consists of 2 equal and opposite parallel
forces whose lines of action do not coincide (重疊).
F
F
d
d/2
F
torque of couple = F x d/2 + F x d/2 = Fd
29
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d/2
F
Short test after each chapter
•Homework (SQ 2, 5, LQ 1)
•Next day 1
•Short test : Moment
•Next day 1
30
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The End
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