1. Structured-type questions (4)
HKASL Physics
Examination format and allocation of marks
Paper
Format
Weighting
1 (1 hour 10 min.)
Structured-type questions (4)
33%
2 (1 hour 50 min.)
MC (25)
30%
Essays (2 out of 4)
22%
3 (2 years)
TAS
15%
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2. 1.1 Statics 靜力學 and Dynamics 動力學
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3. 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
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 q F2 F3
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Vector approach F1 q F2 F3
If an equilibrium system consists of n forces,
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Vector Addition
Triangular law Parallelogram law
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Polygon law
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Zero resultant forces F4 F3 F5 F6 F2 F1
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9. Turning effect of a force ─ rotate about axes
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
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10. Moment ─ the turning effect of a force
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
Moment
＝Force  Moment arm
＝F  d
door hinge (pivot)
Unit of moment: N m
door hinge
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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
Same turning effect
F1  d1 = F2  d2
As d1 > d2,
　　F1 < F2
door hinge
Note: the longer d, the smaller will be the force required.
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12. Which requires a smaller force ? (d2 > d1)
12.1 Turning effect of a force (SB p.172)
Moment
Which requires a smaller force ? (d2 > d1)
force
force
pivot 
case 1
case 2
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13. Moment ─ clockwise or anticlockwise
12.1 Turning effect of a force (SB p.172)
Moment
Moment ─ clockwise or anticlockwise
an anticlockwise moment
a clockwise moment
pivot
anticlockwise
clockwise
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14. Principle of moments Moment of F1 = 10  0.4 = 4 N m (anticlockwise)
12.2 Principle of moments (SB p.174)
Principle of moments
Moment of F1
= 10  0.4
= 4 N m (anticlockwise)
Moment of F2
= 5  0.8 = 4 N m (clockwise)
pivot
Two moments：
same in magnitude, but in opposite direction
cannot turn
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15. Take the mid-point of the ruler as the pivot
12.2 Principle of moments (SB p.175)
Take the mid-point of the ruler as the pivot
= 1.6  10   10  0.4
= 5.6 N m
Total clockwise moment
pivot
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16. Take the mid-point of the ruler as the pivot
12.2 Principle of moments (SB p.175)
Take the mid-point of the ruler as the pivot
= 0.4  10   10  0.4
= 5.6 N m
Total anticlockwise moment
pivot
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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
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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.
400 N
500 N
650 N
pivot d
1 m
1.7 m
Apply the principle of moment,
Clockwise moment = Anticlockwise moment
650 x 1.7 = 500 x x d
d = 1.51 m
Answer
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Find the reaction from the pivot. R
400 N
500 N
650 N
pivot d
1 m
1.7 m
Since the system is in equilibrium, the resultant is zero.
R =
= 1550 N
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20. Take moment about other points
R = 60 N
0.5m
0.2m
0.2m X Y
10 N
20 N
Pivot O
30 N
Take moment about X
Clockwise moment
Anti-clockwise moment
= 30 x x 0.9
= 30 Nm
= 60 x 0.5
= 30 Nm
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21. Take moment about other points
R = 60 N
0.5m
0.2m
0.2m
0.1m X Y
10 N
20 N
Pivot O
30 N
Take moment about Y
Clockwise moment
Anti-clockwise moment
= 60 x 0.5
= 30 Nm
= 10 x x x 1
= = 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.
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Two blocks are placed on a light metre ruler as shown. Find the reactions at X and Y.
3 kg
2 kg X Y
0.25 m
0.75 m
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Step 1: Draw all forces acting on the metre ruler (in equilibrium).
3 kg RX
2 kg RY X Y
20 N
30 N
0.25 m
0.75 m
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Step 2: Apply the principle of moment for objects in equilibrium.
3 kg RX
2 kg RY X Y
20 N
30 N
0.25 m
0.75 m
Take moment about X (any point),
20 x x 0.75 = RY x 1
RY = 27.5 N
Resolve vertically,
RX + RY =
RX = 50 – 27.5 = 22.5 N
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More on moment of force
Find the moment of F about O. q F
d sin 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
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More on moment of force
Find the moment of F about O. (Alternative method)
F cos q
F sin q
Pivot O q d F
Moment of F about O
= Force x perpendicular distance from O
= F sin q x d = Fd sin q
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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
Take moment about A,
(Tsin 30o)(2d) = (20)(d) + (50)(2d)
T = 120 N
Resolve vertically,
Ry + Tsin 30o =
Ry = 10 N
Resolve horizontally,
Rx = T cos 30o
Rx = N
R = sqrt(Rx2 + Ry2) = N
q = tan -1(Ry/Rx) = 5.50o T
T sin 30o
T cos 30o Ry q R
30o A Rx d
hinge
20 N
50 N
Tension is 120 N and the reaction from the hinge is N at an angle of 5.50o to the horizontal.
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29. 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 d F
torque of couple = F x d/2 + F x d/2 = Fd
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30. Short test after each chapter
Homework (SQ 2, 5, LQ 1)
Next day 1
Short test : Moment
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The End
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