manhattan press (h.k.) ltd. © 2001 hkasl physics examination format and allocation of marks...

Manhattan Press (H.K.) Ltd. © 2001

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%


1.1 Statics 1.1 Statics 靜力學靜力學 and and

Dynamics Dynamics 動力學動力學


What is statics?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.


Zero resultant

•If the block is in equilibrium, the resultant force acting on it is zero.

•Resolve horizontally,

F1 cos = F2

•Resolve vertically,

F1 sin = F3

F1

F2

F3


Vector approach

•If an equilibrium system consists of n forces,

F1

F2

F3

0321 FFF

0..021 in FeiFFF


Vector Addition

Triangular law Parallelogram law

A

C

B

ACBCAB

A

B D

C

ADACAB


•Polygon law

A

B C

D

EF

AFEFDECDBCAB


Zero resultant forces

0..021 in FeiFFF

F1

F2

F3

F4

F5

F6


Turning effect of a force ─ rotate about axes

Pivot (or fulcrum) ─ position of axes

12.1 Turning effect of a force (SB p.170)

axis

axis

axis

axis

pivot

pivot

pivot

pivot


Moment ─ the turning effect of a force

Moment arm ─ perpendicular distance between the force and the pivot

Unit of moment: N m

Moment

＝ Force Moment arm

＝ F d

moment arm

door hinge

Moment12.1 Turning effect of a force (SB p.170)

door hinge (pivot)


Note: the longer d, the smaller will be the force required.

MomentMoment of F1 = F1 d1

Moment of F2 = F2 d2

Same turning effect

F1 d1 = F2 d2

As d1 > d2,

F1 < F2


door hinge


Which requires a smaller force ? (d2 > d1)

case 1 case 2


force

pivot

force


Moment ─ clockwise or anticlockwise


an anticlockwise moment a clockwise moment

pivot

pivot

anticlockwise

clockwise


Principle of moments

Two moments ：same in magnitude, but in

opposite direction

cannot turn

Moment of F1

= 10 0.4

= 4 N m (anticlockwise)

Moment of F2

= 5 0.8 = 4 N m (clockwise)

12.2 Principle of moments (SB p.174)

pivotpivot


pivot

Take the mid-point of the ruler as the pivot

= 1.6 10 0.1 + 1 10 0.4

= 5.6 N m


Total clockwise moment


pivot

Take the mid-point of the ruler as the pivot

= 0.4 10 0.2 + 1.2 10 0.4

= 5.6 N m


Total anticlockwise moment


Principle of momentsclockwise moment

anticlockwise moment

When a body is in balance,

Total clockwise moment

＝ Total anticlockwise moment



Answer

Class Practice 1Class 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 Npivot

d

1 m

1.7 m


Apply the principle of moment,

Clockwise moment = Anticlockwise moment

650 x 1.7 = 500 x 1 + 400 x d

d = 1.51 m


•Find the reaction from the pivot.

400 N 500 N 650 Npivot

d

1 m

1.7 m

R

Since the system is in equilibrium, the resultant is zero.

R = 400 + 500 + 650

= 1550 N


Pivot O20 N 10 N

0.5m 0.2m 0.2m

30 N

R = 60 N

XX Y

•Take moment about XX Clockwise moment

Take moment about other points

= 30 x 0.7 + 10 x 0.9 = 30 Nm

Anti-clockwise moment = 60 x 0.5 = 30 Nm


Pivot O20 N 10 N

0.5m 0.2m 0.2m

30 N

R = 60 N

XX Y

•Take moment about YY Clockwise moment

Take moment about other points

= 60 x 0.5 = 30 Nm

Anti-clockwise moment = 10 x 0.1 + 30 x 0.3 + 20 x 1 = 1 + 9 + 20 = 30 Nm

0.1m


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.


•Two blocks are placed on a light metre ruler as shown. Find the reactions at X and Y.

2 kg3 kg

0.75 m

0.25 m

X Y


•Step 1: Draw all forces acting on the metre ruler (in equilibrium).

2 kg3 kg

0.75 m

0.25 m

X Y

30 N20 N

RX RY


•Step 2: Apply the principle of moment for objects in equilibrium.

2 kg3 kg

0.75 m

0.25 m

X Y

30 N20 N

RX RY

Take moment about X (any point),

20 x 0.25 + 30 x 0.75 = RY x 1

RY = 27.5 N

Resolve vertically,

RX + RY = 20 + 30

RX = 50 – 27.5 = 22.5 N


More on moment of force

Pivot O

dF

Find the moment of F about O.

d sin F

Moment of F about O

= Force x perpendicular distance from O

= F x d sin = Fd sin


More on moment of force

Pivot O

dF

Find the moment of F about O. (Alternative method)

Moment of F about O

= Force x perpendicular distance from O

= F sin x d = Fd sin

F cos

F sin


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.

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 + Ry

2) = 104.4 N

= tan -1(Ry/Rx) = 5.50o

hinge

30o

C

A

20 N

50 N

T T sin 30o

T cos 30o

Rx

Ry

d d

R

Tension is 120 N and the reaction from the hinge is 104.4 N at an angle of 5.50o to the horizontal.


Moment and couple

•Couple ─ consists of 2 equal and opposite parallel forces whose lines of action do not coincide (重疊 ).

torque of couple = F x d/2 + F x d/2 = Fd

F

Fd/2 d/2

F

F

dd


Short test after each chapter

•Homework (SQ 2, 5, LQ 1)

•Next day 1

•Short test : Moment

•Next day 1


The EndThe End

manhattan press (h.k.) ltd. © 2001 hkasl physics examination format and allocation of marks...

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