1969-11-26 — Page 22

頁二第張六第二日七十月十年西己歷夏

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三期星日六廿月一十年九六九一曆公年八十五國民華中 育教僑華

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Force

Reaction R

T_Cos30°| T29bn30°

Since the body is in equilibrium

1770英文中學會考試題預習專欄

坚道英文書院主編

cos 30 - T2cos 30%

Therefore,

Solution

物理科

(四)

PHYSICS (4)

27 win30

Exercise 2

1. Problem Procedure Planning.

Step 1

Diagram

Step 2

Isolate the body

Step 3

Indicate forces on it

Step 4

Consider the vector diagram and

tabulate the forget

forques

Step 5 Equate:

and

(Not necessary as all force

are concurrent)

Step 6 Solve the 3 simultaneous equations.

The tension in the cord is also 4 ibiwti

If the problem is solved graphically, an equilateral triangle is drawn.

3. Problem procedure as before. Forces are no

concurrent, the second condition of equil is introduced.

Solution

There is no horizontal forces acting on the trapdoor. The horizontal component on the hinge: is therefore equal to zero,

The weight of the door is mid-way between the tension of the suspending cord and the vertical component of the reaction on the hinge.]

Exercise

100 50.1b,w

(1) Distance covered - 10 miles

Time taken ■ 20 min.

hr.

Solution:

Isolated

TLOSO

Force Tension

Fx

cose

Height

sing

·60

60 lb. vit.

ten

R reaction of the wallon

the bamboo pole

V vertical component of

the reaction of the ground on the pole

He horizontal component of

the reaction of the ground on the pole

Average velocity

30 m.p.h."

h

(ii) The statement "he accelerates his car

uniformly from 30 m.p.h. to 60 m.p.h, in 101 seconds" indicates that the speed is not constant for the whole distance..

(iii) The acceleration in the 10 sec interval is

(88 (44)

= 4.4 ft./sec.

2

For the first condition of equilibrium

(iv) Distance covered in this 10-sac interval

Sut + at

- 44 x 10

440 +

* 4.4 x 10

Reaction

OT

Thrust R

Q:

All units in the table are in lb.w

sin 60

ein260341

137 1b.wt.

137 cos26

120-1b.wt.

(1)

The problem can be solved graphically as follows: Triangle ABC represents the force diagram.

AB, 4 ft, represents 60 lb.wt.

BC, 8

represents R 1b.ut.

2ft, represents 7 lb.wt.

R 60 x

120 1b.wt

W(82

T = 60 x

For the second condition of equilibrium

ΣΤ

= 0

Since only the magnitude of R. is required in this problem, we take moments at

ΣΤΟ

32

12 2 x 8

·20

x 12 x 8 20

1.2

9.6.

Ans; The reaction of the wall is 3.6 lb.wt, Graphical solution suggusted

The lines of action of the weight of the pole, reaction on the wall and the reaction on the ground will be concurrent when produced.

Scale 1 unit: 8 lb.wt.

660 ft.

Neglecting the time taken for sound to travel up the well, the splash heard threa seconda accounts for the flight of stone only. The stone accelerates with g - 32 ft/sec/680

Distance of water loyal below

ut + at

x 32 x

* 16 x 95

144 ft.

The water level is 144 ft. below the point of release.

The block slides down a frictionless plane

undergoes a uniform acceleration of g sine; where is the angle of inclination of the plane. With the item. given by any one of the time intervals in the problem, the magnitude of the acoeleration

can be determined. Take the first second time-interval

ut + bat?

ដ.ន

0 + 2a = 10

20 cm/sec/880

Or, as analysed by the table below

·wnFIqTTInbə jo unŢarpucp sat u 15 x No80

134 1b.wt.

Froblem 'Precedure as befor

Solution

time interval 1st sec 2nd sec 3rd sec 4thsec 5thseo

Tsin30"

The wall is smooth. The only reaction is perpendicular to the wall. The weight 32 lb.wt. sots on the centre of gravity. Both of these lines of force when produced will meet at A. Join AC, a force triangle ABC is formed. AB represents 32 1b.wt. and measures to be 4 units. BC measures 1.2 units and represente

1.2 x 8

9.5/1b.ut

V(cm/sec)

initial vél.

u(om/sec)

final vel

0:

2a

За 4a.

2a

За

·49. За

10

30

50

70 90%

acceleration

a(om/sec2)

20:

·20%

20

20. 20

distance,

s (om):

All the above is calculated by the equations of motion

Veut at

Sa ut: +

It is observed the acceleration is uniform and is 20 cm/sec2.