頁二第張六第二日七十月十年西己歷夏
WAH KIU YAT PO
報日僑單
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.