1969-11-12 — Page 21

頁一第張六第日三初月十年酉己屣夏 WAH KIU YAT PO

EXO英文中學會考試題預習專欄

A force causes a boay to have a motion, while a torque causes a body to have rotary motion.

・道英文書院主編

A torque s

三期星日二十月一十年九六九一曆公年八十五國民華中 育教郓

育敎僑華:

物理科(二)

PHYSICS (2)

FORCE IN EQUILIBRIUM·

CONCURRENT FORCES IN SQUILIBRIUME

If forces act on the same point, or if their lines of action intersect at the same common point, the forces are said to be concurrent. The resultant force of these forcer determines the motion of the body. It may cause a translatory notion --motion straight line, or a rotary motion. It may also experiences no motion at all. Such a situation is known as EQUILIBRIUM.

FIRST CONDITION OF EQUILIBRIUM

ZOTO.

The resultant force acting on a body must be

This also means that the suz-of the I-components must be zero and the sum of the y-components must allo be zero for a body at rest or moving at a constant rate.

If the forces are concurrent, this is the only condition that must be me

FROBLEM PROCEDURE··

Any problem in statice" involving concurrent forces conbe solved simply by applying informatione concerning vectors addition in previous lesson. Certain logical steps should be followed.

1. The problem should be read carefully and a clear

and simple diagram should be drawn to visualize all the explicit and inplicit factors.

2. The body, or portion of the body which is an

equilibrium should be isolated for a detailed: consideration of the forces acting on it. If the forces are concurrent and the body is in equilibrium, then the resultant force is zero, This information helps to evaluate one or more of the unknown forces

3. Indicate by arrows all the forces that act on une

isolated body but not the forces exerted by the body on something else. D

4. Treat these arrows as a vector diagram. Evaluate and tabulate the x- and y- components of all the forces acting on the isolated body.

Foros

·I-component

the product of force and moment.

SECOND CONDITION OF EQUILIBRIUM

If a body is in equilibrium, not only must the forces acting on it add up to zero but also all torques exerted by these forces with respect to any axis whatever must also add up to zero. This 18 known as the second condition of equilibrium?

L-O

where L stands for torque.

PROBLEM PROCEDURES

If the

problem involves nonconcurrent forces, step 71 Summation of all torques about any one axis aquale to zero, must be introduced. Forning a third simultaneous equation and capable to solve thres

Example 2-2.

An advertising sign is suspended as shown. Its weight is 70 1b.wt, and the centre of gravity is assumed to be at the centre of the sign. What is the tension in the supporting cable? Calculate also the vertical and horizontal thrusts on the pivot F.:

Problem Procedure planning

Step 1 Step 2:

Step 31

Diagram.

Isolate the body.

As the diagram may be confusing, separate diagram is made.

Step 4 Indicate forces on it.

They are T, V, H, and W. Consider the vector diegram, Tabulate the force components and

torques.

Step 5

Step 61

Equate

Step 7. Solve the three simultaneous equations.

Solution

45

proper vectors, (b) Determine the forcs by the wall on the pole.

exerted

A uniform trap door in a floor is 4 feet square and is hinged along one edge. If the door weighe 100 1b. and is opened upward by a vertical fores so that it makes an angle of 60 degrees with the horizontal, calculate the vertical and horizontal forces on the hinge.

EXOTCIE

Solution

1. By graphical method,

Scale: 2 on, represents 10 lb.wt.

Vectors A and B are at rt. angle to each other,

A rectangle is constructed with A and B as adjacent sides each represent the two vectors in magnitude and direction. The diagonal GD of the rectangle is measured which is 7.2 cm. The resultant vector is.

72 x 10- 36 lb.wt. 7.203

By calculation:

Resultant in OD

tan

30

400 + 900

OD

W1300

10/13 36 lb.wt.

56°

The resultENT 18

making an ang)

360 with A.

F2008

y-component

Fin

sin

Fzein ez

up the components listed in the table to express the resultant whose value is known to be Zero,

Σ

- Fisin

¿sin 1⁄2 + Fzsin

6. By using the above two simultaneous equations, twÒ

unknown forces may be solved.

Example 2-1

A. 5 1b weight suspended from a 3 ft. cord is pulled to one side by a horizontal force F so that the cord makes an angle of 30 degrees with the vertical. What is the value of the force F, and what is the tension in the cord?

Plan out the problem procedure

Step 1.

Diagram

Step 2. Isolate the body

W

Fores

Tension T

Fx

Toos 45°

Ty

Tsin -0.

T

100

-70

torques about P

4 x 4 sin 45

70.x 2

H: Horizontal component of the reaction

Vi Vertical component of the reaction at P Ws Weight of the sign

For sunmation F

For summation

Toos 45

(1)

Tin

70.

(11)

For summation

4 sin

(111)

From (111)

140

Step 3.

Indicate forces on the body; These are T, F, and w.

4 x sin 45°

Step 41

Stop 5

Step 6.

Solve the similtaneous equation

Consider the vector diagram, Construct table of force components. Equate FO and F0

49.5 lb.

From (11)

From

70

35-35 1b.wt

Tcos 45%

The foros 2 1b.wt. can be resolved into two components namely F and F

- 7 sin 60°

The total force acting on the floor by the blook

F sin601b.wt.

where W 28: the weight of the block.

A reaction force exactly the same magnitude as (W + F sin 60°) lb.wt., exerts by the floor on the block. Therefore, the resultant vertical fores acting on the block is ZERO.

The answer is not Fein60, which will probably be mistakes as the resultant vertical force. (P sim 60o) is only the y-component of F

Solution

Foros

[~Fao#60 | +Tsin60:

Fx Fy

For

F

ེ

quilibrium,

↑ sin 66*

-5

0 ...(2)

From From

Ane

F-0.57

5/0.866

1-15.7.7

F2.89

The tension of the oord is 5.77 10.w

The foron is 2.84 lb.wt.

NONCONCURRENT FORCES AND TORQUES

If the forces acting on a body are non- concurrent, the sum of forces may be zero and cause no translational motion, but the body may not be in equilibrium. A rotary motion mav nocur.

140/4 35 1b.wt.

Ans. The tension in the supporting cable in 49.5 lb.

wt. The vertical and horizontal thrusts on the pivot P, each equals to 35 lb.wts

Exercise 2

1. A 60 lb weight hangs from the end of a horizontal strat protruding 8 ft, from a vertical wall. A tis rod connects the outer end of the strut tova point in the wall 4 fået above the strus. Find tha thrust in the strut and; the tension in the tie rod.

A ploture is supported on a wall by a cord connaoted between two eyelets, one on either side of the picture, and hanging over a pag. At the peg, the cord on either side makes an angle of 30 degrees with the horizontal. What is the

tension in the cord if the picture weighe

A bamboo pole 20 ft. long and weighing 32 lb. rest against a frictionless wall with the foot or the pols 12 ft. from the base of the wall and the: top of the 'pole 16 ft above the ground, The centre of gravity is 8 feet from the lower end, (a) Draw a careful diagram of the pole, representing all forces noting on it by the

8 cas

BC10 newtong.

n6o®

Forse B resolves into 2 components

Boos60

B sin 60

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