CHAPTER 3 – FORCE Presentation4a

8/13/2019 CHAPTER 3 FORCE Presentation4a

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Mass Weight

Mass is the quantity of matter in a body.Weight is the force with which a body is attracted

towards the center of the earth by the gravity.

The mass of an object is constant on

Earth and even in space.

The weight of an object can vary from place to

place and becomes zero at the center of the

earth. It is also zero in places that are far away

from earth. It has no weight without gravity,

m = F/ais the mass of a moving body. W = mg, is the weight of a body.

Mass is a scalar quantity. Weight is a vector quantity.

Mass is a base quantity. Weight is a derived quantity.

The unit of mass in the S.I system is

Kilogram (kg). The unit of weight in S.I system is Newton (N)

1. Mass is an intrinsic property of a

body.

2. It is independent of any external

factor.

1. Weight of an object depends on the mass of

an object that is attracting it.

2. Weight is dependent on the force with which

it is attracted.


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3.4.1 NEWTON'S FIRST LAW


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Newtons third law of motion (Hukum Newton III):

FA

FB

A

B

FA(action) = FB(reaction)


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M = 0

Fx = 0, Fy = 0


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8 N8 N

F = 8N8N = 0

( state of equilibrium)

Equal magnitude, opposite direction


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F2= 4 NF1= 4 N

F4= 2 N

F3 = 2 N

Fx = F1F2 = 44 = 0 N

Fy = F3F4 = 22 = 0 N

( state of equilibrium )

Equal magnitude, opposite direction


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F1

Examples: 3

F2

F3

F1

F2

F3


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3.6.3 PARALLELOGRAM LAW.


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P

QO

(a)

P

QO

R

(b)Figure 3.6.3

180 -

The resultant force, R can be calculated by the cosine rule,

__________________________

R = P2+ Q2- 2PQ cos (180)

___________________

R = P2+ Q2+ 2PQ cos )

[because cos (180) = - cos ]


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8 N

12 NO

600

a

b

1 kN

1.5 kN

O

800


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Solution:

12 N

8 N

O

600 R

A

B

Ca)

Magnitude

R = ( P2+ Q2+ 2PQcos

= [ 82 + 122 + 2(8)(12)cos 600]

= 17.4 N

Angle

sin = ( 8 / 17.4 ) sin 600 = 0.398

= sin-10.398 = 23.5

b)


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b)

1 kN

1.5 kNR

800

O

A

B

C

R = [ 1.52 + 12+ (1.5)(1)cos 800]= 1.9 kN

sin = (1.5/1.9) sin 800 = 0.777

= sin-1 0.777

= 510


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A0

B0

C

0

bc

a

FORCE OPPOSITE

ANGLE

a A0

b B0

c C0


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Proof of Lami's Theorem

draw a closed triangle of force.

Draw the straight line, awith arrowThen draw the parallel line band cDetermine position of the angle


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A0

B0 C0

(180B0

A0

(180- C0)

C0

B0

(180A0)

c

bc

a

a

b

After draw the triangle, use simple trigonometry to solve the problem

By the law of sines,

Where

sin (180A) = sin A

sin (180B) = sin B

sin (180C) = sin C)
http://en.wikipedia.org/wiki/Law_of_sineshttp://en.wikipedia.org/wiki/Law_of_sineshttp://en.wikipedia.org/wiki/Law_of_sineshttp://en.wikipedia.org/wiki/Law_of_sines


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Q

135

20N

105

120P

Example 1

FORCE OPPOSITEANGLE

P 135Q 105

20N 120


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B

150

10 N

110

100

A

Solution


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Solution

According to the lamis theoremFORCE OPPOSITEANGLE

A 150B

110

10N 100


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Example 3:

X

600

90

70

50

Y

900

50 N


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FORCE OPPOSITE ANGLE

X 160Y 150

50N 50

According to the lamis theorem

So that;

X

60

90

70

50

Y

90

50 N


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CHAPTER 3 – FORCE Presentation4a