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EQUILIBRIUM OF RIGID BODIES

KESETIMBANGAN BENDA

TEGAR

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Base on object of Equilibrium

Equilibrium Of POINT( Kesetimbangan Titik)

Equilibrium of Rigid Bodies( Kesetm. Benda Tegar)

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Equilibrium Of POINT( Kesetimbangan Titik)

Point

W

T1T2

Syarat Setimbang:1. Σ Fx = 0

2. Σ Fy = 0

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Equilibrium Of POINT( Kesetimbangan Titik)

Point

T1

W

T2

αβ T1 cos α

T1 sinα

T2 cos β

T2 sinβΣ Fx = 0

T1 cos α- T2 cos β = 0

Σ Fy = 0

T1 sinα + T2 sinβ – W = 0

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Example

Ditermine tension of each string T1 , T2 and T3 !

5 Kg

T1

T2

T3

60o30o

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Answer

3

3

3

T2

T1

T3

30o60o

T2 cos 30o

T2 sin30o

T3 cos60o

T3 sin60oT1 = W = m.g = 50 N

Σ Fx = 0

T2 cos 30o - T3 cos60o = 0

T2 ½ = T3 ½

T3 = T2

3

Σ Fy = 0

T2 sin30o + T3 sin60o – W = 0

T2 ½ + T3 ½ = 50

T2 + T3 = 1003T2+T2.3=100

4 T2=100

T2= 25 N T3= 25 N

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Shortcut Formula

T1T2

T3

α1 α2

α3

3

3

2

2

1

1

sinsinsin TTT

Notes :

Di kwadran 2 berlaku :

Sin ( 180 – α ) = sin α

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Example

Ditermine tension of each string T1 , T2 and T3 !

5 Kg

T1

T2

T3

60o30o

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Answer

3

3

2

2

1

1

sinsinsin TTT

NT

T

T

TT

o

o

252

250

150sin

2

1

50

sin

2

90sin

1

21

2

5 Kg

T1

T2

T3

60o30o

90o

120o150o

T1 = W = 50 NNT

T

TT

TT

TT

o

oo

325

503

160sin

90sin120sin

1sin

1

3sin

3

3

21

3

13

13

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Moment Of Force

sin

.sin

Fl

lF

l

α

F

porosα

l sinα

F sinα

F sin.lF

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Moment of Force is Vector

+

_

As clockwise

Anti clockwise

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Equilibrium Of Rigid Bodies

Prerequisite Of Equilibrium of Rigid Bodies

0F

0

0 xF

0 yF

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A B

2m

100 kg

60 kgX = ?

To becomes equilibrium condition, so where are B object must be placed from O ? X = ?

O

Example No: 1

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2m

wA = 1000N

WB=600NX = ?

O

N

0xF

0 yF

N – WA – WB =0

N – 1000 -600 =0

N = 1600 N

0

WB.x – WA.2=0

600x – 1000.2 =0

600x = 2000

X = 2000/600

X = 3,3 m

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Dua orang A dan B ingin membawa beban 1200 N dengan menggunakan batang homogen yang masanya dapat diabaikan. Panjang batang 4 meter. Dimanakah beban harus diletakkan ( diukur dari B ) agar B menderita gaya 2 kali dari A.

AB

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A B

1200N

NA NB4m

X=?(4-x)

0xF

0 yF

0

NA+NB-w=0

NA+NB=1200

NA+2NA=1200

3NA = 1200

NA= 400 N

NB= 800 N

NA.(4-X)-NB.X=0

400(4-X)-800X=0

1600-400X-800X=0

1600-1200X=0

1600=1200X X= 1,33 m

NB = 2 NA

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From the following picture, how far C must be placed from B so that equilibrium system !

mA = 80 kg

mB = 30 Kg

mC = 20 kg

AO = 1,5 m

OB = 1,2 m

ACB

X =?

o

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0xF

ACB

X =?

o

N

1,5m 1,2m

800N300N 200N

0 yF

0

N – 800 – 300 – 200 = 0N = 1300 N

Poros O

300.1,2 + 200(1,2+x)-800.1,5 = 0

360+240+200x=1200

600 + 200 x = 1200

200 x = 600

X = 600/200

X = 3 meter

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WEIGHT POINT

W

W4

W3

W2

W1

( Xo, Yo)

( X1, Y1)

( X2, Y2)

( X3, Y3)

( X4, Y4)

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W4

W3

W2

W1

( X1, Y1)

( X2, Y2)

( X3, Y3)

( X4, Y4)

W

( Xo, Yo)

W.Xo = w1.x1+ w2.x2 + w3.x3 + w4.x4

Xo = w1.x1+ w2.x2 + w3.x3 + w4.x4

w1 + w2 + w3 + w4

W = w1 + w2 + w3 + w4

W.Yo = w1.y1+ w2.y2 + w3.y3 + w4.y4

Yo = w1.y1+ w2.y2 + w3.y3 + w4.y4

w1 + w2 + w3 + w4

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If we concern about AREA

Xo = A1.x1+ A2.x2 + A3.x3 + A4.x4

A1 + A2 + A3 + A4

Yo = A1.y1+ A2.y2 + A3.y3 + A4.y4 A1 + A2 + A3 + A4

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If we concern about VOLUME

Xo = V1.x1+ V2.x2 + V3.x3 + V4.x4

V1 + V2 + V3 + V4

Yo = V1.y1+ V2.y2 + V3.y3 + V4.y4 V1 + V2 + V3 + V4

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If we concern about LENGTH

Xo = l1.x1+ l2.x2 + l3.x3 + l4.x4

l1 + l2 + l3 + l4

Yo = l1.y1+ l2.y2 + l3.y3 + l4.y4 l1 + l2 + l3 + l4

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Example : 1

Determine the coordinate of weight point, from following area object !

2 6

3

10

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answer

2 6

3

10

x

y

form A (x , y)

I

II(1,5)

(4,1 ½ )

20

12

1 , 5

4 , 1,5

A1.x1+ A2.x2

A1 + A2Xo =

= 20.1+12.4

20+12

= 68

32

= 2,125

yo =A1.y1+ A2.y2

A1 + A2= 20.5 + 12.1,5

20 + 12

= 100 + 18

32

= 118

32

= 3, 688

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Example : 2

Determine the coordinate of weight point, from following volume object !

2R

2R

2R

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Answer :

2R

2R

2R

x

yform volume (x , y )

2πR3

-2/3πR3

2/3πR3

0, R

0, 3/8R

0, ½ R

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Xo = V1.x1+ V2.x2 + V3.x3

V1 + V2 + V3

XO= 0

Yo = V1.y1+ V2.y2 + V3.y3 V1 + V2 + V3