Post on 11-Dec-2015
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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