roghibin's blog equilibrium of rigid bodies kesetimbang an benda tegar
TRANSCRIPT
roghibin's blog
EQUILIBRIUM OF RIGID BODIES
KESETIMBANGAN BENDA
TEGAR
roghibin's blog
Base on object of Equilibrium
Equilibrium Of POINT( Kesetimbangan Titik)
Equilibrium of Rigid Bodies( Kesetm. Benda Tegar)
roghibin's blog
Equilibrium Of POINT( Kesetimbangan Titik)
Point
W
T1T2
Syarat Setimbang:1. Σ Fx = 0
2. Σ Fy = 0
roghibin's blog
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
roghibin's blog
Example
Ditermine tension of each string T1 , T2 and T3 !
5 Kg
T1
T2
T3
60o30o
roghibin's blog
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
roghibin's blog
roghibin's blog
roghibin's blog
roghibin's blog
roghibin's blog
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