1. The document discusses the equilibrium of coplanar, concurrent, parallel, and nonconcurrent force systems. 2. It provides the conditions for equilibrium as the sum of all forces equaling zero and the sum of all moments equaling zero. 3. Methods for analyzing equilibrium include the force triangle method and the method of components. Free body diagrams are drawn to represent the different force systems.

1. T.Chhay
lMnwgénRbBn§½kMlaMgkñúgbøg;
Equilibrium of Coplanar force system
1> lkç½xNнlMnwg Condition of equilibrium
RbBn§½kMlaMgmYy CaRbBn§½kMlaMgmanlMnwgeTA)anluHRtaNaEt kMlaMgpÁÜbénRbBn§½kMlaMgenaHesμIsUnü.
kMlaMgpÁÜbénRbBn§½kMlaMgesμIsUnü RtUveKarBnUvlkçxNнTUeTA 2 KW³
- plbUkBiCKNiténRKb;kMlaMgTaMgGs; b¤bgÁúMénkMlaMgTaMgGs; tamGkS½NamYy b¤tamTisedANamYy
RtUvEtesμIsUnü.
- plbUknBVnþénm:Um:g;énkMlaMgTaMgGs;tamGkS½NamYy b¤Rtg;cMnucNamYyRtUvEtesμIsUnü. b¤eKGacni-
yaymü:ageTotfa plbUknBVnþénm:Um:g;énkMlaMgEdlmanTisRsbRTnicnaLika mantMélesμInwg pl
bUknBVnþénm:Um:g;énkMlaMgEdlmanTisRcasRTnicnaLika.
lkçxNнlMnwg GacsresrCaBiCKNitdUcxageRkam³
∑ F = 0 nig ∑ M = 0
edaysarkñúgRbBn§½kUGredaen manGkS½BIrKW X − Y dUcenHplbUkkMlaMgtamGkS½RtUv)anbMEbkCaBIrKW³
∑ F = 0 nig ∑ F = 0
X Y
mü:ageTotedayGkS½Gab;sIus X manTisedk Horizontal nigGkS½Gredaen Y manTisQr Vertical
dUcenHeBlxøHplbUkkMlaMgtamGkS½TaMgBIrGacsresrmanragCa³
∑ F = 0 nig ∑ F = 0
H V
2> düaRkamGgÁesrI b¤düaRkambnÞúk b¤ düaRkam FBD (Free-body diagram)
düaRkam FBD CaKMnUsbgðajBI³
- GgÁÉeka Edldac;edayElkBIGgÁd¾éT
- kMlaMgxageRkAEdlmanGMeBImkelIGgÁenaH Edl)anBIlT§plénGnþrGMeBIrvagGgÁesrI nigGgÁdéT.
viFIsaRsþkñúgkarKUdüaRkam FBD
- CaTUeTArcnasm<n§½mYyekIteLIgBIbgÁúMénGgÁCaeRcIn. KUsbgÁúMénGgÁTaMgenaHdac;edayELkBIKña.
- KUsbgðajnUvkMlaMgEdlCasmμtikmμ EdlmanGMeBImkelIGgÁenaH edaybgðajBIGaMgtg;sIuet ExSskmμ
nigTisedAénkMlaMgenaH.
- KUsbgðajnUvral;kMlaMgEdlCaGBaØti. CaTUeTAeKsÁal;TisénkMlaMg ÉTisedArbs;vaRtUv)aneKsnμt;.
lMnwgénRbBn§½kMlaMgkñúgbøg; 13

2. T.Chhay
rUbTI1 bgðajBIkUnCBa¢IgBüÜredayExS. sMrab;rcnasm<n§½enHpÁMúeLIgedayGgÁBIr KW ExS nigkUnCBa¢Ig.
düaRkamGgÁesrI)anbgðaj eGayeKeXIjBIkMlaMgsmμtikmμ nigkMlaMgGBaØtiEdlmanGMeBImkelIGgÁesrIenaH.
rUbTI2 bgðajBInimitþsBaØaxøHénRbePTTMrrbs;rcnasm<n§½ nigdüaRkam FBD EdlRtUvKñanwgRbePTTMr
enaH.
lMnwgénRbBn§½kMlaMgkñúgbøg; 14

3. T.Chhay
3> lMnwgénRbBn§½kMlaMgRbsBV Equilibrium of concurrent force system
edIm,IeGayRbBn§½kMlaMgRbsBVkñúgbøg;manlMnwg luHRtaEtplbUkBiCKNiténbgÁúMkMlaMgtamGkS½edk
¬Gab;sIus X ¦ nigGkS½bBaÄr ¬Gredaen Y ¦ mantMélesμIsUnü. Rcasmkvij RbsinebIeKRsaybBa¢ak;eXIj
fa plbUkBiCKNiténbgÁúMkMlaMgtamGkS½edk ¬Gab;sIus X ¦ nigGkS½bBaÄr ¬Gredaen Y ¦ mantMélesμIsUnü
enaH)ann½yfaRbBn§½kMlaMgRbsBVkñúgbøg;manlMnwg.
∑ F = 0 nig ∑ F = 0
Y X
enHedaysarEt RbBn§½kMlaMgRbsBVmanGMeBIrelIGgÁFatumYy Rtg;cMnucmYy EdlCaehtuminbNþal
eGayGgÁFatuenaHmanrgVileT. dUcenH smIkarBIrRKb;RKan;kñúgkaredaHRsayrkkMlaMgEdlCaGBaØti.
]TahrN_³ kUnTMgn;mYyman TMgn; 100 N RtUvBüÜredaydgsÞÚc (boom) dUcbgðajkñúgrUbTI3. kMNt;kMlaMg C
EdlmanenAkñúgdgsÞÚc nigkMlaMgtMnwgExSkab T .
dMeNaHRsay³
eKmanBIrviFIkñúgkaredaHRsaylMhat;enH.
- viFIkMlaMgRtIekaN (The force triangle method)
tamrUbTI3>ckMlaMgTaMgbIbegáIt)anCaRtIekaNmYy.
tamRTwsþIbTsIunUs eK)an
lMnwgénRbBn§½kMlaMgkñúgbøg; 15

4. T.Chhay
100 T C
= =
sin 80o sin 40o sin 60o
dUcenH T=
100
sin 80 o
sin 40o = 65.3N
nig C=
100
sin 80o
sin 60o = 87.9 N
- viFIbgÁúMkMlaMg (The method of components)
edIm,IeGaycMnuc Q manlMnwgluHRtaEt
∑ FX = 0 ⇒ −TH + C H = 0
⇒ −T cos 30o + C sin 40o = 0
sin 40o
⇒T = C
cos 30o
nig ∑ F Y = 0 ⇒ +TV + CV − 100 = 0
⇒ T sin 30o + C cos 40o − 100 = 0
sin 40o
⇒ o
sin 30o C + C cos 40o − 100 = 0
cos 30
100 cos 30o
C= = +87.9 N
(
cos 40o − 30o )
o
sin 40
⇒T = 87.9 = +65.3 N
cos 30o
sBaØabUk bBa¢ak;faTisedAsnμt;RtwmRtUv.
4> lMnwgénRbBn§½kMlaMgRsb Equilibrium of parallel force systems
RbBn§½kMlaMgRsbmYymanlMnwg luHRtaEtvaeKarBnUvlkçxNÐxageRkam³
- plbUkBiCKNiténRbBn§½kMlaMgenaHesμIsUnü ∑ F = 0
- plbUkBiCKNiténm:Um:g;énRbBn§½kMlaMgenaHeFobcMnucNamYykñúgbøg;esμIsUnü ∑ M = 0
]TahrN_³
FñwmmYyrgnUvbnÞúkcMcMnucbBaÄrdUcbgðajkñúgrUbTI4. FñwmenHRTedayTMr pin enAcMnuc A nigRT edayTMr
roller elIépÞedkenAcMnuc B. FñwmEdrRTedayTMrEbbenHRtUv)aneKeGayeQμaHfa FñwmsamBaØ (simple beam)
ÉTMrRtUv)aneKehAfa TMrsamBaØ (simple supports). RbtikmμTMrRtUv)aneKsnμt;manTisRsbnwgTisrbs;kMlaMg.
kMNt;RbtikmμTMrnImYy² edayecalbnÞúkpÞal;rbs;Fñwm.
lMnwgénRbBn§½kMlaMgkñúgbøg; 16

5. T.Chhay
dMeNaHRsay³
kMlaMgRbtikmμRtg; B RtUv)ankMNt;edayeRbIplbUkBiCKNiténm:Um:g;énRbBn§½kMlaMgRsbRtg;cMnuc A
∑ M A = + RB × 12 − 50 × 2 − 200 × 5 − 80 × 9 − 150 × 11 = 0
⇒ RB = +289.2 N
kMlaMgRbtikmμRtg; A RtUv)ankMNt;edayeRbIplbUkBiCKNiténm:Um:g;énRbBn§½kMlaMgRsbRtg;cMnuc B
∑ M B = − RAV × 12 + 150 × 1 + 80 × 3 + 200 × 7 + 50 × 10 = 0
⇒ RAV = +190.8 N
epÞógpÞat;
∑ FV = +190.8 + 289.2 − 50 − 200 − 80 − 150 = 0 OK!
5> lMnwgénRbBn§½kMlaMgminRbsBV Equilibrium of nonconcurrent force systems
RbBn§½kMlaMgminRbsBVkñúgbøg;mYymanlMng luHRtaEtvaeKarBnUvlkçxNÐxageRkam³
w
- plbUkBiCKNiténRbBn§½kMlaMgtamGkS½ X enaHesμIsUnü ∑ F = 0 X
- plbUkBiCKNiténRbBn§½kMlaMgtamGkS½ Y enaHesμIsUnü ∑ F = 0 Y
- plbUkBiCKNiténm:Um:g;énRbBn§½kMlaMgenaHeFobcMnucNamYykñúgbøg;esμIsUnü ∑ M = 0
lMnwgénRbBn§½kMlaMgkñúgbøg; 17

6. T.Chhay
]TahrN_³
kMNt;RbtikmμTMr A nig B rbs;eRKagdMbUldUcbgðajkñúgrUbTI5. vamanTMr roller enAcMnuc A nigmanTMr pin
enAcMnuc B . 200N
dMeNaHRsay³
100N 300N
100N 90°
2m
A B
2m 6m
2 2m
200N 300N 400N
(a ) R o o f tru ss
200N
600
100N 5 300N
2
1
100N 300
2m
5
2
1
RBH
2m 2m 2m 2m
200N 300N 400N
RA R BV
(b ) F re e-b o d y d iag ram
rUb TI5
plbUkBiCKNiténkMlaMgtamGkS½edk
⎛ 1 ⎞
∑ FX = RBH − 100 − 100 − 300⎜ ⎟=0
⎝ 5⎠
⇒ RBH = +334.2 N
kMlaMgRbtikmμRtg; A RtUv)ankMNt;edayeRbIplbUkBiCKNiténm:Um:g;énRbBn§½kMlaMgRsbRtg;cMnuc B
600 300
∑ M B = − RA × 8 + 200 × 6 + 300 × 4 + 400 × 2 + 200 × 4 + 100 × 1 + 100 × 2 + ×2+ ×1 = 0
5 5
⇒ RA = +621.4 N
kMlaMgRbtikmμRtg; B RtUv)ankMNt;edayeRbIplbUkBiCKNiténm:Um:g;énRbBn§½kMlaMgRsbRtg;cMnuc A
600 300
∑ M A = + RBV × 8 − 200 × 2 − 300 × 4 − 400 × 6 − 200 × 4 − 100 × 1 − 100 × 2 − ×6− ×1 = 0
5 5
⇒ RBV = +747 N
lMnwgénRbBn§½kMlaMgkñúgbøg; 18

7. T.Chhay
epÞógpÞat;
600
∑ FV = +621.4 + 747 − 200 − 300 − 400 − 200 − = 0 OK!
5
lMnwgénRbBn§½kMlaMgkñúgbøg; 19

8. T.Chhay
lMhat;
edaHRsayRbtikmμTMr
1> 1300N
A B
2m 1m
2> 300N.m
A B
6m 4m
3> 135N
60°
A B
3m 1m
4> w=2kN/m
A B
4m
5> 10N
2kN/m
45°
A B
2m 1m 1m
6> 2kN/m 3N.m
A B
2m 1m 1m
7> 40N
10N/m
A B
2m 3m 4m
8> 60N
5N/m 120°
A D
B C E
1m 3m 2m 2m
lMnwgénRbBn§½kMlaMgkñúgbøg; 20

9. T.Chhay
9> w
A
l
10> P
A
l
11> m
A
l
12> 30kN
5kN/m
60°
A
1m 1m 1m
13> 15kN.m 15kN/m
A
1m 1m 1m
14> 30kN
15kN.m
60°
A
1m 1m 1m
15> 5N/m
50N 2N/m 5N.m
30°
A B
3m 2m 3m 4m 2m
16> 5N/m
35N
3N/m
60°
A B
3m 2m 2m 3m 2m
lMnwgénRbBn§½kMlaMgkñúgbøg; 21

10. T.Chhay
17> 2N/m
B
2.5m
6N
2.5m
A
4m
18> 3N/m 5N.m
3m
15N
2m
A
B
4m 1.5m
19>
4N/m
3m
4m
20>
4N/m
3m
4m
lMnwgénRbBn§½kMlaMgkñúgbøg; 22