1. DYNAMIC FORCE ANALYSIS
WEEKS-2,3

2. MASS RELATIONS Location of Center of Mass: y m1 m2 G G G x O x1 x2 xnmn O z R
x y
m m mn x O x1 G x2 xn R1 G R2 Rn x dm G

3. MASS RELATIONS Mass Moments and Products of Inertia:
Mass Moments of Inertia Mass Products of Inertia
Inertia Tensor
If there is symmetry wrt three axes: Ixy= Ixz= Iyz=0
(In this case, Ixx, Iyy, Izz are principal moments of inertia)
If there is symmetry wrt two axes: Two of Ixy, Ixz, Iyz=0
If there is symmetry wrt one axis : One of Ixy, Ixz, Iyz=0
Unit: massXdistance2

4. Inertia Forces and D’alembert’s Principle  aG h aG  F G G
FG
MG F2
Newton’s Law:
F=maG (F=FG )
MG=Fh=IG
(h=IG/maG)
h is selected according to the sense of 

5.  (F=FG ) MG D’alembert’s Principle:Fa F  
FGa h aG  aG G G
FG
(F=FG )
TGa
MG
D’alembert’s Principle:
F- maG = 0 F+ FGa = 0  F= (Inertia Force: Fa = FGa = - maG ) (A fictitious force)
MG-IG = 0  MG+TGa =0  M=0 (Inertia Torque: TGa = - IG ) (A fictitious torque)

6. EXAMPLE (Analytical Solution)G3 A
G2,O2
aG3 O4
aG4 FC C B G4
5,1°
18,7°
73°
127° 4 3 2
115° ω2 3 4
AO2=60 mm O4O2=100 mm
BA=220 mm BO4= 150 mm
CO4=CB=120 mm G3A= 90 mm
G4O4=90 mm m3=1.5 kg
m4=5 kg IG2=0.025 kg.m2
IG3=0.012 kg.m2 IG4=0.054 kg.m2
2=0 3=-119k r/s2
4=-625k r/s2 FC=-0.8j N
aG3= 162
aG4= ° m/s2
-73.2 ° m/s2
G4O4C = 20.4°
G4O4B = 36°
G3AB = 30° G

7. FREE BODY DIAGRAMS FaG3=-m3aG3 TaG3=-IG33 4 unknowns G3 3 equations
G2,O2
F12 2
F32
M12 A
F14
F34 FC C O4 G4
FaG4 =-m4aG4
TaG4=-IG44
5 unknowns
3 equations
4 unknowns
3 equations
Total:
9 unkowns
9 equations
A matrix solution can be performed in the computer.
Practically, the most suitable way is to make a coupled
solution of links 3 and 4, for F34 and F43 . Therefore,
the solution is reduced to 2 equations with 2 unknowns.

8. MOMENT AT LINK 4 WRT POINT O4
F14
F34 FC C O4 G4
FaG4 =-m4aG4
TaG4=-IG44
MOMENT AT LINK 4 WRT POINT O4
ΣMO4=0
ΣMO4= O4G4 x FaG4 + TaG4 + O4C x FC + O4B x F34= 0
ΣMO4=0.09[cos( )i + sin25.5j] x 5(104)[cos( )i+sin 53j] + (0.054)(625k)
413°=53°
+0.12(cos 5.1i + sin 5.1j) x (-800j)
+0.15[cos( )i + sin61.5j] x(F34xi+F34y j)=0
-0.125F34x F34y = 37 (1)

9. MOMENT AT LINK 3 WRT POINT A
F23
F43 A B G3
FaG3=-m3aG3
TaG3=-IG33 3
MOMENT AT LINK 3 WRT POINT A
ΣMA=0
ΣMA = AG3 x FaG3 + TaG3 + AB x F43 =0
ΣMA = 0.09[cos( )i +sin48.7j] x
1.5 (162) [cos( )i+sin(106.8)j]
+(0.012)(119k)
+0.22(cos18.7i+sin18.7j) x (-F34xi-F34yj)=0
0.0705F34x-0.208F34y=-20 (2)

10. Finding Forces F34 and F14 -0.125F34x + 0.083F34y = 37FC C O4 G4
FaG4 =-m4aG4
TaG4=-IG44
Finding Forces F34 and F14
-0.125F34x F34y = 37
0.0705F34x-0.208F34y=-20
F34x= F34y=-5,39 F34=-300i-5.39j N
Force Balance at Link 4
ΣF= F14+F34+FC+FaG4=0
F14-300i-5.39j-800j+5(104)(cos53i+sin53j)=0
F14=-13i+ 390j N

11. Finding Force F23 Force Balance at Link 3 ΣF=0 F23+F43+FaG3=0
F23+300i+5.39j+1.5(162)(cos106.8i+sin106.8j)=0
F23=-230i-238j N
F23
F43 A B G3
FaG3=-m3aG3
TaG3=-IG33 3

12. Finding Force F12 and Moment M12
Force Balance at Link 2
ΣF= F12+F32=0
F12=-F32=F23=-230i-238j N
Moment Balance at Link 2
ΣMO2= O2A x F32 + M12 + TaG2=0
0.06(cos115i+sin115j)x(230i+238j)+M12k+0=0
M12=18.6 N.m M12=18.6 k N.m
G2,O2
F12 2
F32
M12 A

13. EXAMPLE (Graphical Solution)
F34O4B B
F14 FC C O4 G4
FaG4 =-m4aG4
TaG4=-IG44
MO4=0
-(F34O4B)O4B-FC(O4C)x - (FG4)ax (O4G4)y +(FG4)ay (O4G4)x+(TG4)a =0
-(F34O4B)(0.15)- (800)(0.12 cos5.1)-
(5)(104cos53)(0.09sin25.5)+
(5)(104sin53)(0.09cos25.5)+(0.054)(625)=0
(F34O4B)=-268.4
MA=0
-(F43AB)AB+ (FG3)ax (AG3)y +(FG3)ay (AG3)x+(TG3)a =0
-(F43AB)(0.22)+(1.5)(-162cos106.8)(0.09sin48.7)+
(1.5)(162sin106.8)(0.09cos48.7)+(0.012)(119)=0
(F43AB)= (F34AB)= 90.9
F23
F43 A B G3
FaG3=-m3aG3
TaG3=-IG33 3
F43AB

14. F34 Link 4: F=0 F34O4B F14 FC C O4 FaG4 =-m4aG4 TaG4=-IG44
!!! (F34O4B) is negative
Force scale (hf ) : 1cm=100N Of
Force scale (hf ) : 1cm=50N
F14
F34AB
92
F34O4B
F34
F34//O4B FC
F34//AB
181
F34
(FG4)a
F34= (6cm)(50 N/cm)=300 N
F14= (3.9cm)(100 N/cm)=390 N
F34= 300 /181 N
F14= 390 /92 N

15. Link 3: F=0 Force scale (hf ) : 1cm=50N Link 2: F=0
G2,O2
F12 2
F32
M12 A
F23
F43 A B G3
FaG3=-m3aG3
TaG3=-IG33 3
Link 3: F=0 Force scale (hf ) : 1cm=50N
Link 2: F=0
F12=-F32= F23= 330 /226 N
226
F43 A
F32
M12
FG3a
F23
F12 h Of O2
MO2=0
M12 – F32 h =0
M12 =18.6 N.m
M12 =18.6k N.m
F23= (6.6cm)(50 N/cm)=330 N
F23= 330 /226 N

16. Example y O2A=3 in. W2=18 lb BA=12 in. W3=3.5 lb
G2O2=1.25in. W4=2.5 lb
G3A=3.5 in
IG2= in.lb.s2
IG3=0.110in.lb.s2 B 4 FB O2 G4 
30 2 2 3 G2 G3 A
2= 3=-3090k r/s FB= ° N
aG2=  ft/s2 aG3= ° ft/s2 aG4= ° ft/s2
O2A sin30 = BA sin =7.18 °
Problem will be solved by including gravity forces.

17. W4
aG4 B
All forces pass through mass center G4 for the moment balance
4// G4 FB
FG4a
Sense is assumed
F34
F14 W4
MA=0
AB x(FB+ F14+W4 +FG4a )+
+ AG3X(W3+FG3a ) + TG3a =0
12(cos7.18i+sin7.18j)X[-800i+F14j-2.5j+
+(2.5/386)(6280)(12)i] +
3.5(cos7.18i+sin7.18j)X[-3.5j+
+(3.5/386)(6130)(12)(-cos158.3i-sin158.3j)]+
+(0.110)(3090)k=0  F14= j lb
3+4//
FG4a FB B
TG3a
aG3
F14 G3 A
FG3a W3
F23

18. 3+4// F=0 F23+ FG3a + W3 +FB+F14 +W4 +FG4a=0
F23+(3.5/386)(6130)(12)(-cos158.3i-sin158.3j)]-3.5j-800i j-2.5j+ (2.5/386)(6280)(12)i=0
F23=-307.8i j lb
4// F=0 F34+FB+F14 +W4 +FG4a=0
F34-800i j-2.5j+(2.5/386)(6280)(12)i=0  F34=311.92i j lb
F12
2// F=0 F32+F12+W2 +FG2a=0
307.8i j+F j+(0.95/386)(2640)(12)(-cos150i-sin150j)=0
F12= i j lb
MO2=0 O2G2X FG2a+ O2G2X W2 +O2AX F32+M12=0
0+1.25(cos30i-sin30j)X(-0.95j)+3(cos30i-sin30j)X(307.8i j)+M12k =0 O2 G2
FG2a
M12
F32 W2 A
M12= k in.lb