DYNAMIC FORCE ANALYSIS

If the acceleration of moving links in a mechanism are running with considerable amount of

linear and/or angular accelerations, inertia forces are generated and these inertia forces also

must be overcome by the driving motor as an addition to the forces exerted by the external

load or work the mechanism does. So, 0=∑Fv

and 0=∑ τv

are no longer applicable.

Governing rules will be:

aFvv

m=∑

ατvv

I=∑

I and m are inertial (bodily) properties. At this stage we need to know the description of the

inertial properties.

Centroid and Mass Centre

Centroid is the point where the resultant of distributed force system is assumed to act and

generate the same dynamic results.

Centroid

Resultant Force

If the distributed force is gravity force acting on each particle of mass, then concentrated force

itself is called the “weight” and the centroid is called the “center of gravity” or “mass center”.

Mass times distance, mr, is called as the first mass moment. This concept of first mass

moment is normally used in deriving the center of mass of a system of particles or a rigid

body. In figure a series of masses are located on a line. The center of mass or centroid is

located at

321

332211

1

1

mmm

mxmxmx

m

mx

xn

i

i

n

i

ii

++

++==

∑

∑

=

=

x3

x1x2

x-m1 m2 m3

G

m1

m2

m3

x-

y-

G

The coordinates of the masses located on a plane can be obtained as:

321

332211

1

1

mmm

mxmxmx

m

mx

xn

i

i

n

i

ii

++

++==

∑

∑

=

= and 321

332211

1

1

mmm

mymymy

m

my

yn

i

i

n

i

ii

++

++==

∑

∑

=

=

This procedure can be extended to masses concentrated in a volume by simply writing an

equation above for the z axis. A more general form of mass center location for three

dimensional body can be obtained by using integration instead of summation. The relations

then become

m

xdmx∫= ,

m

ydmy∫= and

m

zdmz∫=

Mass Moment of Inertia

Mass moment of inertia is the name given to rotational inertia, the rotational analog of mass

for linear motion. It appears in the relationships for the dynamics of rotational motion. The

Mass Moment of Inertia of a solid measures the solid's ability to resist changes in rotational

speed about a specific axis. The moment of inertia for a point mass is just the mass times the

square of perpendicular distance to the rotation axis. The mass moment of inertia for a single

particle is given as:

mrI2

00 =

where m = the mass of the particle, and r = the shortest distance from the axis of rotation to

the particle

r

o

o

m

When calculating the mass moment of inertia for a rigid body, one thinks of the body as a sum

of particles, each having a mass of dm. Integration is used to sum the moment of inertia of

each dm to get the mass moment of inertia of body. The equation for the mass moment of

inertia of the rigid body is

r

o

o

dm

I r dm r dmm

= =∑ ∫2 2

The integration over mass can be replaced by integration over volume, area, or length. For a

fully three dimensional body using the density ρ one can relate the element of mass to the

element of volume. Following are the mathematical equations to calculate the Mass Moment

of Inertia:

x

y

dm

z

rz

ry

rx

( )dmzydmrI xxx ∫ ∫ +== 222

( )dmzxdmrI yyy ∫ ∫ +== 222

( )dmyxdmrI zzz ∫ ∫ +== 222

These three integrals are called the principle mass moment of inertia of the body. Another

three similar integrals are

∫−== xydmII yxxy

∫−== yzdmII zyyz

∫−== xzdmII zxxz

and these three integrals are called the mass products of inertia of the body.

Radius of gyration: Sometime in place of the mass moment of inertia the radius of gyration k

is provided. The mass moment of inertia can be calculated from k using the relation

2mkI =

where m is the total mass of the body. One can interpret the radius of gyration as the distance

from the axis that one could put a single particle of mass m equal to the mass of the rigid body

and have this particle have the same mass moment of inertia as the original body.

G

ICG, m kICG, mk

Original Body Model Model

Parallel-axis theorem: The moment of inertia around any axis can be calculated from the

moment of inertia around parallel axis which passes through the center of mass. The equation

to calculate this is called the parallel axis theorem and is given as

2mdII CG +=

Gd

ICG, m

o o

Example 1 a) What are the mass moments of inertial of a point mass about an axis passing

through itself and about an axis r distance away from it ?

Mass moments of inertial of a point mass about an axis passing through itself

∫ ∫ === 002dmdmrI oo . ANS.

Mass moments of inertial of a point mass about an axis r distance away from it.

Using parallel axis theorem

mrmrmrII xxoo

222 0 =+=+= ANS.

o

o

m

o

o

r m

x

x

b) Find the mass moment of inertia of a slender rod of

length L (slender rod means that it has a length, and

the remaining dimensions are negligible small) about

an axis perpendicular to the rod and passing through

its mass center.

∫=2/

0

2*2

L

oo dmxI

Let density of the material is ρ in kg/m. Then, infinitesimal mass dxdm ρ= . Substituting

this into above equation,

12232|

32*2

23

2/

2/

0

32 mLLx

dxxIL

o

L

oo =

=== ∫ρ

ρρ ANS

Example 2 An uniform steel bar shown in the figure is used as an oscillating cam follower.

Drive the equation of mass moment of inertia of the follower about an axis through O. Use

the density of steel 3/7800 mkg=ρ . Hint: ( )dmyxI ∫ += 22

k=2kN/m

2 cm25 cm 50 cm 25 cm

5 cm

y

x

O

L

L/2

o

o

L

L/2

o

o

dx x

From symmetry

x

y

z

xyr

dx

dzdy

t

w

l

∫= dmrI zz

2

where

dxdydzdm ρ=

Subsituting these into eq. 1

∫∫∫ += dxdydzyxI zz ρ)( 22

∫∫∫∫∫∫ += dxdydzydxdydzxI zz ρρ 22

∫∫∫∫ +=⇒+=2/

0

2

2/

0

222****2****2********

wl

zzzz dyyltdxxwtIdyltydxwtxI ρρρρ

+

=3

2****2

3

2****2

33w

lt

l

wtI zz ρρ 12

****12

****22

wwlt

llwtI zz ρρ +=

since ρ*** twlm = then

1212*

12*

2222wl

mw

ml

mI zz

+=+=

mass moment of inertia about O can be found by parallel axis theorem,

222 yxr +=

48

47

412*

22

2

2222

0

wlm

lm

wlmdmII zz

+=+

+=+= , kgm 8.702.0*05.0*1*7800 == and

mkgI .139.148

05.0*41*78.7

22

0 =+

=

D’Alembert’s Principle:

D’Alembert’s principle permits the reduction of a problem in dynamics to one in statics. This

is accomplished by introducing a fictitious force equal in magnitude to the product of the

mass of the body and its acceleration, and directed opposite to the acceleration. The result is a

condition of kinetic equilibrium. If we subtract a vector of magnitude av

m− from both sides of

the Newton’s second law. The meaning of the equation; i.e. indication of a dynamic case still

holds true, but equation, having zero on right hand side becomes very easy to solve, like that

in a “static force analysis” problem.

0=−=−∑ aaaFvvvv

mmm and similarly for moment

0=−=−∑ ααατvvvv

mmm

Solution of a dynamic problem using D’Alembert’s principle

i) Do an acceleration analysis and calculate the linear acceleration of the mass

centers of each moving link. Also calculate the angular acceleration of each

moving link.

ii) Masses and centroidal inertias of each moving link must be known beforehand.

iii) Add one fictitious force on each moving body equal to the mass of that body times

the acceleration of its mass center, direction opposite to its acceleration, applied

directly onto the center of gravity, apart from the already existing real forces.

iv) Add fictitious torque on each moving body equal to the centroidal inertia of that

body times its angular acceleration, direction or sense opposite to that of

acceleration apart from the already existing real torques.

v) Solve statically.

Example In the figure, a double- slider mechanism working in horizontal plane is shown.

The slider at B is moving rightward with a constant velocity of 1 m/sec. Calculate the

amount of force on the slider B in the given kinematic state. AB=10 cm, AG3=BG3=5 cm,

θ=60o m3=m3=0.5 kg, m3=0.8 kg, I3=0.01 kg.m

2.

BABA VVVvv

+=

smVB

A /1547.1=

b?=AV

ABtoVB

A ⊥= ?

→= smVB /1

↓= smVA /5774.0

BABA aaavvv

+=

BAA aavv

=

t

BA

n

BAA aaa

vvv+=

b?=Aa

BtoAfromsmAB

Va B

An

BA

22

2

/33.131.0

1547.1===

ABtoat

BA

⊥= ?

2/396.15 smaA =

2/698,7 smat

BA

=

4

B

2A

3

G3

x

VB

VA/B

VA

Scale: 4 cm. stands for 1 m/sec

VB

VA/B

VA

Scale: 4 cm. stands for 1 m/sec

n

BA

a

t

BA

a

Aa

3Ga

3G

2/698.72/

3smaa AG ==

3αABat

BA

=

CCWsradAB

at

BA

2

3 /98.781.0

698.7===α

D’Alembert forces and moments

090698.7396.15*5.0 ∠==− NmaA

0901584.6698.7*8.03

∠==− NmaG

CWNmI 7689.098.76*01.03 ==− α

4

B

2A

3

G3

xθ

NmaG 1584.63

=−

Nma A 698.7=−

NmI 7689.03 =− α

BF

4

B

2A

3

G3

x

θ

N1584.6

N698.7

Nm7689.0

BF

y +

14F

12F

∑ =⇒=−= 1212 0;0 FFFFF BBx

NF

FFy

86.13

01584.6698.7;0

14

14

−=

=++=∑

NF

F

M B

11.150867.0

3087.1

060sin*1.0*60cos*05.0*1584.6

60cos*1.0*698.77698.0

;0

12

12

==

=+−

−−

=∑

→= NFB 11.15

Example 3 A robot manipulator

working in the vertical plane is

shown in figure. The payload at

point A is to have a velocity of 2

m/sec with o120∠ and an

acceleration of 5 m/sec2 with

o120∠ . Mass of the payload is 10

kg. Find the forces that the

actuators exert.

m3 = 30 kg, m2 = 30 kg.

AB = 1m, BC = 1m, AG3= 0.5m. BG2= 0.5m

BGA

BGA

aaa

VVVvvv

vvv

==

==

3

3

Link 3 is translating, so any point on this link has the same velocity and acceleration

CBCB VVVvvv

+= C

BCB aaavvv

+=

2

3

G3

G2

AB

C

VB=VA

VC=1 m/s

VB/C=1.73 m/s

A,B

OV60

o

aB=aA

aC=2.5 m/s

aB/C=4.33 m/s

A,B,G3

Oa60

o

C,G2 2

2

g=9.81 m/s 2

aG3=a

A=14.14 m/s 2

10.03o

Acceleration of A and G3 including the effect of gravity becomes:

2

3 /14.14 smaa AG ==

D’Alembert forces

o

o

o

0.0755.2*30

03.102.42414.14*30

03.104.14114.14*10

22

33

∠==−

∠==−

∠==−

Nam

Nam

Nam

G

G

AP

3G3A B FBx

FBy

424.2 N

141.4 N

τB

Vertical actuatorcarries this force

NCosFBy 96.55603.10)2.4244.141( =+= ANS

NSinFBx 51.9803.10)2.4244.141( =+=

NFF BxCx 51.17351.987575 =+=+= ANS

2

G2

B

C

75 N

m2g

Fcx

Fcy

τc

τBFBx

FBy

Horizontal actuatorcarries this force

Example Simple spring mass system is shown in the figure, which

has mass of 2 kg, and spring stiffness is 500 N/m. a) Write the

kinetic energy of this system b) Write the potential energy of this

system c) Drive the equation of motion of this system by using

Lagrange equation. d) Give a solution to the equation of motion. d)

If the system is initially displaced by 0.1m and released from rest, calculate the amplitude of

the resulting vibration.

a) Kinetic energy, 2

2

1xmK &=

b) Potential energy, 2

2

1kxV =

c) 22

2

1

2

1kxxmVKL −=−= &

0=−

x

L

x

L

dt

d

δδ

δδ&

xmx

L&

&=

δδ

, xmx

L

dt

d&&

&=

δδ

, kxx

L−=

δδ

Substituting into above eqn

00 =+⇒=−− kxxm)kx(xm &&&&

d) Soln, tsinBtcosAx nn ωω +=

e) For oxxandt == 0

AxsinBcosAx onno =⇒+= 00 ωω

For 00 == xandt &

0

000

=⇒

+−=⇒+−=

B

cosBsinAtcosBtsinAx nnnnnnnn ωωωωωωωω&

m..BAAmplitude 10010 222 =+=+=

m

k = 500N/m

x

Example Crank AB of the mechanism shown is balanced such that the mass center is at A.

Mass center of the link CD is at its mid point. At the given instant, link 4 is translating

rightward with constant velocity of 5 m/sec. Calculate the amount of motor torque required

on crank AB to keep at the given kinematics state.

14

2

3

A

C

B

D

θ=45

F

AC=10 cm, CB=BD=4 cm, AF=2 cm CG3=5 cm, m2=m3=m4=5 kg, I2=I3=I4=0.05 kg-m2

CBCB VVVvvv

+=

BCto?V

sec/mV

ABto?V

CB

C

B

⊥=

→=

⊥=

5

VC

VB=3.85 m/secVB/C=5.05 m/sec

CBCB aaavvv

+=

0=Ca

n

CB

n

CB

t

B

n

B aaaa +=+vv

BCto?a

CtoBfromsec/m..

.

BC

Va

ABto?a

AtoBfromsec/m..

.

AB

Va

t

CB

CB

n

CB

t

B

Bn

B

⊥=

===

⊥=

===

22

2

222

6637040

055

51920770

853

aB=776 m/sec

aB/C=483 m/sect 2

2t

aBn

aB/Cn

aB

o262800 2 ∠= sec/maB

CWsec/rad.

*BCat

CB

2

33 20000040

800==⇒= αα

CCWsec/rad.

*ABat

B

2

22 100780770

776==⇒= αα

D’Alembert forces and moments

o82400080053 ∠==− N*am B

CCWNm*.I 10002000005033 ==− α

CWNm*.I 5041007805022 ==− α

14

23

A

C

B

D

θ=45

F

504 Nm

1000Nm

T

4000N

4

23

A

C

B

D

θ=45

F

T

FAy

FBy

C

B

AFCx

FCy TA4

504 Nm

1000Nm

4000N

FCx

FCy

FBxFBy

FBx

FAx

FAy4

22°

From FBD of the link 2

∑ =⇒=−= BxAxBxAxx FFFF;F 00 (1)

∑ =⇒=+−= ByAyByAyy FFFF;F 00 (2)

007140028280504

022225040

=−+−

=−++−=∑ByBx

ByBxA

F*.F*.

cos*AB*Fsin*ABFT;M (3)

From FBD of the link 3

069556

08240000

=++

=++=∑.FF

cos*FF;F

BxCx

BxCxx (4)

0073961

08240000

=−+

=−+=∑ByCy

ByCyy

F.F

Fsin*F;F (5)

00282800282801000

0454510000

=−+

=−+=∑CyCx

CyCxB

F*.F*.

cos*BC*Fsin*BCF;M (6)

From FBD of the link 4

∑ =−= 00 Cxx F;F (7)

∑ =−−= 00 4AyCyy FF;F (8)

∑ =+= 00 4 CA*FT;M CyAA (9)

From eq.4

N.FBx 69556−=

From eq.6

N..

FCy 3435355028280

1000==

From eq.5

N...FBy 41393160739613435355 =+=

From eq.3

CCWNm.T

.*.).(*.

933326

041393160714069556028280504

=

=−−+−

Example In the figure a motor geared to a

shaft on which a flywheel is mounted is

shown. The moments of inertia of the

parts are as follows: flywheel, If=3 kg.m2;

flywheel shaft, Ifs=0.02 kg.m2; gear, Ig =

0.2 kg.m2; pinion, Ip= 0.005 kg.m

2; motor,

Im= 0.05 kg.m2. If the motor has starting

torque of 75 N.m, what is the angular

acceleration o the flywheel shaft at the

instant the motor switch is turned on?

Tangential accelerations of the gears at the

contact point are equal to each other. Tangential

acceleration of the gear and flywheel assembly is

gg

t

g ra α= and tangential acceleration of the

motor and pinion assembly is pp

t

g ra α= . Then,

ppgg rr αα = (1)

From the freebody diagram of the motor and

pinion assembly;

ppGm IrFT α2=− (2)

From the freebody diagram of the gear and flywheel assembly;

ggG IrF α3= (3)

Solving these three equations for gα ;

2

23

1773

0250

100550

10

0250223

75sec/rad.

.

.*.

.

.*.

r

rI

r

rI

T

p

g

g

p

m

g =

+=

+

=α

Tm

FG

FG

I2=Ip+Im

I3=Ifs+If+Ig

αp

αg

Example Simple pendulum is shown in the figure, which has mass of 2 kg, and length of

0.5m. a) Write the kinetic energy of this system b) Write the potential energy of this system

c) Drive the equation of motion of this system by using Lagrange equation. d) Give a

solution to the equation of motion. d) If the system is initially displaced by 0.1 rad and

released from rest, calculate the amplitude of the resulting vibration.

a) Kinetic energy, 22

2

1mlIwhere,IK == θ&

b) Potential energy, )cosll(mgV θ−=

c) )cos(mglmlVKL θθ −−=−= 12

1 22 &

0=−

δθδ

θδδ LL

dt

d

&

θθδ

δ &&

2ml

L=

, θ

θδδ &&&

2ml

L

dt

d=

, θ

δδ

sinmglx

L−=

Substituting into above eqn

0

00 22

=+⇒

=+⇒=−−

θθ

θθθθ

sinl

g

sinmglml)sinmgl(ml

&&

&&&&

For small angles

θθ ≅sin

l

gwhere nn ==+ ωθωθ 02&&

d) Soln, tsinBtcosA nn ωωθ +=

e) For oandt θθ == 0

AsinBcosA onno =⇒+= θωωθ 00

For 00 == θ&andt

0

000

=⇒

+−=⇒+−=

B

cosBsinAtcosBtsinA nnnnnnnn ωωωωωωωωθ&

rad..BAAmplitude 10010 222 =+=+=

Example A double slider mechanism working in horizontal plane is shown in figure. The

slider at A is moving upward with a constant velocity of 4 m/sec. Calculate the amount of

forces required on link 2 in the given kinematic state.

AB=7 cm, AG3=3 cm, m2=m4=0,m4=10 kg, I3 =0.05 kg-m2

60°

20°

B

A1

2

3 4

G3

VA

ABAB VVVvvv

+=

ABto?V

sec/mV

or?V

AB

A

B

⊥=

↑=

∠=

4

20020 oo

VA=4m/sec

VB=3.12m/sec

VB/A=5.85m/sec

ABAB aaavvv

+=

0=Aa

t

AB

n

ABB aaa +=

v

ABto?a

AtoBfromsec/m..

.

BC

Va

or?a

t

AB

AB

n

AB

B

⊥=

===

∠=

22

2

9488070

855

20020 oo

aB/AaB/A=582m/s

aB=758 m/s

nt

G32

2

ov20085324 2

3 ∠= s/m.aG

CWs/rad..AB

aB 2

3 5710828070

758===α

D’Alembert forces and moments

o2053248853241033 ∠==− N..*am G

CCWNm..*.I 43541571082805033 ==− α

60°

20°

B

A1

2

3 4

G3

7580 N

541.43 Nm

FA

60°

20°

B

A

2

3 4

G3

3248.5 N

541.43 Nm

FA

NA

NB

x

y+

d2

d1

From the FBD;

∑ =−+= 07020532480 cos*Ncos*.N;F BAx (1)

∑ =++−= 07020532480 sin*Nsin*.F;F BAy (2)

N..

.*..N

.*dN*d.;M

B

BA

156160450

53248023043541

053248435410 21

=+

=

=−+−=∑ (3)

from eq2

N.sin*.sin*.FA 04169070156162053248 =+= upward

Example Each of the two drum and connected hubs of 10cm radius has a mass moment of

inertia of 1 kg-m2. Calculate the angular acceleration of each drum.

r=10 cm

40 N

r=10 cm

40/9.81 kg

Case A Case B

ατ I=

r*T=τ

α*Ir*T =

α** 1100

1040 =

CWsec/rad24=α

ατ I=∑

r*T=∑τ

α*Ir*T = (1)

maF =∑

mgTF +−=∑

mamgT =+− (2)

α*ra = (3)

Subsituting eq 3 into 1 and solving for T

mgmrT +−= α

Subsituting T into eq1 and solving for α

αα Imgrrmr =+− CWsec/rad..

.

.*

mrI

mgr 2

228433

10819

401

1040=

+=

+=α

r=10 cm

40 N

Case A

r=10 cm

40/9.81 kg

Case B

a

T

mg

T

2- (35%) In the figure a slider-

crank mechanism is shown with

the appropriate dimensions. Crank AB of the

mechanism shown is

balanced such that the

mass center is at A. Mass

center of the link 4 is at C.

At the given instant, link 4

is translating rightward

with constant velocity of 5

m/sec. Calculate the

amount of motor torque

required on crank AB to

keep at the given

kinematic state.

Mechanism is working on horizontal plane. Assume no friction between the link 1 and 4.

Mechanism is given is scale. 2

3

2

2432 03.0,01.0,4,6,2 kgmIkgmIkgmkgmkgm =====

CBCB VVVvvv

+=

BCto?V

sec/mV

ABto?V

CB

C

B

⊥=

→=

⊥=

5

CBCB aaavvv

+=

0=Ca

n

CB

n

CB

t

B

n

B aaaa +=+vv

BCtoa

CtoBfrommBC

Va

ABtoa

AtoBfrommAB

Va

t

CB

CB

n

CB

t

B

Bn

B

⊥=

===

⊥=

===

?

sec/2.13505.0

6.2

?

sec/4.774025.0

4.4

22

2

222

o262sec/2.1572 2 ∠= ma B

A

D(6cm, 3cm)

B

C

1

2

3 4

100 N

120o

x

y

CG3

CWradBCat

CB

2

33 sec/1548805.0

4.774* ==⇒= αα

CWradABat

B

2

22 sec/5408025.0

2.135* ==⇒= αα

D’Alembert forces and moments

o1354000055.393*633 ∠==− Nam G

CCWNmI 64.46415488*03.033 ==− α

CCWNmI 08.545408*01.022 ==− α

A

atb/cB

C

2

3

4

100 N

120 o

x

y

B

C

Vc

Vb Vc/b

m3*a3I3*g3

From FBD of the link 2

∑ =⇒=−= BxAxBxAxx FFFF;F 00

∑ =⇒=+−= ByAyByAyy FFFF;F 00

)(.32,20

36,12498*012.0182,3435*022.008,54

012,0*022,0*08,54;0

CWmNT

T

TFFM ByBxA

=

=−+

=++=∑

From FBD of the link 3

m2*a2

CY

CY

BY

CX

BX

BX

CX

I2*α2

N

NF

FF

FFF

Bx

BxCx

BxCxx

182,3435

018,3335

045cos*66,4716;0

=

=+−

=−−=∑

NF

F

FFF

By

By

ByCyy

36,12498

018,333518,9163

045cos*66,4716;0

−=

=++

=++=∑

NF

amIFFM

Cy

gCxCyB

18,9163

0025,0***025,0*044,0*;0 3333

=

=+−+=∑ α

From FBD of the link 4

NFF Cxx∑ == 100;0

NF

NFF

Cy

Cyy

=

=−=∑ 0;0

∑ =++= 0017,0*03,0*;0 4 CxCyAA FFTM

3- (30%) Convert coupler

link of the slider crank

mechanism given in question

2 into a form where only two

point masses at B and C

exist, satisfying the criteria

of equivalency of masses

and location of mass centers.

Then, using your two mass

model, calculate the amount

of motor torque required on

crank AB to keep at the

given kinematic state as in

question 2. Compare your

result with the result of

question 2. Mechanism is

given is scale.

A

D

B

C

2

3

4

100 N

120o

x

y

B

C

BC mmm +=3 mC=mB=3 kg

BBCC mlml = mC=3+4=7kg

lC= lB=0,025m

Link 3 is two force member so FC=FB=115,47

FC=100/cos(30)

For link 2

mB*aB*0,005+FB*0,025+I2*α2=T T=

A

D(6cm, 3cm)

B

C

1

2

3 4

100 N

120o

x

y

CG3

mB*aB

FB

FB

FC

FC

Q-3 (20%) Crank BC of the mechanism shown is balanced such that the mass center is at B.

Mass center of the link AP is at P. Mass of the link 3 is negligible small. At the given instant;

link BC is rotating CCW direction with a constant velocity of 5 rad/sec, link AP is rotating

CCW direction with zero angular velocity and angular acceleration of 25 rad/sec2. Calculate

the amount of motor torque required on crank AP to keep at the given kinematics state.

Mechanism is working on horizontal plane given in scale. m2= m4=5 kg, I2= I4=0.05 kg-m2

2

22 250 sec/rad, == αω

sec/rad54 =ω

←=== 2

2 75125070 sec/m.*.*APaP α

D’Alembert forces and torques become;

→==− N..*am P 75875152

CWNm.*).*.(I 8631250705050 2

22 =+=− α

Link 4 and 2 is two force member,

A

BC

23

4

P

CC

T

1.863 Nm

8.75 Nm

F32 F23 F43F34F14

F12

From freebody diagram of the link 2;

07580 1232 =+−⇒=∑ FF.;Fx (1)

086310707580 32 =++−−⇒=∑ TF..*.;M A (2)

004086310307580 12 =++−−⇒=∑ T.*F..*.;M C (3)

Solving these three equation for CCWNm.T 93335=

A

BC

1

2

3

x

y

4

P