# ME 320 HW #5 Lecture 15 through 18 1. Consider the following four-bar mechanism for guiding a bucket fixed to the coupler link from the position 1 to the position 2. The position loop equation was...

ME 320 HW #5 Lecture 15 through 18
1. Consider the following four-bar mechanism for guiding a bucket fixed to the coupler link from the position
1 to the position 2. The position loop equation was used to solve the coupler and follower rotation angles,
j and  j, at the position 2 of the bucket when the crank rotates CW of j=-12o.
o
o
XXXXXXXXXX
16.8609
j
j

=
=
(1) Validate the coupler (j) and follower rotation (j) angles from the schematics provided below. You
can show those angles in the diagram below.

(2) Calculate the crank torque Tao needed for statically balancing the external force of Fp=(0, -300N)
applied at P1 in bucket position 1 and at P2 in the bucket position 2. Note that j=j=j= at the
initial position 1 and for the position 2 you need to use  j and  j values provided above for j=-12o.
You can use the portion of the program Statics_StampingMachine_Lec15.m posted under
Lecture 15 to calculate crank torques.
• Submit your Matlab code for calculating torques (No code is needed for plotting four bar
linkages and calculation of j and j):
Crank torque Ta0 for Fp1 at position 1
Crank torque Ta0 for Fp2 at position 2
Bucket Pos 1
Bucket Pos 2
Fp1=(0,-300)N
Fp2=(0,-300)N
-12 deg
Ta0
(Watt II mechanism)
2. The following plot shows initial (dashed line) and final configurations (blue solid lines) of Watt II
mechanism. The initial link parameters are shown below the plot. The final configuration with =90
degrees is shown in solid lines and its associated solution of the position loop equations for the final
configuration is
*
*
XXXXXXXXXX
4-bar (left) : XXXXXXXXXXbar (right) :
XXXXXXXXXX
 
 
= = −
= =

(1) Validate the angles * *, , ,    given in the problem. You can mark those angles in the diagram.
(2) Calculate the static crank torque Tao needed to balance an external force
*
1 PF . You can modify the
Matlab program Statics_Watt_II_Lec16.m.
=90o
ao bo
o*
Tao
FP1*=[0, -200]
3. The following is a planar 4 bar mechanism used for a
ake pad from the released position to the applied
position. When the applied position is reached, a static force of 2 (200,0)PF N= is to be applied at the
displaced point P2. Note that the following is link parameters for the released position:

The following is solution of the loop equation for =40 degrees when the pad is fully applied:
=19.9988deg, =34.9987deg
Calculate the torque Ta0 and x and y components of a static force applied at the pin a2 that balances
the applied force Fp2 in the fully applied
ake pad position. You can modify the program
Statics_StampingMachine_Lec15.m posted under Lecture 15.
a0
0
4. For the following two-link robot arm, three vectors R1, R2, and R3 are used to express the locations of the
mass centers C1 and C2. The values of those vectors are provided for the configuration given below for
1 2 0 = = .
( ) ( )
( ) ( )
( ) ( )
1
1
2
60
1
60
2
30
3
XXXXXXXXXX43
XXXXXXXXXX.43
XXXXXXXXXX.25
ii
ii
ii
R e e i
R e e i
R e e i
 


++

= = = − −
= = =
= = = −

(1) The following expression gives the locations of C1 and C2 in terms of R1, R2, R3, and 1 2,  .
Determine the positions of C1 and C2 for 1 2 0 = = .
( )
1
2 1
1 1
XXXXXXXXXXwhere
i
c
i i
c a a
P R e
P P R e P R R e

 
= −
= − = − +
(Eq 1)
(2) Using the expression for 1CP and 2CP shown in (Eq. 1), determine the velocity of mass centers 1CV
and 2CV for
1 2
1 2
0
1.0 1.0
 
 
= =

= =
.
(3) Using the expression for 1CV and 2CV derived in (2), determine the acceleration of mass center 1Ca
and
2Ca for
1 2
1 2
1 2
0
1.0 1.0
2.0 3.0
 
 
 
= =

= =
 = =
.
1=60o
2=-30
o

C1
C
2

R1
R
3

R
2

X
Y
a1
5. The following illustrates a planar four-bar mechanism used to guide a wiper blade. The dimensions for the
mechanism and the link dynamics parameters are provided in the diagram.
Relevant m-file in Canvas: Dynamic_Planar_4Bar1_Lec17.m.

(1) Determine the initial crank, coupler, and follower angles, (  and ) using the dimensional data
provided in the above table on the left in the diagram.
(2) Determine R2, R4, and R5 using the data provided in the table in the above diagram for =0o. Note
that R1, R3, and R6 are given in the table.
(3) Assuming that the crank rotates with a constant speed of 2 rad/s CW, follow the steps to calculate
the crank torque Ta0 needed when the crank rotates 180 = − as shown below:
0

Configuration for =-180
o

CG3
CG2
CG1
Configuration for =0
o

Initial configuration (=0)and dimensions for each link
Mass center

 a0 b0
0
a0
Mass centers and mechanical parameters for each link
Step 1: Calculate angular velocity ,  and acceleration ,  of the coupler and follower for
2180 , 2 rad/s, 0rad/s  = − = − = using the approach used in Lecture 8 Kinematics Four-Bar Part-1.
Note that angular positions for the coupler, (=−), and follower, (=−), are given in the above
diagram for 180 = − .
Step 2: Calculate the positions of CG1, CG2, and CG3 for 180 = − . Note that 1 jR  can be
determined by either 1 1
i
jR R e

 = or 1 1
j j
j j
j
C S
R R
S C
 

 
− 
=  
  
.
1 1 1
XXXXXXXXXX
3 1 6
i j
CG j
i j i j
CG j j
CG j
P R e R
P W e R e W R
P G R

 
 

= − = −
= − = −
= −

Step 3: Calculate the accelerations of CG1, CG2, and CG3 for 180 = − . Refer to the slide 4 of
Lecture 17.
1
2
1 1
2
1 1 1
2
2 3 3
2
3 6 6
i ij j
j j
j j
CG j j j j
CG a j j j j
i W e W e
CG j j
a i R R
a a i R R
a i R R
 
 
 
 
 
 
 
 

= − +
= − +
= − +

Step 4: Calculate the crank toque needed for 180 = − assuming 0 0bT = and no external force, i.e.
Fp=0. You can use inverse dynamics formulation shown below and modify the m file posted in Lecture
17, Dynamic_Planar_4Bar1_Lec17.m.
0
0
0
1 1 2 2
1
1
XXXXXXXXXX
0
1
6 6 5 5
XXXXXXXXXX
XXXXXXXXXX
XXXXXXXXXX
XXXXXXXXXX
XXXXXXXXXX
XXXXXXXXXX
XXXXXXXXXX
XXXXXXXXXX
XXXXXXXXXX
x
y
j j j j
x
y
j j j j x
y
x
j j j j
a
a
a
y x y x
a
a
y x y x
y x y x
T
F
FR R R R
F
F
R R R R F
F
F
R R R R
   
   
   
 
 
 
 − −
 
− − 
 − −
 
 − −
 
 
 
 
− − 
 
1
1
2
2
3
1
1
1
2
2
2
3
3
3 0
1
3
y
CG x
CG y
CG x
CG y
CG x
C
j
j
j
G
y
m
m
I
m
m
I
m
a
a
a
a
I T
F
a
am

 
  
  
  
  
  
  
   
=   
   
   
   
   
   
   − 
  

• Note that R6 and R7 in the slide #6 of Lecture 17 are replaced by R5 and R6 since there is no coupler point P1 in
this problem.
• No external forces applied to the coupler points, i.e. Fpx=Fpy=0.

6. The crank motion used in Problem 5 has a constant acceleration. Formulate the crank motion again using
the same boundary conditions by the cubic spline discussed in class.
1
2
1 1
2 2
0sec
1.2sec
( ) 0, ( ) 0
( ) 40deg, ( ) 1.2rad/s
t
t
t t
t t
 
 
=
=
= =
= =

(1) Determine the coefficient row vector C.
Answered 1 days AfterApr 09, 2022

## Solution

Sathishkumar answered on Apr 11 2022