Notes_06_01
1 of 11
Static Force Analysis for Skid Loader - Scalar
A trunnion mount hydraulic cylinder actuates the arm of a skid steer loader as shown below. At
this position, e = 40 inches,  = 61.131°, e = -12 ips,  = -0.3625 rad/s.
Determine the force on the hydraulic cylinder required to lower an 800 lbf payload attached to
point D by a cable. The payload moves with constant velocity at the position shown. You may
neglect the effects of friction. The weight of the arm and cylinder are small compared to the
payload. Show your work.
sin  / AB = sin  / e
 = 52.01°
2261.9 lbf up/left
FCYLINDER __________________________
 = 77.131°
A
C
4

D
Payload
e


Not to scale
AB = 36 inches
AC = 42 inches
AD = 96 inches
= 16°
F14y A
D 4
 P
C

F14x
FC
∑M on 4 about A CCW+
-(FC sin ) AC +(P sin() AD = 0
FC = 2261.9 lbf
B
What corresponding hydraulic pressure would be required for a cylinder with a 3 inch DIA bore?
320 psi
PCYLINDER __________________________
A =  D2 / 4 = 7.069 in2
Is this value reasonable? Why?
OK, industrial hydraulics often go to 3000 psi
If you include friction between the piston and cylinder wall, will it increase or decrease your
computation for pressure.
pressure pushes up
increase decrease Why?
friction force will be up opposing piston motion
What value would you use for the coefficient of friction between the piston and cylinder wall?
0.1 lubricated
 _________________________________
Why?
Should your analysis be different if the cylinder were retracting at constant velocity instead of
the payload moving at constant velocity?
yes
no
Why?
constant e means  will not be constant means velocity of the payload will
not be constant, therefore must account for acceleration of payload mass
Notes_06_01
2 of 11
Static Force Analysis for Four Bar - Scalar
FP = 150 N
B
FQ = 200 N
P
60
13.151
4
2
T12
A
C
3
77
Q
D
65.173
65
AB = 30 cm
BC = 60 cm
CD = 45 cm
AD = 90 cm
BP = 23 cm
DQ = 24 cm
D 1
1
24.827
F34y
F32x
65
F43y
B
13.151
F32y
T12
60
25
25
F12y
A
AB = 30 cm
65
F12
x
65.173
76.849
FP = 150 N
46.849
F23x
13.151
76.849
B
F23y
C
77
F43x
CD = 45 cm
DQ = 24 cm
CQ = 21 cm
P
BC = 60 cm
BP = 23 cm
CP = 37 cm
M on 3 about B CCW+
- F43x (BC) sin 13.151 + F43y (BC) sin 76.849 + FP (BP) sin 60 = 0
- 13.651 F43x + 58.426 F43y = -2987.78 N
F on 4 up +
F34 x  81.993 N 
 y  

F34  70.295 N 
F34x - FQ cos 37.827 + F14x = 0
F34y - FQ sin 37.827 + F14y = 0
F14x = 75.980 N
F14y = 52.360 N
F43x - FP cos 46.849 + F23x = 0
F43y + FP sin 46.849 + F23y = 0
F23x = 184.580 N
F23y = -39.138 N

F on 3 right +
F on 3 up +
M on 2 about A CCW+
- F32x (AB) sin 65 + F32y (AB) sin 25 + T12 = 0
T12 = - 5514.8 N.cm

F on 2 right +
F on 2 up
F12x + F32x = 0
F12y + F32y = 0
D
F14x
- F34x (CD) sin 65.173- F34y (CD) sin 24.827 + FQ (DQ) sin 77 = 0
40.842 F34x + 18.895 F34y = 4676.98 N
F on 4 right +
37.827
Q
65.173
M on 4 about D CCW+
18.895  F34 x   4676.98 N 
 40.842
 13.651  58.426  y    2987.78 N 

 F34  

FQ = 200 N
C
F34x
F12x = 184.580 N
F12y = -39.138 N
F14y
24.827
Notes_06_01
3 of 11
Static Force Analysis for Four Bar - Superposition
FP = 150 N
B
60
3
C
FQ = 200 N
P
13.151
4
2
T12
A
AB = 30 cm
BC = 60 cm
CD = 45 cm
AD = 90 cm
BP = 23 cm
DQ = 24 cm
77
Q
65.173
65
D
1
1
P
T12
CASE I
Q
T12
CASE I - F34 parallel to BC
CASE II
F32
C

M on 4 about D CCW +
F34
B
101.676
- F34 sin 101.676º CD + FQ sin 77 º DQ = 0
F34 = 106.129 N = F32
77
Q
M on 2 about A CCW +
F14y
+ F32 sin 128.151º AB + T12 = 0
T12 = -2503.75 N.cm
F12y
A
D

M on 3 about B CCW +
- F43 sin 101.676º BC + FP sin 60 º BP = 0
F43 = 50.849 N
FP = 150 N
P
60
13.151
F32//BC
C
3
78.324
F43
B
F on 3 //BC right +
+ F43 cos 78.324º - FP cos 60 º + F32//BC = 0
F32//BC = 64.709 N
- F43 sin 78.324º + FP cos 60 º - F32
F32┴BC = 80.107 N
F12x
F14x
CASE II – F43 parallel to CD
F on 3 ┴BC up +
128.151
T12
FQ
┴BC
F32┴BC
38.151
F32┴BC
=0
B
T12
M on 2 about A CCW +
+ F32//BC sin 51.849º AB + F32┴BC sin 38.151º AB + T12 = 0
T12 = -3011.14 N.cm
T12 = T12(Case I) + T12(Case II) = -5514.89 N.cm
F12y
F12x
A
51.849
F32//BC
Notes_06_01
4 of 11
Static Force Analysis for Four Bar - Matrix
F32x
B
FP = 150 N
F32
T12
46.849
A
M on 2 about A CCW +

F on 3 right +
F on 3 up +
M on 3 about P CCW +
F on 4 right +
F on 4 up +
M on 4 about Q CCW +
1
0

0

0
0

0
0

0
0

0
1
1
0
y
0 rB / A
0
1
0
0
0  rBy / P
0
0
0
0
0
0
0
1
 rBx / A
0
1
rBx / P
0
0
0
F43x
13.151
77
37.827
Q
F14y
F23y
F12x
F on 2 up +
C
F34x
B
65
F on 2 right +
P
FQ = 200 N
C
60
y
F23x
F12y
F34y
F43y
65.173
F12x + F32x = 0
F12y + F32y = 0
- F32x rB/Ay + F32y rB/Ax + T12 = 0
F23x + F43x + FPx= 0
F23y + F43y + FPy= 0
- F23x rB/Py + F23y rB/Px - F43x rC/Py + F43y rC/Px = 0
F34x + F14x + FQx = 0
F34y + F14y + FQx = 0
- F34x rC/Qy + F34y rC/Qx - F14x rD/Qy + F14y rD/Qx = 0
0
0
0
1
0
rCy / P
1
0
 rCy / Q
0
0
0
0
1
 rCx / P
0
1
x
rC / Q
0
0
0
0
0
0
1
0
 rDy / Q
0
0
0
0
0
0
0
1
rDx / Q

FP = 150 N @ 133.151 = - 102.589 + j 109.433 N

FQ = 200 N @ 217.827 = - 157.973 - j 122.656 N
B/A = 30 cm @ 65 = 12.678 + j 27.189 cm
B/P = 23 cm @ 193.151 = -22.396 - j 5.233 cm
C/P = 37 cm @ 13.151 = 36.030 + j 8.418 cm
C/Q = 21 cm @ 114.827 = -8.817 + j 19.059 cm
D/Q = 24 cm @ -65.173 = 10.077 - j 21.782 cm
0  F12x   0 
0  F12y   0 


1 F23x   0 
  

0 F23y   FPx 
  

0 F34x    FPy 

0 F34y   0 
  

0  F14x   FQx 

0  F14y   FQy 
  

0 T12   0 
F14x
D
Notes_06_01
1
0

0

0
0

0
0

0
0

0
1
0
0
0
0
0
1
0
1
0
0
0
0
0 27.189  12.678
0
0
0
0
0
1
0
1
0
0
0
0
0
1
0
1
0
0
0 5.233  22.396 8.418
 36.030
0
0
0
0
0
1
0
1
0
0
0
0
0
1
0
1
0
0
0
 19.059  8.817 21.782 10.077
 F12x   184.59 N 
 y 

 F12   - 39.14 N 
F23x   184.59 N 
 y 

F23   - 39.14 N 
 x 

F34    82.00 N

F x   70.29 N 
 34  

 F14x   75.98 N 
 y 

 F14   52.36 N 
T12  - 5514.89 N.cm
5 of 11
0  F12x  
0

 y 


0  F12  
0


1 F23x  
0
 y  

0 F23   102.589 
  

0 F34x    109.433


0 F34x  
0
  

0  F14x   157.973 
  

0  F14y   122.656 

0 T12  
0
Notes_06_01
6 of 11
Static Force Analysis for Pushups - Matrix
A person doing pushups is represented with wrists A, elbows B, shoulders C, and toes D. Mass
of the torso/legs is 180 lbm.
Use the additional assumptions:
a) All muscular effort is provided by triceps as torque T32 across the elbows.
b) Angular velocity of link 4 is constant at this position
c) Weight of the arms is negligible compared to weight of the torso/legs.
d) Friction is negligible at A, B, C and D.
e) No muscular torque is generated at A, C and D.
AD = 52 inch
AB = 12 in
BC = 14 in
CD = 57.7 in
DG4 = 39 in
2 = 45°
3 = 149.14°
4 = 164.24°
2 = 0.5 rad/sec
3 = -1.435 rad/sec
4 = -0.387 rad/sec
C
C
3
G4
B
3
2
A
4
B
2
4
D
Determine angular velocity across the elbows 2/3 for the position and velocity provided above.
2/3 = 2 - 3 = +1.935 rad/sec
Determine elbow torque T32 for the position and velocity provided above.
T32 = 1351.1 in.lbf CCW
Do the magnitude and direction for your answer seem reasonable? Why?
direction seems OK approximate solution W4 (DG4) ~ F34y (CD), F34y ~ 122 lbf ~ F12y
F12y (AB) ~ T32 ~ 1464 in.lbf
Rate the last four assumptions and state your reasoning.
b) constant 4
1=poor
2=acceptable for an approximation
c) weight of arms is negligible
d) friction is negligible
1=poor
1=poor
e) no muscle force at A, C, D
3=very good
2=acceptable for an approximation
2=acceptable for an approximation
1=poor
3=very good
3=very good
2=acceptable for an approximation
3=very good
Determine 4 of the torso/legs when the forearm is aligned with the upper arms (2 = 3). 4 = 
F43y
59.14
F43x
F23y
30.86
59.14
C
T23
B
F23x
30.86
Notes_06_01
7 of 11
F34y
74.24
15.76
F34x
C
y
F32
T32
G4
45
45
74.24
W4
y
74.24 F14
B F32x
F14x
45
F12y
A
45
D
15.76
F12x
F on 2 right +
F on 2 up +
M on 2 about A CCW +

F on 3 right +
F on 3 up +
M on 3 about B CCW +
F on 4 right +
F on 4 up +
M on 4 about D CCW +









F12x + F32x = 0
F12y + F32y = 0
- (F32x sin 45°) AB + (F32y sin 45°) AB + T32 = 0
F23x + F43x = 0
F23y + F43y = 0
- (F43x sin 30.86°) BC - (F43y sin 59.14°) BC + T23 = 0
F34x + F14x = 0
F34y + F14y - W4 = 0
- (F34x sin 15.76°) CD - (F34y sin 74.24°) CD + (W4 sin 74.24°) DG4 = 0
Notes_06_01
1
0

0

0
0

0
0

0
0

0
1
0
0
0
1
0
1
0
0
0 AB sin 45  AB sin 45
0
0
0
1
0
1
0
0
0
1
0
1
0
0
0
BC sin 30.86
BC sin 59.14
0
0
0
1
0
0
0
0
0
1
0
0
0
 CD sin 15.76  CD sin 74.24
1
0

0

0
0

0
0

0
0

0
1
0
0
0
1
0
1
0
0
0 8.4853  8.4853
0
0
0
1
0
1
0
0
0
1
0
1
0
0
0
7.1812
12.0179
0
0
0
1
0
0
0
0
0
1
0
0
0
 15.6718  55.5309
0
0
0
0
0
0
1
0
0
8 of 11
0
0
0
0
0
0
1
0
0
0
0 0   F12x  

 y 


0
0 0   F12  


0
0 1   F23x  

 y  
0
0 0   F23  

  

0
0 0   F34x   

 y  

0
0  1 F34
  

0
0 0   F14x  

 y  

 W4
1 0   F14  

0 0  T32   ( W4 sin 74.24)DG 4 
0 0   F12x   0 
 
0 0   F12y   0 
0 1   F23x   0 
  

0 0   F23y   0 
  

0 0   F34x    0 

0  1  F34y   0 
  

0 0   F14x   0 
  

1 0   F14y   180 
0 0  T32   6756.1
 F12x    29.3 lbf 
 y 

 F12   129.9 lbf 
 F23x    29.3 lbf 
 y 

 F23   129.9 lbf 
 x 

 F34     29.3 lbf 
 F y   129.9 lbf 
 34  

 F14x   29.3 lbf 
 y 

 F14   50.1 lbf 
T32  1351.1 in .lbf 
Notes_06_01
9 of 11
Static Force Analysis for Slider Crank - Scalar
B
T12
R = AB
L = BC
3
2
C


A
4
1
1
B
T12

+ direction for
impending
VC to right
B

2
A

F23
3
F32

F34
C

F14x
F12x
F43
y
F12
F on 4 right +
+ P - F14x sign(VC) - F34 cos = 0
F on 4 up +
+ F34 sin - F14y = 0
M on 3
F23 = F43
M on 2 about A CCW +
+ T12 - ( F32 sin() ) R = 0
friction
F14x =  F14y
F34 
P
cos   sign (VC )  sin 
F14 
P sin 
 P tan  for   0
cos   sign (VC )  sin 
F14 
P sin 
cos   sign (VC )  sin 
y
x
F12  
x
F12 
y
T12 
P
P cos 
 P for   0
cos   sign (VC )  sin 
P sin 
 P tan  for   0
cos   sign (VC )  sin 
P R sin(   )
cos   sign (VC )  sin 
C 4
F14y
P
Notes_06_01
10 of 11
Static Force Analysis for Slider Crank - Matrix
B
T12
2
A
Q
3

Q

C

4
1
1
F32y
T12 B
A
y
F23


F Qy
F23x
F32x
2
B

Q
F on 2 up +
M on 2 about A CCW +

F on 3 right +
F on 3 up +
M on 3 about Q CCW +
F on 4 right +
F on 4 up +
friction
1
0

0

0
0

0
0

0
0

F43y

C
4
F34x
F43x
3
C
F12y
+ direction for
impending
VC to right
F34y
FQx
F12x
F on 2 right +
P

F14x
F14y
F12x + F32x = 0
F12y + F32y = 0
- (F32x sin) AB + (F32y cos) AB + T12 = 0
F23x + F43x + FQx = 0
F23y + F43y + FQy = 0
-(F23x sin) BQ -(F23y cos) BQ +(F43x sin) CQ +(F43y cos) CQ = 0
F14x + F34x + P = 0
F14y + F34y = 0
F14x = -  abs(F14y) sign(VC)
0
1
0
0
0
1
0
1
0
0
0  AB sin   AB cos 
0
0
0
1
0
1
0
0
0
1
0
1
0  BQ sin   BQ cos   CQ sin   CQ cos 
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
1
1   sign (VC )
0  F12x   0 
 
0  F12y   0 


1 F23x   0 
  

0 F23y   FQx 
  

0 F34x    FQy 

0 F34y   0 
  

0  F14x    P 

0  F14y   0 
0 T12   0 
P
Notes_06_01














11 of 11
  F12x  
 y  
  F12  
  F23x  
 y  
  F23  

x


 F34  
 y  
  F34  
F x  
  14y  
  F14  
 T  
  12  














