1. Outline Kinetics Linear Angular Forces in human motion
Mechanical work, power, & energy
Impulse-momentum
Angular
Torques in human motion

2. Outline Torques in human motion Definitions
External force ----> muscle force (static analysis)
Review of approach
Mechanical advantage
Musculoskeletal complexity
External force ----> muscle force (dynamic analysis)

3. Torque (= moment) angular equivalent of force
Capability of a force to produce rotation
Units: N*m
Importance?
Muscles cause movement by creating torques about joints.

4. Torque (T): Capability of a force to produce rotationMR
T = MR * F
F = force
MR = moment arm (perpendicular distance from the point of rotation to the line of force application)
rF = distance to F
MR and rF are NOT the same!! rF

5. Torque (T): Capability of a force to produce rotationMR
T = MR * F
F = force
MR = moment arm (perpendicular distance from the point of rotation to the line of force application)
rF = distance to F rF

6. What is the Torque (T) due to force F
What is the Torque (T) due to force F? F=100N; distance to F: rF = 1m, q=30o
T = MR * F F
86.6 N
100 Nm
86.6 Nm
50 Nm
50 N MR q rF

7. Torque is a Vector! Right-Hand Thumb Rule Figure 2.4
r: MR (moment arm)
Right-Hand Thumb Rule:
1. align your hand with MR
2. curl your fingers towards F
3. direction of thumb is direction of torque vector

8. Torque and the Coordinate System
Direction of Positive Torque? If using default coordinate system: Use right hand thumb rule Counter-clockwise (CCW) If using flexion/extension terms: Extension is +ve! Be CONSISTENT! y x

9. What is the Torque (T) due to force F
What is the Torque (T) due to force F? F=100N; distance to F: rF = 1m, q=30o y F x MR q
Positive
Negative
It Depends rF

10. Outline Torques in human motion Definitions
External force ----> muscle force (static analysis)
Review of approach
Mechanical advantage
Musculoskeletal complexity
External force ----> muscle force (dynamic analysis)

11. Example
A person holds their elbow at 90° with their forearm parallel to the ground.
Elbow torque?
Step 1: Draw a free body diagram
“system” = the forearm + hand
Upper
arm
Forearm
Elbow

12. Factors affecting Elbow Torque: Weight of forearm (Fw) and position of its COM
From Table in Enoka (BW = 600 N)
Fw (forearm+hand) = 11 N
Distance from proximal end to COM is 0.16 m (MR) Fw MR

13. Elbow torque due to weight of forearm
T=MR * F
T = 0.16m * 11N
T = 1.8 Nm
Direction?
T = -1.8Nm
11 N
0.16 m

14. A person holds their forearm so that it is 30° below the horizontal
A person holds their forearm so that it is 30° below the horizontal. Elbow torque due to forearm weight? Fw=11N; rF = 0.16m
1.76 Nm
1.5 Nm
0.88 Nm
-1.5 Nm
-0.88 Nm Fw q rF

15. A person holds their forearm so that it is 30° below the horizontal
A person holds their forearm so that it is 30° below the horizontal. Elbow torque due to forearm weight? Fw=11N; rF = 0.16m
11 N MR
30°
0.16 m

16. Now let’s look at 2 weights

17. We must consider the effects of 2 forces:
A person is holding a 100N weight at a distance of 0.4 m from the elbow. What is the total elbow torque due to external forces?
We must consider the effects of 2 forces:
forearm (11N)
weight being held (100N)
100 N
11 N

18. T= (-Tarm) + (-Tbriefcase ) T = Tarm+Tbriefcase
A person is statically holding a 100N weight at a distance of 0.4 m from the elbow. What is the elbow torque due to external forces?
100 N
T= (-Tarm) + (-Tbriefcase )
T = Tarm+Tbriefcase
T = (-Tarm) + Tbriefcase
T = Tarm + (- Tbriefcase)
It depends
11 N
0.16
0.4 m

19. A person is statically holding a 100N weight at a distance of 0
A person is statically holding a 100N weight at a distance of 0.4 m from the elbow. What is the elbow torque due to external forces?
100 N
11 N
0.16
0.4 m

20. Torque about shoulder due to external forces when 5 kg briefcase is held with straight arm.
Forearm
force
Upper
arm
force
Briefcase
force

21. Upper arm force Forearm force Briefcase force 20N 0.16 m 15N 15N
None of the above

22. 20N
0.16 m
15N
15N
0.48 m
49N
0.48 m
0.65 m

23. Muscles create torques about joints
Upper
arm
Elbow
flexor
muscle
Biceps
force
Elbow
Forearm T

24. Static and Dynamic Analyses
Statics (acceleration = 0)
 F = 0
 M = 0 (M is moment or torque)
Dynamics (non-zero acceleration)

25. Static equilibrium F1 F2 R2 R1 Teeter- totter Static equilibrium
All accelerations are zero
Three equations for analysis
 Fx = 0
 Fy = 0
 M = 0

26. Static equilibrium F1 F2 R2 R1 a F3
Fx = 0: No forces in this direction
Fy = 0
F3 - F1 - F2 = 0
Ma = 0
T1 – T2 = 0
F1R1 - F2R2 = 0
Convention: Counter Clockwise is positive

27. Upper
arm
Elbow
flexor
muscle Fm Fw
Elbow
Forearm
Question: What muscle force (Fm) is required to support the forearm weight (Fw)?
Free-body diagram - static equilibrium

28. Step 1: Free body diagram.
Joint reaction force (Fj): net force generated between adjacent body segments
External forces
Muscle forces
Upper
arm Fj Fm Fw Fm Fw
Forearm
Forearm
Elbow
Elbow

29. Segmental Free body diagrams
System = forearm+hand
Weight
Other external forces
Muscle force
Joint reaction force (Fj): net force generated between adjacent body segments
Direction?
If you are unsure of the
direction a force is acting,
draw a POSITIVE vector!! Fw Fj Fm Fw
Forearm+hand
Elbow

30. Upper arm Fj Fm Fw Fm Fw Forearm Forearm Elbow Elbow
Question: What muscle force (Fm) is required to support the forearm weight (Fw)? Step 1: Free body diagram. Givens: Fw = 11 N, Rw = 0.16 m, Rm = 0.03 m Step 2: Apply appropriate equation
Upper
arm Fj Fm Fw Fm Fw
Forearm
Forearm
Elbow
Elbow

31. Solve for Muscle force, Fm
Fj,y Fm Fw
Solve for Muscle force, Fm Rm
Static equilibrium
Melbow = 0
Fj creates no moment at elbow
-(Tw) + (Tm) = 0
-(Fw * Rw) + (Fm * Rm) = 0
Fm = (Fw * Rw) / Rm
Substitute: Fw = 11 N, Rw = 0.16 m, Rm = 0.03 m
Fm = 59 N Rw

32. Ignore weight of forearm Information
If a person holds a 100 N weight in their hand 0.4 m from the elbow, what is the elbow flexor muscle force?
Ignore weight of forearm
Information
Rm = 0.03 m
Rext = 0.4 m
Step 1: Free body diagram
Upper
arm
Muscle
Fext
Elbow
Forearm

33. Solve for Elbow flexor force
Rm = 0.03 m
Fext = 100 N
Rext = 0.4 m
Solve for Elbow flexor force
Rext Fm Rm
Fext
Fj,y
Fj,x

34. When Rm < Rext, muscle force > external force
Fm = Fext (Rext / Rm)
Last example
Fext = 100N
Rext > Rm
Fm = 1333 N
Upper
arm
Biceps
brachialis
Fext Rm
Elbow
Rext

35. Free body diagram Apply equations Fj,y Fext Fm Fj,x Rm
If a person holds a 100 N weight in their hand 0.4 m from the elbow, what is the joint reaction force? (ignore weight of forearm).
Free body diagram
Apply equations
Fj,y
Fext Fm
Fj,x Rm
Rext

36. Free body diagram Apply equations: Fy = 0 Fj,y Fext Fm
If a person holds a 100 N weight in their hand 0.4 m from the elbow, what is the joint reaction force? (ignore weight of forearm).
Free body diagram
Apply equations:
Fy = 0
Fj,y + Fm - Fext = 0
Fm = 1333 N, Fext = 100 N
Fj,y = N
Fx = 0
Fj,x = 0 N
Fj,y
Fext Fm
Fj,x Rm
Rext

37. What is the muscle force when a 5 kg briefcase is held with straight arm?Fm Fj
Forearm
force
Upper
arm
force
Briefcase
force

38. 20N
0.16 m
15N
15N
0.48 m
49N
0.48 m
0.65 m
T = (49N * 0.65m) + (15N * 0.48m) + (20N * 0.16m)
T = (31 Nm) + (7.2 Nm) + (3.2 Nm)
T = 41 Nm

39. Fm Fj 20N 0.16 m 15N 0.48 m 49N 0.65 m Rm = 0.025 m, Fm = ??? -1640 Nm
None of the above

40. Fm Fj
20N
0.16 m
15N
49N
0.48 m
0.65 m
Does Fjx = 0?
Yes No
It depends

41. Step 1: Find moment arm of Fg,x (MRx) & Fg,y (MRy) about ankle.
At the end of stance phase while running, Fg,x under the right foot is 200N and Fg,y is 350N. Fg is applied to the foot 0.2 m from the ankle. What is the ankle torque due to Fg?
Step 1: Find moment arm of Fg,x (MRx) & Fg,y (MRy) about ankle.
0.2 m
30°
200N
350N

42. At the end of stance phase while running, Fg,x under the right foot is 200N and Fg,y is 350N. Fg is applied to the foot 0.2 m from the ankle. What is the ankle torque due to Fg?
Step1
MRx = 0.2 sin 30° = 0.10 m
30°
350N
200N
MRx
0.2 m

43. At the end of stance phase while running, Fg,x under the right foot is 200N and Fg,y is 350N. Fg is applied to the foot 0.2 m from the ankle. What is the ankle torque due to Fg?
30°
350N
200N
MRx
MRy
0.2 m
Step 1
MRx = 0.2 sin 30° = 0.10 m
MRy = 0.2 cos 30° = 0.17 m
Step 2
T = (Tx) + (Ty)
T = (Fg,x *MRx) + (Fg,y * MRy)
T = (200 * 0.10) + (350 * 0.17)
T = 79.5 Nm

44. At the end of stance phase while running, Fg,x under the right foot is 200N and Fg,y is 350N. Fg is applied to the foot 0.2 m from the ankle. What is the ankle extensor muscle force? MRmusc = 0.05m
30°
350N
200N
MRx
MRy
0.2 m

45. Outline: Torque External force ----> muscle force (statics)
Review of approach
Mechanical advantage
Musculoskeletal complexity
External force ----> muscle force (dynamics)

46. Mechanical advantage (MA)
Fext = Fm * MA
MA = Rm / Rext
MA = 1
Fm = Fext
MA < 1
Fm > Fext
MA > 1
Fm < Fext
Upper
arm
Muscle
Fext Rm
Rext

47. MA < 1 Rmuscle < Rext Shank Fm Rm Foot Rext Ankle Fext = Fg
Fmuscle > Fext
Shank Fm Rm
Foot
Rext
Ankle
Fext = Fg

48. MA < 1 Fm
Briefcase
force

49. MA > 1 Rmuscle > Rext Fmuscle < Fext Fmuscle (splenius
capitis)
Fext (Fw)

50. MA > 1 MA = Ractive / Rext Ractive > Rext Factive < Fext

51. Joint torques during standing
Fg,y vector closely aligned with joints (knee, hip, lumbar inter-vertebral joints)
Joint torques are almost zero
Fg,y is not aligned with ankle
soleus muscle counteracts it
Fg,y

52. Lifting heavy objects B C A
200 N
200 N
200 N

53. Outline: Torque External force ----> muscle force (statics)
Review of approach
Mechanical advantage
Musculoskeletal complexity
External force ----> muscle force (dynamics)

54. Muscle moment arms change with joint angle
D: It depends… B C A
Data from Krevolin et al. 2004; Kellis and Baltzopoulos 1999.

55. Table 3.2
Must know muscle moment arm

57. Point of Failure during push-ups
A: Extended
Fmuscle = 714 N
MR extended = 2.81cm
MR flexed = 2.04cm
Fmuscle = 714 N
T extended = 20 Nm
T flexed = 14.6 Nm
B: Flexed

58. Elbow Fw Fw Muscle co-activation Elbow extensor flexor (antagonist)
2 Agonists
Agonist-antagonist
Elbow
flexor
(agonist)
extensor
(antagonist) Fw
Elbow Fw
biceps brachialis & brachialis

59. Multiple muscles about a joint: indeterminant problem
Melbow = 0
(Fw * Rw) - (F1 * R1) - (F2 * R2) = 0
Known
Mmus = - (F1 * R1) - (F2 * R2) Or
Assume F1 / A1 = F2 / A2 Fw 1 2 Fj

60. Muscle force can be directly measured
“Tendon buckle”: placed on tendon & force is measured.
Achilles tendon
Skeletal
muscle
Tendon

61. Muscle force change with joint angle and velocity
Force-length relationship
Force-velocity relationship

62. Musculoskeletal complexity
Muscle moment arms
Muscle force sharing
Muscle length
Muscle velocity

63. Outline External force ----> muscle force (statics)
Statics approach
Mechanical advantage
Musculoskeletal complexity
External force ----> muscle force (dynamics)

64. Statics vs. Dynamics Statics: Acceleration = 0
 F = 0
 M = 0
Dynamics: Acceleration ≠ 0
 F = ma
 M = I 
I = moment of inertia
 = angular acceleration
Only about COM, or static pivot point!!

65. Linear versus angular acceleration
Linear acceleration in y direction
∑ Fy = m ay
Angular acceleration about the y axis
∑ My = Iy y y

66. Moment of inertia (I)
Resistance of an object to an angular change in its state of motion.
body segment
Units of moment of inertia: kg * m2

67. Moment of inertia: depends on distribution of mass relative to the axis of rotation
Iy =  miri2 y
i = 1 r1
m = mass
r = distance 1 2 n

68. Axes in body angular motion
“Twist”: rotate about longitudinal axis
Hammill and Knutzen

69. Axes in body angular motion
“Somersault”: rotate in sagittal plane
Hammill and Knutzen

70. Twist
Icm = 4.1 kg * m2
Somersault: tuck
Icm = 3.8 kg * m2

71. Icm = 12.5 kg * m2
Somersault: layout position
Icm = 4.1 kg * m2
Somersault: tuck

73. Segmental moment of inertia
I for each body segment rotating about its COM (Enoka, Table 2.3)
Examples
Somersault axis
Foot: ICOM = kg • m2
Trunk: ICOM = 1.09 kg • m2
Twist axis
Foot: ICOM = kg • m2
Trunk: ICOM = 0.38 kg • m2

74. Often body segments rotate about either their proximal or distal end
C.O.G.
Bicep curl:
forearm rotates around its proximal end
Iprox > ICOM

75. Often body segments rotate about either their proximal or distal end
Icom Iprox Idistal Icom is always minimum

76. Parallel axis theorem Iprox = ICOM + mr2 ICOM from published values
m = segment mass
r = distance from COM to proximal end
Proximal
axis
COM
axis r

77. What is the moment of inertia of the forearm about the elbow
What is the moment of inertia of the forearm about the elbow? Given: ICOM = kg * m2 m =1.2kg & COM is 0.2m distal to elbow
Iprox = ?
a) kg m2
b) kg m2 c)
d) I have no idea
Elbow
C.O.M.
0.2m

78. Statics vs. Dynamics Statics: Acceleration = 0
 F = 0
 M = 0
Dynamics: Acceleration ≠ 0
 F = ma
 M = I 
I = moment of inertia
 = angular acceleration
Only about COM, or static pivot point!!

79. Overview of dynamics problems
Draw free body diagram + CS
Use equations:
 F = ma
 M = I 
Calculate “ma” or “I ”
 Mcom = Icom 
 Mo = Io  --> where O is a fixed point
Sum the forces or moments
Solve for unknown

80. ( ICOM = 0.0065 kg * m2 ; Iprox = 0.054 kg * m2)
What is Fmusc needed to accelerate the forearm at 20 rad/s2 about the elbow?
( ICOM = kg * m2 ; Iprox = kg * m2)
Step 1: Draw free body diagram.
Step 2: :  M = I  
Elbow Fm Fj Fw Rm Rw

81. ( ICOM = 0.0065 kg * m2 ; Iprox = 0.054 kg * m2)
What is Fmusc needed to accelerate the forearm at 20 rad/s2 about the elbow?
( ICOM = kg * m2 ; Iprox = kg * m2)
Step 1: Draw free body diagram.
Step 2: :  M = I 
a)  Melbow = Iprox 
b)  Melbow = Icom 
c)  Mcom = Iprox 
d) I’m lost 
Elbow Fm Fj Fw Rm Rw

82. ( ICOM = 0.0065 kg * m2 ; Iprox = 0.054 kg * m2)
What is Fmusc needed to accelerate the forearm at 20 rad/s2 about the elbow?
( ICOM = kg * m2 ; Iprox = kg * m2)
Step 1: Draw free body diagram.
Step 2: :  M = I 
 Melbow = Iprox 
 Melbow = * 20 = 1.1 Nm
Step 3: Find moments due to each force on forearm
(Fm * Rm) - (Fw * Rw) = 1.1 N m
Fw = 11 N, Rw = 0.16 m, Rm = 0.03 m
Fm = 95 N 
Elbow Fm Fj Fw Rm Rw

83. (Iprox = 0.06 kg * m2, Fw = 15 N, Rw = 0.2 m)
What is net muscle moment needed to accelerate the forearm at 20 rad/s2 about the elbow?
(Iprox = 0.06 kg * m2, Fw = 15 N, Rw = 0.2 m)
Net muscle moment? 
Elbow
Fflex Fj Fw
Fext

84. Net muscle moment: net moment due to all active muscles
Mmus =- (Fm,ext*Rm,ext) + (Fm,flex*Rm,flex)
Fm,flex
Fm,ext 
Elbow Fj Fw

85. (Iprox = 0.06 kg * m2, Fw = 15 N, Rw = 0.2 m)
What is net muscle moment needed to accelerate the forearm at 20 rad/s2 about the elbow?
(Iprox = 0.06 kg * m2, Fw = 15 N, Rw = 0.2 m)
Step 1: Free body diagram
Step 2:  Melbow = Iprox 
 Melbow = (0.06)(20) = 1.2 N m
Step 3: Find sum of the moments about the elbow
 Melbow = 1.2 = Mmus - (Fw * Rw)
Mmus = 4.2 N • m Fj
Mmus Rw Fw

86. Another sleepy day in 4540
Keeping your head upright requires alertness but not much muscle force. Given this diagram/information, calculate the muscle force. Head mass = 4 kg.
a) 2.4 N
b) N
c) N
d) I’m lost

87. I-70 Nightmare
While driving, you start to nod off asleep, and a very protective reaction kicks in, activating your neck muscles and jerking your head up in the nick of time to avoid an accident.

88. I-70 Nightmare
Calculate the muscle force needed to cause a neck extension acceleration of 10 rad/s2. The moment of inertia of the head about the head-neck joint is 0.10 kg m2.
a) N
b) -73 N
c) N
d) I’m lost