Outline• Kinetics

– Linear• Forces in human motion• Mechanical work, power, & energy• Impulse-momentum

– Angular• Torques in human motion• Mechanical work, power, & energy• Impulse-momentum

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)

Torque (= moment) angular equivalent of force

Capability of a force to produce rotationUnits: N*m

Importance?Muscles cause movement by creating torques about joints.

T = MR * FF = forceMR = 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!!

Torque (T): Capability of a force to produce rotation

MRF

rF

T = MR * FF = forceMR = moment arm (perpendicular distance from the point of rotation to the line of force application)rF = distance to F

Torque (T): Capability of a force to produce rotation

MR

F

rF

T = MR * F

What is the Torque (T) due to force F?F=100N; distance to F: rF = 1m, q=30o

MR

F

rF

q

a) 86.6 Nb) 100 Nmc) 86.6 Nmd) 50 Nme) 50 N

Torque is a Vector!Right-Hand Thumb Rule

Figure 2.4

r: MR (moment arm)Right-Hand Thumb Rule:

1. align your hand with MR2. curl your fingers towards F3. direction of thumb is direction of torque vector

Torque and the Coordinate SystemDirection of Positive Torque?

If using default coordinate system:Use right hand thumb ruleCounter-clockwise (CCW)

If using flexion/extension terms:Extension is +ve!

Be CONSISTENT!

x

y

What is the Torque (T) due to force F?F=100N; distance to F: rF = 1m, q=30o

MR

F

rF

q

x

y

A) PositiveB) NegativeC) It Depends

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)

ExampleA 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 + handElbow

Upperarm

Forearm

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)

MR

Fw

Elbow torque due to weight of forearm

T=MR * FT = 0.16m * 11NT = 1.8 Nm

Direction?T = -1.8Nm

0.16 m

11 N

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

rF = 0.16m

Fw

rF

a) 1.76 Nmb) 1.5 Nmc) 0.88 Nmd) -1.5 Nme) -0.88 Nm

q

11 NMR30°

0.16 m

A person holds their forearm so that it is 30° below the horizontal. Elbow torque due to forearm weight?

Fw=11N; rF = 0.16m

Now let’s look at 2 weights

We must consider the effects of 2 forces:

forearm (11N) weight being held (100N)

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?

100 N

11 N

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?

A) T= (-Tarm) + (-Tbriefcase )B) T = Tarm+Tbriefcase

C) T = (-Tarm) + Tbriefcase

D) T = Tarm + (- Tbriefcase)E) It depends

0.4 m

100 N

0.16

11 N

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?

0.4 m

100 N

0.16

11 N

Torque about shoulder due to external forces when 5 kg briefcase is held with straight arm.

Briefcaseforce

Forearmforce

Upperarmforce

49N

0.65 m

0.48 m

20N0.16 m15N

0.48 m

15N

Briefcaseforce

Forearmforce

Upperarmforce

a) 31 Nmb) 20.6 Nmc) - 41Nmd) 41 Nme) None of the above

49N

0.65 m

0.48 m

20N0.16 m15N

0.48 m

15N

Muscles create torques about joints

Elbowflexormuscle

Elbow

Upperarm

Forearm

Bicepsforce

T

Statics (acceleration = 0) F = 0 M = 0 (M is moment or torque)

Dynamics (non-zero acceleration)

Static and Dynamic Analyses

Static equilibrium

Static equilibrium All accelerations are zero

Three equations for analysis Fx = 0 Fy = 0 M = 0

F1 F2R1 R2 Teeter-

totter

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

Static equilibrium

R1

F1

R2

F2

F3

a

Fm Fw

Elbowflexormuscle

Elbow

Upperarm

Forearm

Question: What muscle force (Fm) is required to support the forearm weight (Fw)?

Free-body diagram - static equilibrium

Joint reaction force (Fj): net force generated between adjacent body segments

External forcesMuscle forces

Fm FwFm FwFj

Elbow

Upperarm

Forearm Elbow Forearm

Step 1: Free body diagram.

System = forearm+hand

1. Weight2. Other external forces3. Muscle force4. 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!!

Fm FwFj

Elbow Forearm+hand

Segmental Free body diagrams

Fw

Fm FwFm FwFj

Elbow

Upperarm

Forearm Elbow Forearm

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

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

Solve for Muscle force, Fm

Rw

FwFm

Rm

Fj,y

Ignore weight of forearm Information

Rm = 0.03 m

Rext = 0.4 m

Step 1: Free body diagram

If a person holds a 100 N weight in their hand 0.4 m from the elbow, what is the elbow flexor muscle

force?

Muscle

Elbow

Upperarm

Forearm

Fext

Solve forElbow flexor force

Rm = 0.03 m

Fext = 100 N

Rext = 0.4 m

Rext

Fm

Rm

FextFj,y

Fj,x

When Rm < Rext,muscle force > external force

Bicepsbrachialis

Elbow

Upperarm

Fext

Rext

Rm

Fm = Fext (Rext / Rm)

Last example Fext = 100N

Rext > Rm

Fm = 1333 N

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 diagramApply equations

Rext

Fm

Rm

Fj,y Fext

Fj,x

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 diagramApply equations: Fy = 0

Fj,y + Fm - Fext = 0

Fm = 1333 N, Fext = 100 N

Fj,y = -1233 N

Fx = 0

Fj,x = 0 N

Rext

Fm

Rm

Fj,y Fext

Fj,x

What is the muscle force when a 5 kg briefcase is held with straight arm?

Briefcaseforce

Forearmforce

Upperarmforce

Fm

Fj

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

49N

0.65 m

0.48 m

20N0.16 m15N

0.48 m

15N

Rm = 0.025 m, Fm = ???

49N

0.65 m

0.48 m

20N0.16 m

15N

Fm

Fj

a) -1640 Nmb) 1640 Nmc) 1640 Nd) None of the above

Does Fjx = 0?

49N

0.65 m

0.48 m

20N0.16 m

15N

Fm

Fj

a) Yesb) Noc) It depends

Step 1: Find moment arm of Fg,x (MRx) & Fg,y (MRy) about ankle.

30°

350N

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?

200N

0.2 m

Step1MRx = 0.2 sin 30° = 0.10 m

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

MRx0.2 m

Step 1

MRx = 0.2 sin 30° = 0.10 m

MRy = 0.2 cos 30° = 0.17 mStep 2T = (Tx) + (Ty)

T = (Fg,x *MRx) + (Fg,y * MRy)

T = (200 * 0.10) + (350 * 0.17)T = 79.5 Nm

30°

350N

200N

MRx

MRy

0.2 m

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

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

Outline: Torque• External force ----> muscle force (statics)

Review of approach Mechanical advantage Musculoskeletal complexity

• External force ----> muscle force (dynamics)

Mechanical advantage (MA)

• Fext = Fm * MA MA = Rm / Rext

• MA = 1Fm = Fext

• MA < 1 Fm > Fext

• MA > 1Fm < Fext

MuscleUpperarm

Fext

Rm

Rext

MA < 1 • Rmuscle < Rext

Fmuscle > Fext

Ankle

Shank

Foot

Fext = Fg

Fm

Rm

Rext

MA < 1

Briefcaseforce

Fm

MA > 1

• Rmuscle > Rext

• Fmuscle < Fext

Fmuscle

Fext (Fw)(spleniuscapitis)

MA > 1

• MA = Ractive / Rext

• Ractive > Rext

• Factive < Fext

Fext

Factive

Ractive

Rext

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

Lifting heavy objects

200 N 200 N 200 N

A B C

Outline: Torque• External force ----> muscle force (statics)

Review of approach Mechanical advantage Musculoskeletal complexity

• External force ----> muscle force (dynamics)

Muscle moment arms change with joint angle

Data from Krevolin et al. 2004; Kellis and Baltzopoulos 1999.

A

B

C

D: It depends…

Table 3.2Must know muscle moment arm

Point of Failure during push-ups

Fmuscle = 714 NMR extended = 2.81cmMR flexed = 2.04cm

Fmuscle = 714 NT extended = 20 NmT flexed = 14.6 Nm

A: Extended

B: Flexed

Muscle co-activation

Agonist-antagonist

Elbowflexor(agonist)

Elbow

Elbowextensor

(antagonist)

Fw

2 Agonists

Elbow Fw

biceps brachialis & brachialis

Multiple muscles about a joint: indeterminant problem

∑ Melbow = 0

(Fw * Rw) - (F1 * R1) - (F2 * R2) = 0

Fw

1 2

Fj

Known

Mmus = - (F1 * R1) - (F2 * R2) OrAssume F1 / A1 = F2 / A2

Muscle force can be directly measured

• “Tendon buckle”: placed on tendon & force is measured.

• Achilles tendon

SkeletalmuscleTendon

Muscle force change with joint angle and velocity

• Force-length relationship• Force-velocity relationship

Musculoskeletal complexity

• Muscle moment arms• Muscle force sharing• Muscle length• Muscle velocity

Outline• External force ----> muscle force (statics)

Statics approach Mechanical advantage Musculoskeletal complexity

• External force ----> muscle force (dynamics)

Statics vs. Dynamics• Statics: Acceleration = 0

F = 0 M = 0

• Dynamics: Acceleration ≠ 0 F = ma M = I

I = moment of inertia = angular accelerationOnly about COM, or static pivot point!!

Linear versus angular acceleration

Linear acceleration in y direction

∑ Fy = m ay

Angular acceleration about the y axis

∑ My = Iy y

y

Moment of inertia (I)Resistance of an object to an angular change in its state of motion.

body segmentUnits of moment of inertia: kg * m2

Moment of inertia: depends on distribution of mass relative to the

axis of rotation

y Iy = miri2

i = 1

n

m = massr = distance

1 n

r1

2

Axes in body angular motion

• “Twist”: rotate about longitudinal axis

Hammill and Knutzen

• “Somersault”: rotate in sagittal plane

Axes in body angular motion

Hammill and Knutzen

Icm = 3.8 kg * m2

Twist Icm = 4.1 kg * m2

Somersault: tuck

Icm = 4.1 kg * m2

Somersault: tuckIcm = 12.5 kg * m2

Somersault: layout position

Segmental moment of inertia• I for each body segment rotating

about its COM (Enoka, Table 2.3)• Examples

– Somersault axis• Foot: ICOM = 0.003 kg • m2

• Trunk: ICOM = 1.09 kg • m2

– Twist axis• Foot: ICOM = 0.0007 kg • m2

• Trunk: ICOM = 0.38 kg • m2

ProximalC.O.G.

Often body segments rotate about either their proximal or distal end

Bicep curl:

forearm rotates around its proximal end

Iprox > ICOM

Icom

Iprox

Idistal

Icom is always minimum

Often body segments rotate about either their proximal or distal end

Parallel axis theorem

Iprox = ICOM + mr2

ICOM from published valuesm = segment massr = distance from COM to proximal end

Proximalaxis

COMaxis

r

Iprox = ?a) 0.0415 kg m2

b) 0.0545 kg m2

c) 0.2465d) I have no idea

What is the moment of inertia of the forearm about the elbow?

Given: ICOM = 0.0065 kg * m2 m =1.2kg & COM is 0.2m distal to elbow

ElbowC.O.M.

0.2m

Statics vs. Dynamics• Statics: Acceleration = 0

F = 0 M = 0

• Dynamics: Acceleration ≠ 0 F = ma M = I

I = moment of inertia = angular accelerationOnly about COM, or static pivot point!!

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

Step 1: Draw free body diagram.Step 2: : M = I

What is Fmusc needed to accelerate the forearm at 20 rad/s2 about the elbow?

( ICOM = 0.0065 kg * m2 ; Iprox = 0.054 kg * m2)

Elbow

FmFj

Fw

Rm

Rw

Step 1: Draw free body diagram.Step 2: : M = I

a) Melbow = Iprox b) Melbow = Icom c) Mcom = Iprox d) I’m lost

What is Fmusc needed to accelerate the forearm at 20 rad/s2 about the elbow?

( ICOM = 0.0065 kg * m2 ; Iprox = 0.054 kg * m2)

Elbow

FmFj

Fw

Rm

Rw

Step 1: Draw free body diagram.Step 2: : M = I

Melbow = Iprox

Melbow = 0.054 * 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

What is Fmusc needed to accelerate the forearm at 20 rad/s2 about the elbow?

( ICOM = 0.0065 kg * m2 ; Iprox = 0.054 kg * m2)

Elbow

FmFj

Fw

Rm

Rw

Net muscle moment?

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)

Elbow

FflexFj

Fw

Fext

Net muscle moment: net moment due to all active muscles

Mmus =- (Fm,ext*Rm,ext) + (Fm,flex*Rm,flex)

Fm,flexFm,ext

Elbow

Fj

Fw

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

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)

Fj

FwRw

Mmus

Another sleepy day in 4540Keeping 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) 23.544N

c) 0.0589 N

d) I’m lost

I-70 NightmareWhile 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.

I-70 NightmareCalculate 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) 123.1 N

b) -73 N

c) -0.117 N

d) I’m lost