Post on 22-Dec-2015
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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