Transcript Torque_Part1

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 rotation
Units: N*m
Importance?
Muscles cause movement by creating
torques about joints.
Torque (T): Capability of a force to
produce rotation
F
MR
T = MR * F
rF
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!!
Torque (T): Capability of a force to
produce rotation
F
MR
T = MR * F
rF
F = force
MR = moment arm (perpendicular distance from the
point of rotation to the line of force application)
rF = distance to F
What is the Torque (T) due to force F?
F=100N; distance to F: rF = 1m, q=30o
T = MR * F
F
a)
b)
c)
d)
e)
86.6 N
100 Nm
86.6 Nm
50 Nm
50 N
MR
q
rF
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
Torque and the Coordinate System
Direction of Positive Torque?
y
x
If using default coordinate system:
Use right hand thumb rule
Counter-clockwise (CCW)
If using flexion/extension terms:
Extension is +ve!
Be CONSISTENT!
What is the Torque (T) due to force F?
F=100N; distance to F: rF = 1m, q=30o
y
F
x
MR
q
A) Positive
B) Negative
C) It Depends
rF
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)
Example
Upper
arm
Forearm
Elbow
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
Factors affecting Elbow Torque: Weight of
forearm (Fw) and position of its COM
Fw
MR
From Table in Enoka (BW = 600 N)
Fw (forearm+hand) = 11 N
Distance from proximal end to
COM is 0.16 m (MR)
Elbow torque due to weight of forearm
11 N
T=MR * F
T = 0.16m * 11N
T = 1.8 Nm
Direction?
T = -1.8Nm
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
Fw
q
rF
a)
b)
c)
d)
e)
1.76 Nm
1.5 Nm
0.88 Nm
-1.5 Nm
-0.88 Nm
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
Now let’s look at 2 weights
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
We must consider the
effects of 2 forces:
forearm (11N)
weight being held (100N)
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
A)
B)
C)
D)
E)
T= (-Tarm) + (-Tbriefcase )
T = Tarm+Tbriefcase
T = (-Tarm) + Tbriefcase
T = Tarm + (- Tbriefcase)
It depends
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
Torque about shoulder due to external
forces when 5 kg briefcase is held with
straight arm.
Briefcase
force
Forearm
force
Upper
arm
force
Briefcase
force
Forearm
force
Upper
arm
force
20N
0.16 m
15N
15N
0.48 m
0.48 m
49N
0.65 m
a)
b)
c)
d)
e)
31 Nm
20.6 Nm
- 41Nm
41 Nm
None of the above
20N
0.16 m
15N
15N
0.48 m
0.48 m
49N
0.65 m
Muscles create torques about joints
Upper
arm
Elbow
Elbow
flexor
muscle
Biceps
force
Forearm
T
Static and Dynamic Analyses
Statics (acceleration = 0)
F=0
 M = 0 (M is moment or torque)
Dynamics (non-zero acceleration)
Static equilibrium
F1
R1
R2
Static equilibrium
All accelerations are zero
Three equations for analysis
 Fx = 0
 Fy = 0
M=0
F2
Teetertotter
Static equilibrium
F1
R2
R1
F2
a
F3
Fx = 0: No forces in this direction
Fy = 0
F3 - F1 - F2 = 0
Ma = 0
T 1 – T2 = 0
F1R1 - F2R2 = 0
Convention: Counter Clockwise is positive
Upper
arm
Elbow
Elbow
flexor
muscle
Fm
Fw
Forearm
Question: What muscle force (Fm) is required to
support the forearm weight (Fw)?
Free-body diagram - static equilibrium
Step 1: Free body diagram.
Joint reaction force (Fj): net force generated between
adjacent body segments
External forces
Muscle forces
Upper
arm
Fm
Elbow
Fw
Forearm
Fj Fm
Elbow
Fw
Forearm
Segmental Free body diagrams
System = forearm+hand
1.
2.
3.
4.
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!!
Fj Fm
Fw
Fw
Elbow Forearm+hand
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
Fm
Elbow
Fw
Forearm
Fj Fm
Elbow
Fw
Forearm
Solve for Muscle
force, Fm
Static equilibrium
Fj,y Fm
Rm
Rw
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
Fw
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
Upper
arm
Rm = 0.03 m
Rext = 0.4 m
Muscle
Fext
Step 1: Free body diagram
Elbow
Forearm
Solve for
Elbow flexor force
Fm
Fj,y
Fj,x
Rm
Rext
Rm = 0.03 m
Fext = 100 N
Rext = 0.4 m
Fext
When Rm < Rext,
muscle force > external force
Upper
arm
Biceps
brachialis
Fext
Rm
Elbow
Rext
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).
Fj,y Fm
Fj,x
Rm
Rext
Fext
Free body diagram
Apply equations
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).
Fj,y Fm
Fj,x
Rm
Rext
Fext
Free body diagram
Apply equations:
Fy = 0
Fj,y + Fm - Fext = 0
Fm = 1333 N, Fext = 100 N
Fj,y = -1233 N
 Fx = 0
Fj,x = 0 N
What is the muscle force when a 5 kg
briefcase is held with straight arm?
Fm
Fj
Forearm
Briefcase force
force
Upper
arm
force
20N
0.16 m
15N
15N
0.48 m
0.48 m
49N
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
Fm
20N
15N
0.48 m
49N
0.65 m
Rm = 0.025 m, Fm = ???
a)
b)
c)
d)
-1640 Nm
1640 Nm
1640 N
None of the above
Fj
0.16 m
Fm
20N
15N
0.48 m
49N
0.65 m
Does Fjx = 0?
a) Yes
b) No
c) It depends
Fj
0.16 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?
Step 1: Find moment arm of Fg,x
(MRx) & Fg,y (MRy) about ankle.
0.2 m
30°
200N
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?
Step1
MRx = 0.2 sin 30° = 0.10 m
MRx
0.2 m
30°
200N
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?
Step 1
MRy
MRx
0.2 m
30°
200N
350N
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
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
MRy
MRx
0.2 m
30°
200N
350N
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 = 1
Fm = Fext
• MA < 1
Upper
arm
Muscle
Fm > Fext
• MA > 1
Fm < Fext
Rm
Rext
Fext
MA < 1
• Rmuscle < Rext
Fmuscle > Fext
Shank
Fm
Rm
Ankle
Foot
Rext
Fext = Fg
MA < 1
Fm
Briefcase
force
MA > 1
• Rmuscle > Rext
• Fmuscle < Fext
Fmuscle
(splenius
capitis)
Fext (Fw)
Factive
Ractive
Rext
Fext
MA > 1
• MA = Ractive / Rext
• Ractive > Rext
• Factive < Fext
Joint torques during standing
• Fg,y vector closely aligned with
joints (knee, hip, lumbar intervertebral joints)
– Joint torques are almost zero
• Fg,y is not aligned with ankle
– soleus muscle counteracts it
Fg,y
Lifting heavy objects
B
A
200 N
200 N
C
200 N
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
D: It depends…
B
A
Data from Krevolin et al. 2004; Kellis and Baltzopoulos 1999.
C
Must know
muscle
moment arm
Table
3.2
Point of Failure during push-ups
A: Extended
B: Flexed
Fmuscle = 714 N
MR extended = 2.81cm
MR flexed = 2.04cm
Fmuscle = 714 N
T extended = 20 Nm
T flexed = 14.6 Nm
Muscle co-activation
Agonist-antagonist
2 Agonists
biceps brachialis &
brachialis
Elbow
Fw
Elbow
extensor
(antagonist)
Elbow
flexor
(agonist)
Elbow
Fw
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
Fj
12
Fw
Muscle force can be directly measured
Tendon
Skeletal
muscle
• “Tendon buckle”:
placed on tendon &
force is measured.
• Achilles tendon
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 acceleration
Only about COM, or static pivot point!!
Linear versus angular acceleration
y
Linear acceleration in y
direction
∑ Fy = m ay
Angular acceleration about the
y axis
∑ My = Iy y
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
Moment of inertia: depends on
distribution of mass relative to the
axis of rotation
n
Iy =
y
r1
1 2
n

miri2
i=1
m = mass
r = distance
Axes in body angular motion
• “Twist”: rotate about
longitudinal axis
Hammill and Knutzen
Axes in body angular motion
• “Somersault”: rotate in sagittal plane
Hammill and Knutzen
Twist
Icm = 3.8 kg * m2
Somersault: tuck
Icm = 4.1 kg * m2
Somersault: layout position
Icm = 12.5 kg * m2
Somersault: tuck
Icm = 4.1 kg * m2
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
Often body segments rotate about
either their proximal or distal end
Bicep curl:
Proximal
C.O.G.
forearm rotates
around its proximal
end
Iprox > ICOM
Often body segments rotate about
either their proximal or distal end
Icom
Iprox
Idistal
Icom is always
minimum
Parallel axis theorem
Proximal
axis
r
COM
axis
Iprox = ICOM + mr2
ICOM from published
values
m = segment mass
r = distance from COM to
proximal end
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
Iprox = ?
a) 0.0415 kg m2
b) 0.0545 kg m2
c) 0.2465
d) I have no idea
Elbow
C.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 acceleration
Only 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
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)
Step 1: Draw free body diagram.
Step 2: :  M = I 
Fj
Fm
Rm
Elbow
Rw

Fw
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)
Step 1: Draw free body diagram.
Step 2: :  M = I 
a) Melbow = Iprox 
b) Melbow = Icom 
c) Mcom = Iprox 
d) I’m lost
Fj
Fm
Rm
Elbow
Rw

Fw
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)
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
Fj
Elbow
Fm
Rm
Rw

Fw
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?
Fext
Fj F
flex

Elbow
Fw
Net muscle moment: net moment due
to all active muscles
Mmus =- (Fm,ext*Rm,ext) + (Fm,flex*Rm,flex)
Fm,ext
Fj
Fm,flex

Elbow
Fw
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
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) 23.544N
c) 0.0589 N
d) I’m lost
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.
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) 123.1 N
b) -73 N
c) -0.117 N
d) I’m lost