Lecture outline
Muscle and Joint
Forces
• components of joint force
• applications of static analysis:
• mechanics of the elbow during lifting
• mechanics of the shoulder during lifting
• mechanics of the hip during single legged
stance
Ozkaya and Nordin
Chapters 4 and 5 (page 49-115)
Stephen Robinovitch, Ph.D.
KIN 201
2007-1
Lecture 6: Joint Force 1
1
Definitions: joint forces
Mechanical advantage of a lever
joint contact force: “bone on bone” or
“cartilage on cartilage” force
mechanical
moment arm of effort force
advantage =
moment arm of resistance force
(MA)
axis
!
effort
resistance
!
2
ligament force: tensile force in ligament
that connects two bones. Typically
significant only at extremes of motion.
If MA < 1, the lever
amplifies motion.
Most muscle-joint
systems are 1st class
or 3rd class levers
that amplify motion.
joint reaction force: sum of joint contact
force + ligament force
muscle force: tensile force in the
muscle. Transmitted to bone via the
muscle tendon.
if MA > 1, then the
lever amplifies force
!(external forces + joint reaction
force + muscle force) = 0
3
4
How do I determine the direction of
unknown forces?
How do I determine the direction of
unknown forces? (Cont’d)
• Both the magnitude and sense of joint reaction forces
FJx and FJy are often unknown*
• What if, in setting up the free body diagram, incorrect
directions are assumed for FJx and FJy?
• THIS IS NOT A PROBLEM!
• An incorrect assumption regarding the direction of FJx
or FJy will be signified by a minus sign in the solution
i.e., a minus sign indicates that the force is directed
opposite from that originally assumed
Steps to determining the sense of unknown forces:
• * unlike muscle forces, which are always tensile and directed
outward from the border of the FBD
(1) while creating your FBD, make an educated guess
about the sense of FJx and FJy, and draw these forces
acting in the assumed directions
(2) when writing out the equation !Fx = 0, insert FJx as
positive if it has been drawn (i.e., assumed) to act in the
positive x direction, and negative if it has been drawn to
act in the negative x direction
(3) same for FJy in the equation !Fy = 0
5
How do I determine the direction of
unknown forces? (Cont’d)
6
Solve for FJx, FJy, and FM for cases (A) and (B)
100 N
FJy
y
(A)
(4) in the equation !MO = 0 , refer to the moment
generated by a force* as positive if it tends to cause
rotation about point O in the defined positive
direction (typically CCW)
200 N
O
1
(5) if, upon solving for FJx or FJy , one of these
forces turns out to be a negative, the true sense of
this force is opposite to that assumed in setting up
the FBD
(B)
x
FJx
FM
3
100 N
FJx
O
1
7
200 N
x
FJy
FM
*NOTE: we often take moments about the joint centre,
where FJx or FJx contribute no moment
y
3
8
Mechanics of the Elbow
a
FJ
! Fy = 0 :
load
muscle in
tension
"FJ + FM " W " WO = 0
FJ = FM - W - WO
c
(1)
! MO = 0 :
FM a " Wb " WOc = 0
# b&
# c&
FM = W % ( + WO % (
$ a'
$ a'
WO
FM
(2)
Insert (2) into (1) :
"b %
"c %
FJ = W $ ! 1' + WO $ ! 1'
#a &
#a &
W
b
force (N)
9
1200
1000
800
600
400
200
0
-200
FM (WO = 10 N (2.2 lb))
FJ (WO = 10 N (2.2 lb))
FM (WO = 100 N (22.2 lb))
FJ (WO = 100 N (22.2 lb))
Remarks - elbow mechanics during
support of a hand-held weight
W = 20 N
b/a = 3
0
2
4
6
ratio c/a
8
10
• the joint reaction force is approximately 8-fold
greater than the weight of the held object
• vertical muscle force exceeds vertical joint
contact force by an amount (W+WO)
• supporting a load of 67 N (15 lb) requires a
muscle force of approximately 650 N (body
weight for a 145 lb individual)
10
typically, a = 4 cm, b = 15 cm,
c = 35 cm, and c/a = 8.75
11
12
Rotational and stabilizing components of
muscle force
Mechanics of the Shoulder
• Muscle forces can be resolved into
rotational and stabilizing components.
muscle in
tension
• The relative magnitude of each depends
on the angle " between the line of action
of the the muscle force FM and the long
axis of the bone upon which it inserts.
• If " is 90 deg, then FM has only a
rotational effect.
• If " is less than 90 deg, then FM has only a
rotational and stabilizing effect.
• If " is greater than 90 deg, then FM has a
rotational and dislocating effect.
load
a
c
WO
FM
FJ
W
b
13
14
Mechanics of the Hip
Known: a = 15 cm, b = 30 cm, c
= 60 cm, ! = 15 deg, W = 40 N,
and WO = 60N
Task: determine FM, FJ, and "
load
muscle in
tension
a
c
WO
FM
15
FJ
16
Review Questions
Example problem
Estimate the hip joint
reaction force FJ generated
during single-legged stance,
as a function of body weight
W. Use the free body
diagram shown at right, and
assume that W2 = 5/6*W,
(c/a) = 2.6, and ! = 70 deg.
• What is the difference between joint contact force and
joint reaction force?
• For a “type III lever” joint (e.g., elbow) is FJ greater or
smaller than FM? What about for a “type I lever” joint
(e.g., hip)?
• What are typical (c/a) ratios for the elbow, shoulder,
hip, and spine?
• What is the magnitude of hip reaction force during
single legged stance?
• How can I tell whether the direction I initially assume
for FJ (in my FBD) is correct?
a
c
17
18