KIN 840
Static Equilibrium
© 2006 by Stephen Robinovitch, All Rights Reserved
2006-1
Outline
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Characteristics of forces and moments
Equations of static equilibrium
Two and three force members
Free body diagrams
Muscle forces versus muscle moments
Statically indeterminate systems
Components of force
Example. A force of
magnitude 100 N is
directed as shown.
Determine
corresponding x, y,
and z components of
force.
Vector addition of forces
Example. Determine
the magnitude and
direction of the
resultant force c.
70º
30º
How much force must be applied to move
an object against friction?
Fm=µsN
W
Fk = µ k N
P
frictional force F
Fm
N
Fk
Static
equilibrium
Motion
applied tangential force P
F
Frictional force F
equals applied
tangential force
P...up to a limit Fm
Fm depends on:
• coefficient of static
friction µs
• normal force N at
interface
Moment is force times distance
the moment that a force
produces about a fixed
point O is equal to the
force times the
perpendicular distance
from O to the line of
action of the force
r
v
M O = Fd !
d ! = " moment arm"
Right Hand Rule
• the direction of MO is
– perpendicular to the plane formed
by F and O
– determined by the “right hand rule”
• right hand rule: identify the direction
that the moment tends to cause the
body to rotate about point O
• orient your right hand so this
direction is illustrated by the act of
curling your fingers into increased
flexion
• your thumb will be pointing in the
direction of the moment
representation that
shows the direction
of MO
r
F
r
M
Muscles Create Moments
The ability of muscles to
stabilize the body under
external loads, and produce
human body movements, is
due their ability to create
moments (or torques) about
joints.
If the line of action of a
muscle force passes through
a joint axis of rotation, it
cannot stabilize or rotate that
joint.
Muscles have small moment arms
Compared to external loads, muscles have
small moment arms with respect to joint
they span.
Advantage: given the limited deformation
that muscles can undergo, their close
proximity to joints allows for large ranges
of joint rotation
Disadvantage: high muscle forces (and
therefore joint forces) are required to
balance external loads
Multiple forces can be replaced with a
force and a moment
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Multiple forces applied to a rigid body can be replaced by a
single resultant force and a single moment, acting at any
location on the body.
The location will affect the resultant moment, but not the
resultant force.
Multiple
external loads
Equivalent
resultant load
and moment
Newton’s First Law of Motion
If the resultant force acting on a particle is zero,
the particle will remain at rest, or remain moving
with constant speed in a straight line.
Whiplash example: hit
from behind, but head
snaps backward
Newton’s Second Law of Motion
It the resultant force F acting on a particle of mass
m is not zero, the particle will accelerate linearly
in the direction of the force according to the law
F=ma.
It the resultant moment M acting on a body of
mass moment of inertia I is not zero, the body will
have an angular acceleration α, defined by M=Iα.
In component
form:
ΣFx = max
ΣFy = may
ΣFz = maz
ΣMx = Iαx
ΣMy = Iαy
ΣMz = Iαz
Equations of equilibrium for
static conditions.
If inertial forces (i.e., products ma and Iα) are
zero or neglible, then Newton’s equations of
motion reduce to:
ΣFx = 0
ΣFy = 0
ΣFz = 0
ΣMx = 0
ΣMy = 0
ΣMz = 0
Example. Two-force member in
static equilibrium.
"F
x
"F
y
vertical reaction
force FBy
=0:
=0:
If only two forces FA and
FB are applied to a body,
then static equilibrium can
only exist if FB = - FA.
B
A
horizontal
reaction
force FBx
Y
30 deg
applied force FA
X
Example. Two-force member in
static equilibrium.
+
"M
A
=0:
FB
150 deg
A
B
If only two forces FA and
FB are applied to a body,
then static equilibrium can
only exist if the line of
action of FA passes through
point B, and the line of
action of FB passes through
point A.
B
A
30 deg
d
FA
Three-force member in static
equilibrium
"F = 0 :
indicates that FA , FB , and FC must
be in the same plane (coplanar).
"M = 0:
indicates that FA , FB , and FC must
interest at a common point O, or have
parallel lines of action
If only three forces FA, FB, and FC are
applied to a body, then static equilibrium
can only exist if the forces are coplanar,
and intersect at a common point O (or have
parallel lines of action).
FC
FB
C
B
O
A
FA
How do we apply Newton’s Laws to
solve a specific problem?
• need a model of the system of
interest
• model needs to account for
- external forces
- gravitational forces
- internal (muscle and joint
reaction) forces
•
• must know point of application
of forces
Common assumptions involved in
musculoskeletal models
• inertial forces are negligible
(allows for static analysis)
• one muscle (or a “lumped”
muscle group) acts to support the
external load
• the force in antagonist muscles is
zero (no co-contraction)
• lines of action of muscle forces
are known
Free Body Diagrams (FBDs)
Free body diagram (FBD):
visual representation of the
musculoskeletal system of
interest, that guides us in
applying Newton’s second
law to solve for unknowns
(aid to model development).
Isolates the system under
consideration and shows all
the forces acting on it
What’s involved in developing an FBD
1. use geometric boundary to isolate
system of interest from outside
world
2. consider all forces acting on the
system that cross the boundary,
and gravitational forces
3. apply equations of static
equilibrium. In 2D, these are:
ΣFx = 0; ΣFy = 0; ΣMO = 0
Forces to consider in FBD’s
At the FBD border:
– contact forces applied on the system by the external
world (e.g., ground reaction force (acts up), weight
of a carried object (acts down))
– joint reaction forces, often (but not necessarily)
assumed compressive, and directed inward from the
border
– forces in muscles, always tensile and directed
outward from the border
Inside the FBD border:
– gravitational forces associated with weight of system
components, located at appropriate centres of mass
Example. Biceps force to carry a
hand-held load
load
muscle in
tension
a
FJ
c
WO
FM
W
b
Solve for the joint reaction force FJ and muscle force
FM.
"F
y
"M
=0:
O
=0:
W = 20 N
b/a = 3
force (N)
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))
2
4
6
8
ratio c/a
10
typically, a = 4 cm, b = 15 cm,
c = 35 cm, and c/a = 8.75
Remarks - biceps force to carry a
hand-held load
• the joint reaction force is approximately 8-fold
greater than the weight of the held object
• the muscle force exceeds the 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)
Example. Deltoid force to support
a hand-held load
muscle in
tension
load
a
c
WO
FM
FJ
W
b
Determine FM, FJ, and
β.
Known: a = 15 cm, b =
30 cm, c = 60 cm, θ =
15 deg, W = 40 N, and
WO = 60N
Example. For the loading
condition shown, calculate
the erector spinae muscle
force FM and the
compressive and shear
components of joint
0.05 m
reaction force (FJC and FJS)
at the L5/S1 vertebrae (red
square).
Is FJC greater than the
maximum safe value of 3.4
kN recommended by the
U.S. National Institute for
Occupational Safety and
Health (NIOSH)?
35 deg
450 N
O
FJS
0.25 m
FM
FJC
35 deg
0.4 m
Y
200 N
X
Concept of “muscle moment”
Recall that multiple forces applied to a rigid body can
be replaced by a single resultant force and a single
moment, acting at a specific location on the body.
In joint mechanics, we use this concept to define:
net joint force = vector sum of bone-on-bone joint
reaction force and muscle forces spanning the joint
muscle moment = moment at joint centre created by all
muscle forces spanning the joint
Example. What is the muscle moment and net
joint force at the hip during single-legged stance?
=
0.83W
0.83W
MM =
0.11W Nm
2.17W
0.13 m
2.96W
swing
leg (not
contacting
ground)
0.83W
swing
leg (not
contacting
ground)
Example. What is the muscle moment and net joint
force at the L5/S1 vertebrae (red square)?
35 deg
450 N
0.05 m
O
FJS
0.25 m
450 N
=
MM
O
FNET 0.25 m
FM
FJC
0.4 m
200 N
0.4 m
200 N
Example. What is the muscle moment and net joint
force at the ankle during the final (push-off) stage
of stance?
FNET
FJ
a
MM
FM
b
=
W
W
Pros and cons of using muscle moments
versus muscle forces
MUSCLE FORCES
advantage:
• allows estimate of muscle force
and joint reaction force (often
required to determine injury risk,
etc.)
disadvantage:
• works only for models that
include one muscle
• requires knowledge of muscle
geometry, including moment arm
of the muscle
MUSCLE MOMENTS
advantage:
• allows for an estimate of net
moment caused by contraction of
muscles spanning the joint,
without having to account for
muscle geometry
• compatible with strength
measures
disadvantage:
• allows for an estimate net joint
force, but cannot determine joint
reaction force or muscle force
Determining muscle moments in
multi-segment systems
Example: Specify the
direction of muscle
moments at the
ankle, knee, and hip
for the individual
shown at right.
W
W-WS+F
W-WT+S+F
MH
MK
W-WFOOT
WS+F
WT+S+F
MA
WF
W
W
W