Ind strial Engineering and Management
Industrial
Human Factors Engineering: Fundamentals
Biostatic Mechanics
Wh t is
What i Biomechanics
Bi h i
Biomechanics is the study of mechanics as it applies to
biological systems
Biomechanics can be subdivided into three disciplines:
Biostatics is the science of the structure of living
organisms in relation to the forces with which they
interact
Bi
Biodynamic
d i studies
t di ththe nature
t off th
the movementt off
living organisms
Bioenergetics is the study of the energy transformation
Wh t is
What i Biostatic
Bi t ti
Biostatic mechanics represent the engineering
fundamentals (statics) necessary for the human factors
engineer
1. Review the basics of static equilibrium
2. Define the specific system to be studied and the
analytical model to be used
3. Apply 1 and 2 in order to obtain useful information for
the human factors engineer
St ti off Rigid
Statics Ri id Bodies
B di
Static Equilibrium Equations
A body may be treated as a combination of a large
number of particles
The analysis of a body will have to consider the size as
well as the forces that will act upon different particles so
that the external forces will be applied at different points
St ti off Rigid
Statics Ri id Bodies
B di
Static Equilibrium Equations
The forces that act upon rigid bodies are divided into two
types:
External forces: represent the action of those forces
outside the rigid body system. These forces are
responsible for the external behavior of the body
IInternal
t l forces:
f are those
th inside
i id the
th rigid
i id b
body
d th
thatt h
hold
ld
the constituent particles together
St ti off Rigid
Statics Ri id Bodies
B di
Static Equilibrium Equations
Newton’s first law: when the resultant force acting upon a
particle is zero, that particle will remain at rest if it was
originallyy at rest or will move with a constant velocityy in a
straight line if it was originally in motion.
For 2-D motion
ΣFx = 0
ΣFy = 0
St ti off Rigid
Statics Ri id Bodies
B di
Static Equilibrium Equations
Newton’s first law: in order for a rigid body to be in static
equilibrium, that body must be not only in translational
equilibrium but in rotational equilibrium.
In order for a rigid body to be in rotational equilibrium,
equilibrium the
net external moment at any point in the body must be
equal to zero
ΣMo = 0
St ti off Rigid
Statics Ri id Bodies
B di
Static Equilibrium Equations
The procedure for analyzing the forces and moments
acting upon a rigid body:
1. Draw the free-bodyy diagram
g and indicate all the known and
unknown external forces
2. Establish an x- and yy- coordinate system
y and indicate
positive directionality for the translational and rotational
movement
3. Apply the required equation for translational and rotational
q
equilibrium
4. Solve the above equations for the unknown parameters
St ti off Rigid
Statics Ri id Bodies
B di
Simply Supported Structures
Two specific connection and support devices used in
mechanical systems are used when performing a static
y
equilibrium analysis of the human musculoskeletal system
y
The hinge joint is a connection device
The cable is a support device
St ti off Rigid
Statics Ri id Bodies
B di
Simply Supported Structures
A hinge joint:
connects beam to another beam of ground
constrains a beam both in x and y directions
permits no translational movement of the beam
cannot constrain rotational forces
rotates if it experiences a net moment
In to satisfy the condition of static equilibrium, the
e moment
net o e abouabout a hinge
ge jo
joint must
us be zeroeo
St ti off Rigid
Statics Ri id Bodies
B di
Simply Supported Structures
A cable:
is used to connect various members of a mechanical
system together
transmits forces from one member to the other
is
i a flflexible
ibl element
l t off th
the mechanical
h i l system
t
sustains only tensile forces
Thee cab
cable
eoof a mechanical
ec a ca sys
system
e will be treated
ea ed as
functionally similar to human skeletal muscle
St ti off Rigid
Statics Ri id Bodies
B di
Example
A horizontal lamp pole which is a uniform horizontal beam
of weight, Wp = 100N, is hinged to the wall at point A. the
other end of the lamp pole ((B)) is supported byy a cable that
makes an angle θ = 50o with the horizontal. The length of
pp
the lamp pole ((beam)) is 1.33m. The lamp p itself has a
weight, WL = 25N, and its center of mass is located at C.
Solve for the force in the cable ((F)) and for the horizontal
and vertical reaction forces at the hinge joint (Rx and Ry).
St ti off Rigid
Statics Ri id Bodies
B di
Solution
Draw the free body diagram
Assign the sign convention
Since A is a hinge joint
joint, it cannot sustain a rotational force
force,
so we will sum the moments about this point
Th
The center
t off mass off pole
l is
i located
l t d att its
it
midpoint
The mass of the lamp is located at the end
of the pole θ
A B
St ti off Rigid
Statics Ri id Bodies
B di
Solution
F
A C θ
Rx B
Ry Wp WL
Refine the FBD to indicate all known parameters
Fy θ = 50o
A C Fx
Rx B
AB = 1.33m
AC = 0.665m
0 665m
Ry (100N) (25N)
Wp WL
St ti off Rigid
Statics Ri id Bodies
B di
Solution
ΣMA = 0 -Wp(AC) - WL(AB) + Fy(AB) = 0
-100(0.665) – 25(1.33) + F (0.766)(1.33) = 0
F = 99.8/1.02 = 97.8N
ΣFx = 0 Rx - Fx = Rx - F (cos θ) = 0
Rx = 97.8(0.643) = 62.9N
ΣFy = 0 Ry - Wp - WL + Fy = Ry - Wp - WL + F (sin θ) = 0
Ry = 100 + 25 – 97.8(0.766) = 50.1N
St ti off Rigid
Statics Ri id Bodies
B di
The Musculoskeletal System
For the purpose of a human factors engineering, the
specific anatomical system is the musculoskeletal system
Human factors engineers employ biostatic analytical
models is two different ways:
1. External forces act upon the human body
2. Internal forces are generated by contracting muscle
St ti off Rigid
Statics Ri id Bodies
B di
The Musculoskeletal System
Having defined the anatomical system of interest and
having developed a suitable analytical model, the HFE
must then identifyy a suitable application in order to obtain
useful information for the design and development
p
process.
U
Upper E
Extremity
t it andd Hand
H d
Sh ld and
Shoulder d Arm
A
The anatomical system for the shoulder and arm is:
1. Proximal segment: Scapula (shoulder bone)
2. Distal segment: Humerus (arm bone)
3. Joint: Shoulder joint
4 Muscle:
4. M l D Deltoid
lt id ((upper shoulder
h ld muscle)
l )
U
Upper E
Extremity
t it andd Hand
H d
Shoulder and Arm
As human skeletal muscles perform a demanding task
over a prolonged, the muscle undergoes fatigue
Muscle fatigue is physically characterized as a progressive
decrease in its ability to generate force
Modifying the task to reduce muscle force can increase
endurance and prolong the time period before onset of
fatigue
U
Upper E
Extremity
t it andd Hand
H d
Example
A volunteer emergency worker is holding a “Detour Right” sign.
U
Upper E
Extremity
t it andd Hand
H d
Example
a. A volunteer emergency worker (of height H = 1.75m and weight W =
700N) mustt continuously
ti l extend
t d his
hi arm straight
t i ht outt so that
th t the
th arm is
i
at right angle (θ = 90o) to the side body. You may neglect the elbow
angle. The “Detour
Detour Right”
Right sign that represents a load weight, WL =
18N. Determine the force exerted by the deltoid muscle (FM) assuming
that it inserts on the arm bone at an angle, α = 30o. Also determine the
vertical reaction force of the body (Ry) and the horizontal reaction
force of the body (Rx) at the shoulder joint.
U
Upper E
Extremity
t it andd Hand
H d
Example
U
Upper E
Extremity
t it andd Hand
H d
Solution
H = 1.75m
W = 700N
WL = 18N
θ = 90o, α = 30o
AB = 0.14m, AC = 0.35m, AD = 0.7m
ΣFy = 0 Fm (sin30) + Ry – 0.05(700) – 18 = 0
ΣFx = 0 -Fm (cos30) + Rx = 0
ΣMA = 0 Fm (sin30)(0.14) – 35(0.35) – 18(0.7) = 0
Fm = 24.85/0.07 = 355N
Ry = 125.5N
Rx = 307N
U
Upper E
Extremity
t it andd Hand
H d
Example
b. As time passes by and fatigue sets in, the extended arm sags to an
angle
l (θ = 65o),
) Find
Fi d th i α = 25o).
the new FM (assuming
( ) H=1 1.75m,
75 W =
700N, WL = 18N
Fm
Fmy
A α
B
θ
Fmx
C
D
35N
18N
U
Upper E
Extremity
t it andd Hand
H d
Fm
Fmy
Solution
A α
θ = 65o, α = 25o, H = 1.75m, θ
B
Fmx
β C
W = 700N, WL = 18N
AB = 0.14m, AC = 0.35m, AD = 0.7m β D
35N
18N
β = 90 – θ = 90 – 65 = 25o,
Fmy = Fm (sin(α+β)) = 0.766 Fm Fmx = Fm (cos(α+β)) = 0.643 Fm
ΣMA = 0
Fmy (AB cos β) – Fmx (AB sin β) – 35(AC cos β) – 18(AD cos β) = 0
0.766Fm (0.14)(0.906) – 0.643 Fm (0.14)(0.423) – 35(0.35)(0.906) –
18(0 7)(0 906) = 0
18(0.7)(0.906)
Fm = 22.5/0.0594 = 379N
U
Upper E
Extremity
t it andd Hand
H d
Example
c. Realizing the extended arm is sagging, the volunteer
b d th
bends the elbow
lb andd also
l re-extends
t d the th upper arm
straight outward so that θ = 90o, the sign, hand, and
lower arm are now held vertically upward at a 90o angle
to upper arm. Find the new FM (assuming α = 30o). H =
1.75m, W = 700N, WL = 18N
0.02 W + WL
0.03 W
0.1 H
0.2 H
U
Upper E
Extremity
t it andd Hand
H d
D
Solution Fm
18N
θ= 90o, α= 35o, H = 1.75m, WL = 18N A α E
C
B
AB = 0.14m, AC = 0.35m, AE = 0.175m θ
21N 14N
ΣMA = 0
Fm (sin 30o)(0.14) – 21 (0.175) – 32(0.35) = 0
Fm = 14.88/0.07 = 213N
Modifying the task by bending the elbow upright,
reduces the deltoid muscle force by 40%
U
Upper E
Extremity
t it andd Hand
H d
Elbow and Forearm
The anatomical system for the elbow and forearm is:
1. Proximal segment: arm bone
2 Distal segment: forearm bone
2.
3. Joint: Elbow joint
4. Agonist Muscle: Triceps
5. Antagonist Muscle: Biceps
U
Upper E
Extremity
t it andd Hand
H d
Elbow and Forearm
HFE Application
Real muscles always act in agonist-antagonist pairs
Many tasks require the interplay of the agonist-antagonist
muscle pairs
D
During
i motion
ti iin one di
direction,
ti th
the agonist
i t iis th
the prime
i
mover in that direction
The muscle generating the predominant force may
alternate between the actual agonist and its opposing
antagonist depending on the nature of the task performed
U
Upper E
Extremity
t it andd Hand
H d
Example
An overhead roller door (hd = 2.1m) has two opposing forces acting upon
it As
it. A the
th ddoor iis llowered,
d th
there iis an upward
d spring
i fforce (Fs) th
thatt
decreases, also during lowering there is a downward door force due to its
weight (FD). A person (H = 1.67m, and W = 550N) is pulling the lower
edge downward with the forearm and hand.
U
Upper E
Extremity
t it andd Hand
H d
Example
a. If the angle of the elbow θ = 25o when the vertical door drop; h =
0 78 ffor a d
0.78m, door weight
i ht Wd = 120N.
120N Fi
Find
d th
the fforce exerted
t dbby th
the
biceps muscle (FB), the force exerted by the triceps muscle (FT), and
the reaction force at the elbow (Ry). upward spring force (Fs) door
force due to its weight (FD), H = 1.67m, and W = 550N.
U
Upper E
Extremity
t it andd Hand
H d
Solution
θ = 25o, hd = 2.1m, h = 0.78m, Wd = 120N, H = 1.67m, and W = 550N.
AB = 0.025m, BC = 0.025m, BD = 0.167m, BE = 0.334m
Fs = 120 – (120/2.10)h = 120 – 57.1(0.78) = 75.4N
Fd = (120/2.10)h = 57.1(0.78) = 44.5N Fs
E
FB
Ry D
Fd
FT 25o
C
B 11N
A
U
Upper E
Extremity
t it andd Hand
H d
Fs
Solution
AB = 0.025m, BC = 0.025m, BD = 0.167m, BE = 0.334m E
FB
FN = Fs – Fd = 75.4 – 44.5 = 30.9N (upward) Ry
D
FT Fd
25o
FB = 0 (the muscle is not stretched)
C
ΣFy = 0 FT – Ry – 11 + 30.9 = 0 A
B 11N
Ry = FT + 19.9
ΣMB = 0 – FT (AB sin25) – 11(BD sin25) + 30.9(BE sin25) = 0
– FT (0.0106) – 0.7771 + 4.3656 = 0
FT = 3.5885/0.0106 = 339N
Ry = 339 + 19.9 = 359N (downward)
U
Upper E
Extremity
t it andd Hand
H d
Example
b. If the angle of the elbow θ = 142o when the vertical door drop; h =
1 41 ffor a d
1.41m, door weight
i ht Wd = 120N.
120N Fi
Find
d th
the fforce exerted
t dbby th
the
biceps muscle (FB), the force exerted by the triceps muscle (FT), and
the reaction force at the elbow (Ry). upward spring force (Fs) door
force due to its weight (FD), H = 1.67m, and W = 550N.
U
Upper E
Extremity
t it andd Hand
H d
Solution
θ = 142o, hd = 2.1m, h = 1.41m, Wd = 120N, H = 1.67m, and W = 550N.
AB = 0.025m, BC = 0.025m, BD = 0.167m, BE = 0.334m
Fs = 120 – (120/210)h = 120 – 57.1(1.41) = 39.5N
Fd = (120/210)h = 57.1(1.41) = 80.5N
FT
Ry
142o FB
A
B D
C Fs
E
11N
Fd
U
Upper E
Extremity
t it andd Hand
H d
Solution
AB = 0.025m, BC = 0.025m, BD = 0.167m, BE = 0.334m
FT
Ry
FN = Fs – Fd = 39.5 – 80.5 = 41N (downward)
142o FB
FT = 0 (the muscle is not stretched) A
B D
ΣFy = 0 FB – Ry – 11 – 41 = 0 C Fs
Ry = FB – 52
E
11N
Fd
ΣMB = 0 FB (BC cos52) – 11(BD cos52) – 41(BE cos52) = 0
FB (0.01546) – 1.1316 – 8.4355 = 0
FB = 9.5671/0.0154 = 621N
Ry = 621 – 52 = 569N (downward)
U
Upper E
Extremity
t it andd Hand
H d
Wrist and Hand
The anatomical system for the wrist and hand is:
1. Proximal segment: forearm bones
2 Distal segment: other wrist bones and the hand bones
2.
3. Joint: Wrist joint
4. Muscle (action): flexion of the fingers towards the palm
of the hand
U
Upper E
Extremity
t it andd Hand
H d
Wrist and Hand
HFE Application
Gripping strength of the hand is a function of the length of
the finger flexor muscles (which run the length of the
forearm)
The optimal length for these muscles (strongest gripping
force) is when the wrist is straight (as when we shake
hands) Wrist flexion
Fm 0.015H
A
B C
AB = 0.15H θ= 90o
BC = 0.055H
Wrist extension
U
Upper E
Extremity
t it andd Hand
H d
Wrist and Hand
HFE Application
Bending the wrist either toward the palm (wrist flexion) or
away of the palm (wrist extension) lowers the grip strength
by under-lengthening or over-lengthening the finger flexor
muscles for their optimal length
Wrist flexion
Fm 0.015H
A
B C
AB = 0.15H θ= 90o
BC = 0.055H
Wrist extension
U
Upper E
Extremity
t it andd Hand
H d
Example
The gripping force (Fg) is a function of wrist joint angle θ. For a person
(f whom
(for h Fg = 150N whenh θ = 0o) whom
h th
then proceedsd tto lift a floor
fl lamp
l
(of weight, WL) by gripping its polished brass pole (with coefficient of
sliding friction, ρ = 0.3). Find the maximum lamp weight (WL) that can be
held before the lamp will slip in the person’s hand
Fg (N)
a. If the wrist is flexed inward toward the p
palm at 150
an angle of 60o (with respect to the wrist straight)
100
b. When the wrist is in straight alignment with
the forearm 50
c. When the wrist is extended outward away from
θ
the palm at an angle of 40o (fl i ) 90
(flexion) 0 00 900 (extension)
( i )
Fg = 150 (cosθ)
U
Upper E
Extremity
t it andd Hand
H d
Solution Ff
Fg = Human grip force Rg = Lamp pole reactions force
WL = Weight of lamp Ff = Gripping frictional force Rg Fg
WL
a. The
Th wrist
i t is
i fl
flexed ) ρ = 0.3
d (θ = 60o), 03
ΣFx = 0 → Rg – Fg = 0 ΣFy = 0 → Ff – WL = 0 Fg (N)
150
Fg = 150 cos60
60o = 75N
ρ = Ff / Fg 100
Ff = ρ Fg = 0.3(75)
0 3(75) = 22
22.5N
5N 50
θ
WL = Ff = 22.5
22 5 (fl i ) 900
(flexion) 00 900 (extension)
( i )
Fg = 150 (cosθ)
U
Upper E
Extremity
t it andd Hand
H d
Solution Ff
Fg = Human grip force Rg = Lamp pole reactions force
WL = Weight of lamp Ff = Gripping frictional force Rg Fg
WL
b. The wrist is straight (θ = 0o), ρ = 0.3
Fg = 150 cos0o = 150N Fg (N)
150
Ff = ρ Fg = 0.3(150) = 45N
WL = Ff = 45N 100
c. The wrist is extended (θ = 40o), ρ = 0.3 50
Fg = 150 cos40o = 115N
θ
Ff = ρ Fg = 0.3(115) = 34.5N (fl i ) 900
(flexion) 00 900 (extension)
( i )
Fg = 150 (cosθ)
WL = Ff = 34.5N
L
Lower Extremity
E t it andd Foot
F t
Hip and Leg
The anatomical system for the hip and leg is:
1. Proximal segment: Pelvis (pelvic bones)
2 Distal segment: Femur (thigh bones)
2.
3. Joint: Hip joint
4. Muscle (action): (abduction of leg: extending the
straightened leg out from the side of the body, as seen
in from the front of a person when kicking a ball
sideways with the side of the foot)
L
Lower Extremity
E t it andd Foot
F t
Hip and Leg
L
Lower Extremity
E t it andd Foot
F t
Hip and Leg
HFE Application
Walking or climbing a stairs require the alternate shifting of
body weight onto one leg and then the other as that leg
contacts the floor (while the other leg swings through the
air)
Thi
This generates
t high
hi h fforces att th
the hi
hip jjoint
i t and
dbbe
physically demanding for people with weak hip muscles or
hi jjoint
hip i didisorders
d
U
Upper E
Extremity
t it andd Hand
H d
Example
A person (H=1.6m and W=500N) is descending a series of steps that do
nott have
h ahhandrail.
d il Th
The person hhas th
the right
i ht lleg ffully
ll extended
t d d and
d iin
ground contact while the left leg swings
through the air. Find Fm (the hip abductor
muscle force) and Rh (the resultant hip joint
reaction force)
The center of mass of the body must be
aligned directly over the right foot
U
Upper E
Extremity
t it andd Hand
H d
Solution
H = 1.6m, W = 500N, AB = 0.064m,
AD = 0.16m, AC = ½ AD = 0.08m,
ΣFy = 0 → Fm(sin70) – Ry – 75 + 500 = 0
Ry = (0.94)Fm + 425
ΣFx = 0 → Fm(cos70) – Rx = 0
Rx = (0.342)Fm
ΣMB = 0
– Fm(sin70)(0.064) – 75(0.08 – 0.064)
+ 500(0.16-0.064)
500(0 16 0 064) = 0
Fm = 46.8/0.0602 = 777N
U
Upper E
Extremity
t it andd Hand
H d
Solution
Ry = (0.94)Fm + 425
Ry = 1155N
Rx = (0.342)Fm
Rx = 266N
Rh = √ Rx2 + Ry2 = 1185N
at an angle = tan-1 (Ry/Rx) = 77o
U
Upper E
Extremity
t it andd Hand
H d
Example
A person (H=1.6m and W=500N) is
d
descending
di a series
i off steps
t ffor which
hi h a
handrail has been provided. The person
now grips the handrail with the left hand
and offloads 15% of
body weight (Rhr). Find Fm and Rh
The center of mass of the body
is at L distance from the right
foot
U
Upper E
Extremity
t it andd Hand
H d
Example
A person (H=1.6m and W=500N) is descending a
series of steps for which a handrail has been
provided. The person now grips the handrail with the
left hand and offloads 15% of
body weight (Rhr). Find Fm and Rh
The center of mass of the body
is at L distance from the right
f t
foot
U
Upper E
Extremity
t it andd Hand
H d
Solution
H = 1.6m, W = 500N,
RH = 0.15 W = 75N at a distance 0.17H = 0.27m
ΣFy = 0 → Rg + 75 – 500 = 0
Rg = 425N
ΣMO = 0 → – Rg(L) + RH(0.17 H) = 0
L = 20.25/425 = 0.048m
U
Upper E
Extremity
t it andd Hand
H d
Ry
Fm
Solution
A Rx
H = 1.6m, W = 500N,
B
AB = 0.064, AE = 0.16m, DE = 0.048m,
AD = 0.112m, AC = ½ AD = 0.056m
where E is the midline of the body C
75N
ΣFy = 0 → Fm(sin70) – Ry – 75 + 425 = 0
Ry = (0.94)Fm + 350
E
D
ΣFx = 0 → Fm(cos70) – Rx = 0 425N
Rx = (0.342)Fm
U
Upper E
Extremity
t it andd Hand
H d
Solution
Ry
Fm
ΣMB = 0
A Rx
– Fm(sin70)(0.064) + 75(0.064 – 0.056) + (425)
B
(0.112 – 0.064) = 0
FM = 21/0.0602 = 349N
C
75N
Ry = (0.94)Fm + 350 Rx = (0.342)Fm
Ry = 678N Rx = 119N
E
D
Rh = √ Rx2 + Ry2 = 688N at an angle = tan-1 (Ry/Rx) = 80o 425N
The hip reaction force is reduced by 42% using a hand rail
L
Lower Extremity
E t it andd Foot
F t
Knee and Foreleg
The anatomical system for the hip and leg is:
1. Proximal segment: thigh bones
2 Distal segment: shin bones
2.
3. Joint: knee joint
4. Muscle (action): with the flexed (back heel toward the
back of the thigh), extended the foreleg (shin bone)
forward (as in kicking a football)
L
Lower Extremity
E t it andd Foot
F t
Knee and Foreleg
U
Upper E
Extremity
t it andd Hand
H d
Example
A person (H=1.83m) has shifted all his
weight W = 670N onto his right foot at the
weight,
point where the left foot is h = 1cm off the
ground. If the step rise R = 22cm, find the
force exerted by the quadriceps muscle
(Fm), and find the reaction forces at the
k
knee jjoint
i Ry andd Rx.
U
Upper E
Extremity
t it andd Hand
H d
Solution
H = 1.83m, W = 670N, h = 1cm, R = 22cm,
AC = 0
0.439m,
439m CD = 0.531m,
0 531m
AE = 0.53H + 0.01 = 0.98m
CE = 0.751m,
0 751m AF = AE-CE = 0.229m
0 229m
F
C
75N
D
h = 0.01m R = 0.22m
E
U
Upper E
Extremity
t it andd Hand
H d
Solution
A
α
AC = 0.439m, CD = 0.531m, AE = 0.98m φ
570N
CE = 0.751,
0 751 AF = 0 229 Δy = 0.055m,
0.229, 0 055m AB = 0
0.22m
22m B Fm
Δy
67N
F θ Rx
C
θ = sin-1 (AF/AC) = 31
31.4
4o φ = tan-1 (Δy/AC) = 7.1
7 1o
Ry
α = θ – φ = 24.3o
Ry = Rg – 0.05W
0 05W
ΣFy = 0 → – Fm(sin24.3) – 570 – 67 + Ry = 0
Ry = (0.412)F
(0 412)Fm + 637
ΣFx = 0 → Fm(cos24.3)
(cos24 3) – Rx = 0
Rx = (0.911)Fm
U
Upper E
Extremity
t it andd Hand
H d
Solution
A
α
AC = 0.439m, CD = 0.531m, AE = 0.98m φ
570N
CE = 0.751,
0 751 AF = 0 229 Δy = 0.055m,
0.229, 0 055m AB = 0
0.22m
22m B Fm
Δy
67N
θ = 31.4o φ = 7.1o α = 24.3o F θ Rx
C
Ry
ΣMC = 0
Fm(sin24.3)(0.439
(sin24 3)(0 439 0.055)(cos31.4)
0 055)(cos31 4) + 570(0
570(0.439
439 )(cos31
)(cos31.4)
4)
–Fm(cos24.3)(0.439 0.055)(sin31.4) + 67(0.22 )(cos31.4) = 0
FM = 226.3/0.054
226 3/0 054 = 4191N this is 6 times the body weight
Ry = (0.412)Fm + 637 Rx = (0.911)Fm
Ry = 2364N Rx = 3818N this force is applied between the kneecap
and knee joint
U
Upper E
Extremity
t it andd Hand
H d
Example
The person now modifies his step climbing
by springing off with the left foot and
bending forward at the waist, find the force
exerted by the quadriceps muscle (Fm), and
find the reaction forces at the knee joint Ry
and Rx.
(H = 1.83m, W = 670N,
h = 10cm, R = 22cm)
0 25 W
0.25
0.7 W
U
Upper E
Extremity
t it andd Hand
H d
Solution
H = 1.83m, W = 670N, h = 10cm, R = 22cm,
AC = 0.439m, CD = 0.531m,
AE = 0.53H + 0.10 = 1.07m
CE = 0.751m, AF = 0.319m
F
0.25 W
C
0.7 W
75N
D
h = 0.1m R = 0.22m
E
U
Upper E
Extremity
t it andd Hand
H d
A
Solution α
φ
0.167N
AC = 0.439m, CD = 0.531m, AE = 1.07m Δy
B Fm
CE = 0.751,
0 751 AF = 0 319 Δy = 0.055m,
0.319, 0 055m AB = 0
0.22m
22m F
469N
θ Rx
θ = sin-1 (AF/AC) = 46
46.6
6o φ = tan-1 (Δy/AC) = 7.2
7 2o Ry
α = θ – φ = 39.4o
ΣFy = 0 → – Fm(sin39.4) – 469 – 167 + Ry = 0
Ry = (0.636)F
(0 636)Fm + 636
ΣFx = 0 → Fm(cos39.4)
(cos39 4) – Rx = 0
Rx = (0.772)Fm
U
Upper E
Extremity
t it andd Hand
H d
A
Solution α
φ
0.167N
AC = 0.439m, CD = 0.531m, AE = 1.07m Δy
B Fm
CE = 0.751,
0 751 AF = 0 319 Δy = 0.055m,
0.319, 0 055m AB = 0
0.22m
22m F
469N
θ Rx
θ = sin-1 (AF/AC) = 46.6o φ = tan-1 (Δy/AC) = 7.2o C
α = θ – φ = 39.4
39 4o Ry
ΣMC = 0
Fm(sin39.4)(0.439)(cos46.64)
(sin39 4)(0 439)(cos46 64) + 570(0
570(0.439
439 )(cos46
)(cos46.6)
6)
–Fm(cos39.4)(0.439)(sin46.6) + 67(0.22 )(cos46.6) = 0
121 3/0 054 = 2250N this is 3
FM = 121.3/0.054 3.5
5 times the body weight
Ry = (0.412)Fm + 637 Rx = (0.911)Fm
Ry = 2070N Rx = 1740N this force is applied
between the kneecap and knee joint
L
Lower Extremity
E t it andd Foot
F t
Ankle and Foot
The anatomical system for the hip and leg is:
1. Proximal segment: shin bone
2 Distal segment: ankle bone
2.
3. Joint: ankle joint
4. Muscle (action):
U
Upper E
Extremity
t it andd Hand
H d
Example
The force required to press a pedal = 50N. Identify the weight of the foot
(Wf),
) the angle at the ankle (θ)
(θ), and the force exerted by the muscle (FM)
muscle (Fm), and find the reaction forces at the ankle Ry and Rx.
(H = 1.64m, W = 600N, h = 10cm)
FM (N)
150
100
50
θ
1360 1000 640
FM = (150) sin[2.5(θ-64)]
U
Upper E
Extremity
t it andd Hand
H d
Solution
H = 1.64m, W = 600N, h = 10cm, F = 50N
50 = (150) sin[2.5(θ
sin[2 5(θ – 64)] θ = 179.5/2.5
179 5/2 5 = 71
71.8
8o
α = 90 – 71.8 = 18.2o AD = AC + CD = 0.15 H = 0.264m
HFP = 0.001
0 001 + AD sin
sinα = 0.078m
0 078m
FM (N)
150
100
α
50
θ
1360 1000 640
FM = (150) sin[2.5(θ-64)]
U
Upper E
Extremity
t it andd Hand
H d
Solution
H = 1.64m, W = 600N, h = 10cm, FFP = 50N, BC = 0.098m, AC = 0.197m
0 049m φ = 10o, α = 18.2
CD = 0.049m, 18 2o
ΣFy = 0 → Fm(cos 10) + 50 – 9 + Ry = 0 Ry = – (0.985)Fm – 41
ΣFx = 0 → – Fm(sin 10) + Rx = 0 Rx = (0.174)F
(0 174)Fm
FM (N)
150
100
Rx
α
50
Ry
θ
1360 1000 640
FM = (150) sin[2.5(θ-64)]
U
Upper E
Extremity
t it andd Hand
H d
Solution
H = 1.64m, W = 600N, h = 10cm, FFP = 50N, BC = 0.098m, AC = 0.197m
0 049m φ = 10o, α = 18.2
CD = 0.049m, 18 2o
ΣMC = 0 Fm(cos10)(CD)(cos18.2) + 9(BC)(cos18.2) –
50(AC)(cos18 2) – Fm(sin10)(CD)(sin18.2)
50(AC)(cos18.2) (sin10)(CD)(sin18 2) = 0 Fm = 8.52/0.043
8 52/0 043 = 198N
Ry = 236 N (downwards) Rx = 34.5N (rightward)
FM (N)
150
100
Rx
α
50
Ry
θ
1360 1000 640
FM = (150) sin[2.5(θ-64)]
B di
Bending, Lifting,
Lifti andd Carrying
C i
W = body weight
WL = carried load
B di
Bending, Lifting,
Lifti andd Carrying
C i
The horizontal and vertical reaction
forces (Rx and Ry) at the lower joint
translated to:
Ra axial reaction force along the
central axis of the spine.
p Excessive
axial force can result in the fracture of
vertebral bones and/or injury of
intervertebral disc
Rs shear reaction force perpendicular
t the
to th axisi off the
th spine.
i Excessive
E i
shear force can result in the
dislocation of adjacent vertebral bones
also injury to the intervertebral disc
B di
Bending, Lifting,
Lifti andd Carrying
C i
Example:
A person is carrying a single load WL = 0.2W on the back, determine θ at
which the exterior muscle force Fe = 0. Calculate the reaction force ((Ra)
and shear force (Rs)
B di
Bending, Lifting,
Lifti andd Carrying
C i
Solution:
WL = 0.2W, Fe = 0
BE = 0 12H BF = AB cos θ = 0.15H
0.12H, 0 15H cos θ,
θ EF = 0
0.12H 0 15H cos θ
12H – 0.15H
ΣFy = 0 → Ry – 0.36W – 0.2W – 0.18W = 0
Ry = 0.74W
0 74W
ΣFx = 0 → – Fe(cos(θ – 13) + Rx = 0 Rx = 0
F
ΣMA = 0
0.2W(0.12H – 0.15H cos θ) – 0.36W (0.15H cos θ)
– 0.18 (0 3H cos θ) = 0
0 18 W (0.3H
Divided by W and H
0 3 cos θ + 0.054
0.3 0 054 cos θ + 0.054
0 054 cos θ = 0.024
0 024
cos θ = 0.024/0.138 = 0.174 θ = 80o
B di
Bending, Lifting,
Lifti andd Carrying
C i
A person is lifting a load WL = 0.2W,
0 2W calculate the extersor muscle force
Fe directed at angle 13o, the reaction force (Ra) and shear force (Rs) for:
a. θ = 35o
b. θ = 75o
B di
Bending, Lifting,
Lifti andd Carrying
C i
Calculate the extersor muscle force Fe directed at angle 13o, the reaction
force (Ra) and shear force (Rs) for (the additional load is zero):
a. θ = 85o
b. θ = 75o
c. θ = 60o
