 Group of structures connecting arm to

the thorax
 Function as a unit in bearing loads &
absorbing shock
 Large moment arms⇒ large torques:
counteracted by shoulder muscles
 remarkable range of motion, most
mobile joint
 Can abduct, adduct, rotate, be raised in
front of and behind the torso and move
through a full 360° in the sagittal plane
 Extremely unstable, far more prone to
dislocation and injury than other joints
 Soft tissue envelope encircling the
glenohumeral joint& attached to the
scapula, humerus, and head of the biceps
(Capsule)
 Glenohumeral joint
 main joint of the shoulder
 the articulation between the head of the
humerus and the lateral scapula
 multiaxial ball & socket
 most mobile, supports greater loads

 Acromioclavicular joint
 between the acromion process of the
scapula (part of the scapula that forms the
highest point of the shoulder) and the
distal end of the clavicle
 irregular diarthrodial joint-limited motion
in all 3 planes
 Rotation during arm elevation
 Sternoclavicular joint  Coracoclavicular joint
 at the proximal clavicle with the  Syndesmosis with the coracoid
process of the scapula and the
manubrium inferior surface of the clavicle bound
 provides major axis of rotation for together by the coracoclavicular
movements of the clavicle and scapula ligament
 Permits little movement
 modified ball& socket
 shrugging the shoulders, elevating the
arms above the head, swimming

 Scapulothoracic joint
 false joint where the shoulder blade
glides against the thorax (the rib cage)

 Synchronous, simultaneous
contributions from each joint
⇒ Wide range of motion
1. If the weight of the arm is 33N, the Fm
moment arm for the total arm 3 cm
segment is 30cm, and the moment R
30 cm
arm for the deltoid muscle(Fm) is
3 cm, how much force must be 33N
supplied by the deltoid to
maintain the arm in this position?
What is the magnitude of the joint Ʃ MS= 0:
reaction force(R)? (Fm x 30) N-mm= (33 x 300) N-mm
Fm = 330 N

Ʃ Fx = 0:
R = Fm
 2. Repeat problem 1 assuming that 3. The medial deltoid attaches to the
the person is holding a ball of humerus at an angle of 150 as shown
weight 5 kg in the outstretched below. What are the sizes of the
hand, at a distance 60 cm from the rotary and stabilizing components of
shoulder joint. muscle force when the total muscle
Fm force is 500N?
3 cm
R
30 cm
60 cm

33N 5 kg
Ʃ MS= 0:
(Fm x 30) N-mm= [(33 x 300)+(50x600)] N-mm
*Taking g =10 m/s2
Fm = 1330 N
 Rotary component : Fm Sin 150
 Stabilizing component : Fm Cos150
 Calculate the force needed by deltoid muscles, fixed at an angle 160
with humerus, to hold up the arm in the position as shown in figure
below. Given weight of the arm=68N, weight in hand=45N.
Also calculate the reaction force at shoulder.
 Ball& and socket joint with a high degree
of stability and excellent range of
movements
 Articulation between spherical head of
the femur & concave acetabulum of the
pelvis
 Pelvic gurdle : two hip bones + sacrum
->rotate forward, backward, and laterally
 Flexion/extension/hyperextension
Abduction/adduction
Lateral/medial rotation
 Neck of femur ~1350 to the shaft
 3 DOF with 3 mutually perpendicular axes intersecting
the geometric center of rotation of the spherical head
Transverse axis: lies in the frontal(coronal) plane and
controls flexion/extension
Anterior/posterior axis: lies in sagittal plane, controls
adduction/abduction
Vertical axis: coincides with long axis of the limb when the
hip joint in neutral position, controls movements of internal
and external rotation
 Major wt bearing joint:
1/3W supported at each hip during upright standing
~W during swing phase of walking
3-4 W during support phase
5.5 W during fast walking, jugging
8.7 W during stumbling
 Body wt, tension in large, strong hip muscles & impact forces
translated upward add to compression on the joint
 The hip abductors act during stance to oppose the hip adductor
moment that gravity produces, and to keep the pelvis level
 Both gravity (body weight) and abductor muscles produce forces
that compresses the hip joint
 As abductor muscle’s force exceeds that of body weight, most of the
compression that the hip joint surface experience during single limb
stance, walking, or running actually comes from muscle action
 F x  Rx  Px  Rx  2W sin   0  Rx  W (  300 )

F y  Ry  Py  B  Ry  2W cos   5W / 6  0
 Ry  (2 cos 300  5 / 6)W  2.57W

R  ( Rx 2  Ry 2 )1/ 2  2.75W
c
b Direction : tan 1 ( Ry / Rx )  68.70
Ѳ

P
 M  cP  bB  0
5b 5W
R B P  (b / c) B  W ...( B  )
6c 6
 P  2W (b / c  2.4)

B = weight of the body

(minus the amputated leg)
P = abductor muscle force
R = joint reaction force
Ѳ = angle that the abductor muscle line of
#Assume the person is standing on one leg that has been
action makes with y-axise
amputated to reveal the force vectors acting there-the other leg
not touching the ground.
 Horizontal force of the femur on the pelvis is equal to body wt,
and the vertical force is 2.5 times as much

 Force required in the abductor muscles to balance the body on

the head of the weight bearing femur during the stance phase of
gait is twofold body wt

 Magnitude of the forces in the hip joint depends on the ratio of

the body weight moment arm to the abductor muscle moment
arm, and the angle that the abductor muscle force makes with the
vertical.
 Supposethat the distance b in women’s pelves is 10% greater
than men’s because of their greater pelvic width. How does this
change their abductor muscle force magnitude and the
magnitude & direction of their hip joint reaction force, assuming
the moment arm of the abductor muscles is independent of
gender?
Men Women
b b’=1.1 b
P P’=?
R R’=?
 1. How much compression acts on the hip
during two-legged standing, given that
the joint supports 250N of body weight
and the abductor muscles are
producing 600N of tension?
Ɵ=200, B=250N
P=2.4 B= 600N
Rx=Psin20=205.2N, Ry=Pcos20=813.8N
R=sqrt (Rx2+Ry2]=839.28 N

700 Graphical Method

P
P Cosine Law:
F
R2=P2+B2-2PBcosƟ
B R Ɵ= ?
Ɵ
= 1800-200=1600
B
=> R= 839 N
2. What is the magnitude of the reaction force at the hip when
tension in the hip abductors is 750N and 300N body weight is
supported?

3. How much compression acts on the hip during two-legged

standing for a 70kg person if the abductor muscles are
producing 600N tension?
*Hint: B=W/3
 Bears tremendous loads-most heavily stressed joint
 Provides mobility& stability
 Very stable during complete extension, very mobile during flexion
 Tibiofemoral & Patellofemoral articulations
 Shape of the femoral surfaces complementary to the tibial plateaus
 Combination of rolling and gliding between femur& tibia
 During flexion, wt bearing surfaces move backward on the tibial
plateaus& become progressively smaller
 Geometry of patellofemoral articular surfaces remains relatively
constant as the knee flexes
 Patellofemoral contact area < tibiofemoral contact area
 Flexion/extension in the sagittal plane: 0-1300
 Axial rotation around the long axis of the tibia~100
 Rolling/gliding motion of patella during flexion
 ccw rolling between 0-900 , cw between 90-1200
 Mean amount of patellar gliding:
6.5mm/100 flexion(0-800)
 4.5mm/100 flexion(80-1200)
 Quadriceps- Straighten the knee: Extension
 Hamstrings- Bend the knee: Flexion
Patellofemoral joint Tibiofemoral Joint
 Compressive force:  Compressive force:
0.5B, normal walking gait  ≥3*B during stance phase(medial tibial
≤ 3B, stair climbing plateau bearing most of the load)
 upto 7.6B, squat exercise up to 4*B during stair climbing
 Compression increases with knee  Menisci
flexion:
distributes loads over a broader area,
 Increase in knee flexion increases the
reducing the magnitude of joint stress
compressive component of force acting
at the joint(figure)  bears 45% of the total load
 Larger amount of quadriceps  Articular cartilage on the medial
tendon required to prevent the plateau=3* thicker than on lateral plateau
knee from buckling against gravity  Shear force increases for flexion>900
1. How much compression acts on
the patellofemoral joint when the
quadriceps exert 300N of tension
and the angle between the
Compression at the patellofemoral joint is the
quadriceps and the patellar 
vector sum of tension in the quadriceps and
tendon is a)1600 b)900 the patellar tendon

2. Repeat the problem if the

quadriceps exert 400N of tension
and the angle between the
quadriceps and the patellar
tendon is a)1400 b)1000

• Patellofemoral compression increases with knee

flexion during weight bearing
1. How much compression acts on the patellofemoral joint when
the quadriceps exert 300N of tension and the angle between the
quadriceps and the patellar tendon is a)1600 b)900

 Parallelogram Law :
R2= P2+Q2+2PQcosƟ Fm

C
C2=Fm2+Ft2+2FmFtcosƟ
Fm =Ft =300N
Ft
C=104.2 N for Ɵ =1600
C= 424.26 N for Ɵ =900
3.The upper leg muscle
(quadriceps) exerts a force of
1250N, which is carried by a tendon
over the kneecap (patella) at the
angles shown in fig. Find the
direction and magnitude of the
force exerted by the kneecap on the R= ?

upper legbone (femur).

Use Parallelogram law.

Direction: ф =tan-1(QSinƟ/(P+QcosƟ))