Biomedical Engineering Reference
In-Depth Information
oscillates forward and backward, seeming to not have decided where to put the line
of action of the weight force which is along the center of gravity, whether on the
right foot or on the left foot.
We are talking about walking which deals with dynamics, but the forces
involved can be analyzed, considering that the body is in static equilibrium
momentarily, while one of the feet is on the ground.
Example 6.4
Consider a person with 60 kg mass standing erect on her or his right
foot. The mass of each set thigh-leg-foot is of 9.0 kg. Use Fig.
6.1b
and determine
the intensities: (a) of the hip abductor muscle force
F
, exerted by the gluteus muscle
which makes an angle of 70
to the horizontal on the great trochanter of the femur,
and (b) of the contact (reaction) force
C
, exerted by the acetabulum (socket of the
pelvis) on the femur head, as well as the direction of the contact force in relation to
the horizontal.
We locate the rotation axis O, where the contact force
C
is applied. Next we
decompose the forces
F
and
C
in their orthogonal components:
F
sin 70
¼
F
cos 70
¼
F
y
¼
F
0.940 and
F
x
¼
F
0.342,
C
y
¼
C
sin
α
and
C
x
¼
C
cos
α
.
Now we apply the condition that the net torque is zero about the axis of rotation
at O:
0
¼
-
F
y
(0.07 m) -
W
(0.03 m) +
N
(0.11 m),
0
¼
-
F
0.940
(0.07 m) - (90 N)(0.03 m) + (600 N)(0.11 m); from this
equation we obtain:
F
¼
962 N.
329.0 N.
Now we apply the condition that the sum of
y
and
x
components of forces must
be equal to zero:
0
Hence,
F
y
¼
904.3 N and
F
x
¼
¼
F
y
-
C
y
-
P
+
N
¼
(904.3 N) -
C
y
- (90 N) + (600 N),
C
y
¼
1,414.3 N,
0
F
x
-
C
x
,
C
x
¼
¼
329.0 N.
Then, doing the vector sum:
C
y
2
+
C
x
2
C
2
(1,414.3 N)
2
+ (329.0 N)
2
¼
¼
C
1,452 N; as expected,
C
is greater than twice the body weight of this person.
Knowing that (1,452 N)cos
¼
76.9
.
α ¼
329.0, we obtain:
α ¼
arc cos 0.226;
α ¼
Exercise 6.4
Repeat Example 6.4, assuming that the mass of the person has
doubled, i.e., 120 kg. The mass of the set thigh-leg-foot is now 20.5 kg. All the
other values are the same. Discover the consequences of this overweight.
Analyzing the results of Example 6.4 and of Exercise 6.4, we observe that the
relation between the intensities of each of the forces
F
and
C
and the body weight
force
W
is maintained constant. That is:
F
/
W
¼
1.6 and
C
/
W
¼
2.4.
