Biomedical Engineering Reference
In-Depth Information
washbasin in which no weight is being lifted, we verify that when the column is just
curved, huge muscle forces are involved. These forces are stronger the more the
column is bent, which happens with taller persons, since the height of the washbasin
is not personalized, but standardized. Besides, the height of everything that
surrounds us is standardized and can be the source of low back pain as is the case
of the height of desks, of chairs, of kitchen sinks or of toilet seats from which tall
and old persons have extreme difficulty in getting up.
Nowadays, many persons are learning to behave correctly, with posture reedu-
cation, in order to avoid pain that results from incorrect posture.
In Example 6.5 we will calculate the value of the forces involved in a spinal
column during its flexion, without bending the knees, using a simplified model. The
obtained results show that they are very reasonable, in spite of the simplified model,
compared to the complexity of the human body.
Example 6.5
Consider the situation of a person bending over to lift a heavy object
from the floor, with straight legs, i.e., without bending of the knees, as shown in the
Fig.
6.2
The column, considered as a rigid body, makes an angle
of 30
to the
horizontal. The weight
W
1
of the trunk of this person is of 300 N and is applied at
the middle of the column. The weight
W
2
of the head + two (arms/forearms/hands)
of 150 N added to the weight of an object of 200 N acts on the upper part of the
column. The muscle force
F
exerted by the erector muscles acts at 2/3 of the
column length that is 70 cm, forming an angle of 12
with the column. The contact
force
C
compresses the intervertebral disc between the sacrum and the fifth lumbar
vertebra. Assuming that the body is in equilibrium, determine the intensities of
F
,
C
and the direction of
C
. The given data refer to a standard Caucasian adult of 70 kg.
We begin solving the problem by decomposing all of the forces in their
components orthogonal to the bar that simulates the column. As the column
makes an angle of 30
with the horizontal, and the sum of internal angles of a
triangle is 180
, the angle of the weight forces
W
1
and
W
2
with the bar is 60
.We
also know that:
cos 60
¼
0.500; sin 60
¼
0.866; cos 12
¼
0.978; sin 12
¼
0.208,
W
1
y
¼
Φ
W
1
sin 60
;
W
1
x
¼
W
1
cos 60
;
W
2
y
¼
W
2
sin 60
;
W
2
x
¼
W
2
cos 60
,
F
sin 12
;
F
x
¼
F
cos 12
,
F
y
¼
C
y
¼
C
sin
α
;
C
x
¼
C
cos
α
.
We apply the condition of total torque equal to zero about the axis of rotation,
conveniently chosen to be in the fifth lumbar vertebra, since the torque due to the
contact force
C
which is unknown becomes zero. We remember that only the
perpendicular component produces rotation:
-
W
1
y
(0.35 m) -
W
2
y
(0.70 m) +
F
y
(0.467 m)
¼
0.
F
sin 12
.
The only unknown term in the above equation is
F
y
¼
3,123.6 N.
Now we apply the condition that the resultant of all of the applied forces on the
column in static equilibrium is equal to zero. Hence, the resultant of their orthogo-
nal components must also be zero.
Hence,
F
¼
