Biomedical Engineering Reference
In-Depth Information
FIGURE 11.6
y
Free-body diagram for a hip joint.
F
H
F
A
φ
x
O
9 cm
35 cm
88 cm
W
L
W
Substituting these two relationships into the third equation,
ð
5
:
785 cm
ÞF
H
cos
ðφÞ
2
ð
6
:
89 cm
ÞF
H
sin
ðφÞ
52
60
;
864 Ncm
5
:
785 cm
ð
0
:
259
F
A
Þ
2
6
:
89 cm
ð
0
:
966
F
A
1
2600 N
Þ
52
60
;
864 Ncm
ð
1
:
5cm
ÞF
A
2
ð
6
:
566 cm
ÞF
A
2
17
;
914 Ncm
60
;
864 Ncm
52
ð
2
5
:
066 cm
ÞF
A
52
42
;
949 Ncm
F
A
5
8478 N
F
H
cos
ðφÞ
5
0
:
259
ð
8478 N
Þ
5
2196 N
F
H
sin
ðφÞ
5
0
:
966
ð
8478 N
Þ
1
2600 N
10
;
790 N
5
q
ðF
H
cos
q
ð
2
2
2
2
F
H
5
ðφÞÞ
1
ðF
H
sin
ðφÞÞ
Þ
1
ð
;
Þ
5
2196 N
10
790 N
5
11011 N
10
790 N
2196 N
;
tan
2
1
5
φ
5
78
:
5
We show this analysis here because by using some simple assumptions, it is easy to cal-
culate the forces that arise within a joint. From force, we can calculate the stress that acts
on the cartilage and the strain of the cartilage. This is useful because the rate of synovial
fluid formation from the cartilage, during loading conditions, is dependent on the com-
pression of the cartilage. With a higher compression, more synovial fluid will be released
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