Biomedical Engineering Reference
In-Depth Information
(a) Using (
7.12
) and remembering that the variation in velocity is 4.47 m/s, since it
decreases from this value to zero when she stops, we obtain:
F
¼ð
60 kg
Þð
4
:
47m
=
s
Þ=ð
0
:
005 s
Þ;
F
¼
53
;
640 N
;
which is about 90 times the weight of her own body. This is the force applied on
both feet. Hence, on each foot the applied force would be half:
F
¼
26
;
820N
:
In Example 7.4 we found that a compressive force of 50,000 N breaks the
tibia of a person. Therefore, in this case, the woman is within a safety factor.
However, if she were to fall on one leg stiffly, she would not have a safety factor
and the fracture would occur. To decrease this force, it is necessary to dampen
the fall, for example, by bending the knees during the landing, which increases
the deceleration time. That is what athletes do who practice jumping or even
football players who know how to fall, rolling to dampen the landing.
(b)
d
¼
gH
/(
Δ
v
/
Δ
t
)
¼
0.011 m;
d
¼
1.1 cm
According to Dr. A.S. Iutaka, the bone that fractures more easily, when subject
to compression during a fall from a certain elevation with stiff legs, is the calcaneus
of the sole of the foot. It is a bone of the trabecular type with a layer of compact
bone on its surface.
Exercise 7.7
Find the maximum height from which a person with 100 kg mass and
a rigid outstretched leg can fall without breaking the calcaneus. Suppose that in
such a condition the damping distance traveled during the collision with a hard
ground to a stop is 1.0 cm. Consider the compressive strength of the calcaneus the
same as that of a vertebra. Calculate also the duration of the collision. Consider for
this person that the area of calcaneus region that touches the ground is 5.5 cm
2
.
Exercise 7.8
Find the average deceleration force exerted by the ground on each
foot of a person of 70 kg mass during a running activity, in which the height of the
fall of one of the feet before it touches the ground is 10 cm. Consider the case of a
person who does not know how to dampen the fall and the other who knows. In the
first case the interval time of damping is 0.01 s and in the second 0.10 s.
7.10 Answers to Exercises
Exercise 7.1
(a)
T
¼
10 kN; (b)
ε ¼
0.17 %; (c)
T
¼
20 kN.
10
8
Pa; (c) shortening
Exercise 7.2
(a)
r
¼
2.82 cm; (b)
σ ¼
2.8
¼
0.93 cm;
10
5
N.
(d) Weight
¼
5
