Biomedical Engineering Reference
In-Depth Information
10
3
kg/m
3
,
g
the
acceleration of gravity, and
h
the height of the water column. Equation (
1.6
) shows
that the pressure increases linearly with the height of the water column. As
is the density
2
of water
1 g/cm
3
1,000 kg/m
3
where
ρ
¼
¼
¼
and
g
are constants, if the height of water column is doubled, the pressure is also
doubled.
Example 1.7
Determine the absolute pressure on the body of a person who dives in
a lake to a depth of 10 m.
Using (
1.6
),
p
ρ
¼
(1,000 kg/m
3
)(10 m/s
2
)(10 m)
¼
10
5
Pa which is practically
equal to 1 atm. Therefore, when diving and a depth of 10 m is reached, the absolute
total pressure will be 2 atm, which is the sum of the atmospheric pressure of 1 atm
(at sea level) plus the pressure exerted by the 10 m of the water column.
As
p
is directly proportional to the height
h
, if the diver reaches 20 m, he or she
will be subjected to an absolute pressure of 3 atm.
Example 1.8
Two children are playing on a seesaw, whose arm can incline at a
maximum of 30
relative to the horizontal. The mass of one child is 20 kg and the
other is 21 kg. They are playing well, giving small push offs from the ground. At a
given moment, the child with the smaller mass was up and at rest. Find the pressure
exerted by this child on the board, remembering that her or his contact area with the
board is 300 cm
2
0.03 m
2
.
The weight
W
of the child is (20 kg)(10 m/s
2
)
¼
200 N. The gravitational force
is always perpendicular to the ground. Therefore, we have to find the normal
component
W
y
of the gravitational force on the plane of the seesaw:
¼
W
cos 30
¼
W
y
¼
ð
200 N
Þ
0
:
866
¼
173
:
2N
:
173.2 N/0.03 m
2
;
p
Therefore,
p
¼
¼
5,773.3 Pa.
Example 1.9
Find the pressure on the ground exerted by each foot of a child with
mass of 20 kg, when she or he is standing on two feet. Consider the area of each foot
as being 60 cm
2
.
The weight to consider is
W
100 N, which is the half of the weight of the
child, pressing the ground with the sole of each foot:
¼
100 N/0.0060 m
2
;
p
p
¼
¼
16,667 Pa.
Note the increase in this pressure compared to that of the Example 1.8, since the
area of contact was decreased.
Exercise 1.11
Consider now that the child of 20 kg is standing on one foot only.
Find the pressure exerted by this foot on the ground. Estimate now the pressure on
the ground in case the child stands on the toe of one foot, with a contact area of
8cm
2
?
2
Density of a substance
m
/
V
, that is, the ratio between the mass
m
of substance and volume
V
which contains the mass
m
.
ρ ¼
