fluid_mechanics

62 Chapter 2 ■ Fluid Statics
is the weight of the gate and and are the horizontal and
vertical reactions of the shaft on the gate. We can now sum moments
about the shaft
and, therefore,
a M c 0
M F R 1y R y c 2
11230 10 3 N210.0866 m2
1.07 10 5 N # m
O x O y
0.5
(Ans)
y R – y c , m
0.4
0.3
0.2
0.1
(10m, 0.0886 m)
0
0 5 10 15
h c , m
20 25 30
F I G U R E E2.6d
E XAMPLE 2.7
Hydrostatic Pressure Force on a Plane Triangular Surface
GIVEN An aquarium contains seawater 1g 64.0 lbft 3 2 to a
depth of 1 ft as shown in Fig. E2.7a. To repair some damage to
one corner of the tank, a triangular section is replaced with a new
section as illustrated in Fig. E2.7b.
FIND Determine
(a) the magnitude of the force of the seawater on this triangular
area, and
(b) the location of this force.
SOLUTION
(a) The various distances needed to solve this problem are
shown in Fig. E2.7c. Since the surface of interest lies in a vertical
plane, y c h c 0.9 ft, and from Eq. 2.18 the magnitude
of the force is
F R gh c A
164.0 lbft 3 210.9 ft2310.3 ft2 2 24 2.59 lb
(Ans)
0.3 ft
1 ft
COMMENT Note that this force is independent of the tank
length. The result is the same if the tank is 0.25 ft, 25 ft, or 25 miles
long.
(b) The y coordinate of the center of pressure 1CP2 is found from
Eq. 2.19,
0.3 ft
0.9 ft 2.5 ft
(b)
and from Fig. 2.18
y R I xc
y c A y c
y
x
y c
Median line
y R
1 ft
δ A
0.2 ft
c
c
0.1 ft
CP
CP
x R
0.1 ft 0.15 ft 0.15 ft
(c)
(d)
F I G U R E E2.7b–d
F I G U R E E2.7a
of Tenecor Tanks, Inc.)
(Photograph courtesy