1.
Award: 50 out of 50.00 points
A room in the lower level of a cruise ship has a 40-cm-diameter circular window. If the midpoint of the window is 2
m below the water surface, determine the hydrostatic force acting on the window and the pressure center. Take
the specific gravity of seawater to be 1.025.
The hydrostatic force acting on the window is 2527
N.
The vertical distance from the free surface to the pressure center is 2.005
m.
References
Worksheet
Difficulty: Medium
A room in the lower level of a cruise ship has a 40-cm-diameter circular window. If the midpoint of the window is 2
m below the water surface, determine the hydrostatic force acting on the window and the pressure center. Take
the specific gravity of seawater to be 1.025.
The hydrostatic force acting on the window is
2527 ± 10% N.
The vertical distance from the free surface to the pressure center is
2.005 ± 10% m.
Explanation:
It is assumed that the atmospheric pressure acts on both sides of the window, and thus it can be ignored in
calculations for convenience. The average pressure on a surface is the pressure at the centroid (midpoint) of the surface and is determined to
be
Pave = PC = ρghC = (1025 kg/m3 )(9.81 m/s2 )(2 m) (
1N
2)
1 kg⋅m/s
= 20, 111 N/m2
Then the resultant hydrostatic force on each wall becomes
FR = Pave A = Pave πD
4
2
FR = (20, 111 N/m2 )
π(0.4 m)
4
2
= 2527 N
The line of action of the force passes through the pressure center, whose vertical distance from the free surface is
determined from
yP = yC +
Ixx,C
yC A
= yC +
πR4 /4
y C πR2
= yC +
2
R
4y C
=2m+
2
(0.2 m)
4(2 m)
= 2.005 m 2.
Award: 50 out of 50.00 points
A water trough of semicircular cross section of radius 0.9 m consists of two symmetric parts hinged to each other
at the bottom, as shown in the figure. The two parts are held together by a cable and turnbuckle placed every 3 m
along the length of the trough. Calculate the tension in each cable when the trough is filled to the rim.
The tension in each cable when the trough is filled to the rim is 11,923
N.
References
Worksheet
Difficulty: Medium
A water trough of semicircular cross section of radius 0.9 m consists of two symmetric parts hinged to each other
at the bottom, as shown in the figure. The two parts are held together by a cable and turnbuckle placed every 3 m
along the length of the trough. Calculate the tension in each cable when the trough is filled to the rim.
The tension in each cable when the trough is filled to the rim is
11,919 ± 10% N.
Explanation:
The following assumptions have been made here:
1. Atmospheric pressure acts on both sides of the trough wall, and thus it can be ignored in calculations for
convenience.
2. The weight of the trough is negligible. To expose the cable tension, we consider half of the trough whose cross section is a quarter-circle. The
hydrostatic forces acting on the vertical and horizontal plane surfaces as well as the weight of the liquid block are
as follows:
The horizontal force on the vertical surface is
FH = Fx = Pave A = ρghc A = ρg(R/2)A
FH = (1000 kg/m3 )(9.81 m/s2 ) ( 0.9
m) (0.9 × 3 m) (
2
1N
2
1 kg⋅m/s
)
FH = 11919 N
The vertical force on the horizontal surface is zero, since it coincides with the free surface of water. The weight of
fluid block per [tension(2)]-m length is
FV = W = ρgV = ρg(w ×
πR 2
)
4
FV = (1000 kg/m3 )(9.81 m/s2 )[
(3 m)π(0.9 m) 2
] 1N 2
4
1 kg⋅m/s
FV = 18723 N
Then, the magnitude and direction of the hydrostatic force acting on the surface of the 3-m-long section of the
trough become
−−−−−−−−−−−−−−−−−−−−
−−−−−−−
FR = √FH2 + FV2 = √(11919 N)2 + (18723 N)2 = 22195 N
tan θ =
FV
FH
=
18723 N
11919 N
= 1.57 → θ = 57.52°
Therefore, the line of action passes through the center of the curvature of the trough, making 57.52° downward
from the horizontal. Taking the moment about point A where the two parts are hinged and setting it equal to zero
give
ΣMA = 0 → FR R sin(90 - 57.52)° = TR
Solving for T and substituting, the tension in the cable is determined to be
T = FR sin (90 - 57.52)° = (22,195 N)sin (90 - 57.52)° = 11,919 N
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