College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 390
Fluid Mechanics
Spring 2008 Number: 11971 Instructor: Larry Caretto
Exercise Two Solutions – Forces on Submerged Objects and Buoyancy
1. A rectangular gate that is 2 m wide is located in the vertical
wall of a tank containing water as shown in the figure at the
right copied from the text. It is desired to have the gate open
automatically when the depth of the water above the top of
the gate reaches 10 m. (a) At what distance, d, should the
frictionless horizontal shaft be located? (b) What is the
magnitude of the force on the gate when it opens? (Problem
2.63 in text.)
In this problem the horizontal shaft is located in the middle of the
gate. Part of the gate is above the shaft and part is below it. If
the net pressure below the location of the shaft is greater than
that above the shaft the gate will remain closed because of the
block on the right side of the bottom of the gate. However, if the
net pressure above the shaft is greater than the net pressure
below the shaft, the gate will open because there is no blockage
in the gate.
The figure at the left shows the gate with the
coordinates of the centroid and the the
center of pressure. (In this figure yCP is
denoted as yR.) If the hinge is located at the
center of pressure, yR, the net pressure
forces on the top and bottom of the shaft will
be exactly the same. Any increase in depth
will cause the gate to open.
In the figure as shown the coordinate of the
centroid of the gate, yc = 10 m + 2 m = 12 m
relative to the surface of the liquid. The area
of the gate is (4 m)(2 m) = 8 m 2. The 4 m
side is in the vertical direction that is used in
the formula for the moment of inertia about
the centroid: Ixc = (horizontal side) (vertical
side) / 12. Using the equation for the center of pressure gives the following result.
2 m4 m3
yR 
I xc
12
 yc 
 12 m  12.11 m
yc A
12 m8 m2
From the diagram we see that the location of yR equals d + 10 m so that d = yR – 10 m; d = 2.11 m
Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
Exercise two solutions
ME 390, L. S. Caretto, Spring 2008
Page 2
To determine the force we can use our usual equation for the resultant force with an angle of 90 o
and the specific weight of water.
FR  yc A sin  


9.80 kN
12 m  8 m 2 sin 90o
3
m
FR = 941 kN
2. The closed vessel shown at the right
(copied from the text) contains water with
an air pressure of 10 psi at the water
surface. One side of the vessel contains
a spout that is closed by a 6-inch circular
diameter gate that is hinged along one
side as illustrated. (Note that the slope of
the gate is that of the hypotenuse of a
right triangle with sides of 3 and 4 as
shown.) The horizontal axis of the hinge
is located 10 ft below the water surface.
Determine the minimum force that must
be applied at the hinge to hold the gate shut. Neglect the weight of the gate and the
friction at the hinge. (Problem 2.67 in text.)
Here we have the usual computation of the forces on a submerged surface with an additional
force, the air pressure at the top of the water. When we analyzed the force on a submerged
object in class, we computed the force as the integral dF = PdA = (p0 + ysin)dA with p0 = 0.
In this case, p0 = 10 psi so it must be considered in the integral. The result is simple: p0dA = p0A.
Furthermore, when we consider the point where this equivalent force due to p 0 acts, it will be
exactly at the centroid,
not at the centre of
pressure. Thus we have
to consider two separate
forces on the gate: F1 is
the force of the pressure
at the surface of the
liquid, p0, which acts at
the centroid, and F2 is
the usual force from the
pressure of the fluid,
which acts at the center
of pressure. This is
shown in the diagram at
the left.
The force acting on the gate due to the pressure at the top of the water, p0 is simply p0 times the
area, A, of the gate.
F1  p0 A 
10 lb f 
6 in 2  283 lb f
2
4
in
The pressure force from the liquid is found by the usual formula, F R = hcA = ycAsin; in this case
 = tan-1(3/4) = 0.6435 so sin= sin(0.6453) = 0.6. The value of hc is the 10 ft from the water
surface to the top of the liquid, plus sin times the (6 in)/2 = 0.25 ft distance from the top of the
circular gate to the centroid in the center of the circular gate. This gives h c = 10 ft + (0.5 ft)sin =
10 ft + (0.25 ft)(0.6) = 10.15 ft. So, the pressure force due to the water is found to be
Exercise two solutions
F2  hc A 
ME 390, L. S. Caretto, Spring 2008
62.4 lb f
ft 3
10.15 ft   6 in 2
4
Page 3
ft 2
 124 lb f
144 in 2
To compute the location of the resultant pressure force we need the distance yc from the top of
the water to the centroid along the slanted line. We can find this as yc = hc/sin = (10.15 ft)/(0.6)
= 16.916667 ft. For a circle of radius R, the moment of inertia about the centroid is R4/4. The
location of the center of pressure is then found as follows.

yR 
0.25 ft 4
I xc
4
 yc 
 16.916667 ft  16.91759 ft
yc A
16.916667 ft  0.25 ft 2
The net moment about the hinge is the sum of the two moments from the two forces. The value of
yR found above is essentially the same as the centroid so we could assume that both forces act at
the same point, the centroid of the circular gate, 3 in = 0.25 ft from the hinge. The solution below
illustrates the general approach where we have to determine the distance from the hinge, which
is equal to 16.91759 ft – (10 ft)/sin, to get the correct moment arm. (This calculation gives a
result of 0.2509 ft; so, using the centroid coordinate, 0.25, would give an error of only 0.4%.)
Setting the sum of the moments about the hinge to zero gives the following equation for the
resulting hinge moment, C.
10 ft 

C  F1 1  F2  2  238 lb f 0.25 ft   124 lb f 16.91759 ft 

0.6 

C = 102 ft▪lbf
Exercise two solutions
ME 390, L. S. Caretto, Spring 2008
Page 4
3. A plate of negligible weight
closes a 1-ft diameter hole
in a tank containing air and
water as shown in the
figure at the right (copied
from the text.) A block of
concrete (specific weight =
150 lbf/ft3), having a
volume of 1.5 ft3, is
suspended from the plate
and is completely
immersed in the water. As
the air pressure is
increased the differential
reading, h, on the
inclined-tube mercury manometer increases. Determine h just before the plate starts to
lift off the hole. The weight of the air has a negligible effect on the manometer reading.
(Problem 2.88 in text.)
When the plate is just about to lift off the hole, the pressure force
on the plate is just equal to the weight of the concrete less the
buoyancy force on the concrete. (See the free body diagram at
the left.) In equation form this gives pA = W – FB. (Note that the
use of a gage pressure gives the net resultant force of the
pressure inside the tank acting on the bottom of the plate and the
atmospheric pressure outside the tank acting on the top of the
plate.) The weight of the block is simply the volume of the block
times the specific weight of the concrete:
W = concreteV =
150 lb f
ft
3
1.5 ft   225 lb
3
The buoyancy force is the weight of the displaced water which iswaterV =
62.4 lb f
ft
3
f
1.5 ft   93.6 lb
3
The area of the plate on which the pressure acts is A = D2/4 = (1 ft)2/4 = 0.7854 ft2.
Applying these values in the force balance gives 225 lbf = 93.6 lbr + p(0.7854 ft2) so that p = (225
lbf – 93.6 lbr) / (0.7854 ft2) = 167 lbf/ft2. This air pressure causes the manometer displacement of
h so that the vertical height difference of the mercury column is h sin30o. This elevation
difference is caused by the difference between the (gage) air pressure inside the tank and the
open side of the manometer where the gage pressure is zero. Applying the manometer equation
gives
167 lb f
p air  p open   Hg h
 h 
p air  p open
 Hg

p air  p open
SG Hg  water
h = 0.394 ft
0
ft 2

62.4 lb f
13.6
ft 3
f