College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 390
Fluid Mechanics
Spring 2008 Number: 11971 Instructor: Larry Caretto
February 12 Homework Solutions
2.49
A rectangular gate having a width of 5 ft is located in the
sloping slide of a tank as shown in the figure on the right
(copied from the text.) The gate is hinged along its top
edge and is held in place by the force P. Friction in the
hinge and the weight of the gate can be neglected. Find
the force, P.
The force P has to offset the force created by the water
pressure at the gate. This pressure creates an average
resultant force, FR, that acts at the center of pressure, yCP,
and by taking moments about the hinge we can find the force
P. In the equations for the resultant force, FR = Aycsin, the
centroid coordinate, yc, is measured from the top of the water. Since the centroid of a rectangle is
at its midpoint (half of the 6 ft. height of the gate) yc = 10 + 6/2 = 13 ft. The area is (6 ft)(5 ft) = 30
ft2. We can then find the resultant force as follows.
FR  yc A sin  
62.4 lb f
ft
3
13 ft 30

ft 2 sin 60 o  2.11x10 4 lb f
In order to find the location of the center of
pressure we need to find the moment of inertia
about the x axis, Ixc. As shown in the diagram at
the left, Ixc = ba3/12 where a is the dimension along
the y axis. In our case a = 6 ft and b = 5 ft so Ixc =
(5 ft)(6 ft)3/12 = 90 ft4. We then use the formula for
the center of pressure to find yR = yCP as follows.
yCP 


I xc
90 ft 4
 yc 
 13 ft  13.23 ft
yc A
13 ft  30 ft 2


We can now use a moment balance around the hinge as
shown in the free-body diagram to the right. The moment
balance gives P(6 ft) = FR(yCP – 10 ft). Note that we have to
correct the value of yCP to make it relative to the location of
the hinge instead of the value we have which is relative to the
top of the wall. We can thus find P = (2.11x104 lbf) (3.23 ft)
/(6 ft).
P =1.14x104 lbf
Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
February 12 homework solutions
2.52
ME 390, L. S. Caretto, Spring 2008
Page 2
A homogenous 4-ft wide, 8-ft long rectangular gate,
weighing 800 lbf is held in place by a horizontal
flexible cable as shown in the figure in the right
copied from the text. Water acts against the gate
which is hinged at point A. Friction in the cable is
negligible. Determine the tension in the cable.
This problem is similar to problem 49, except that the
gate extends above the top of the water. Since the
distance along the gate where water is present is 6 ft,
the centroid of the area where there is water pressure
is yc = 3 ft. The area impacted by the water pressure
is A = (6 ft)(4 ft) = 24 ft2. We can find the resultant force by the usual equation.
FR  yc A sin  
62.4 lb f
ft
3
3 ft 24

ft 2 sin 60 o  3.89 x103 lb f
In order to find the location of the center of pressure we need to find the moment of inertia about
the x axis. As shown in the diagram for problem 49, Ixc = ba3/12 where a is the dimension along
the y axis that is exposed to the fluid. In this case a = 6 ft and b = 4 ft so Ixc = (4 ft)(6 ft)3/12 =
72 ft4. We then use the formula for the center of pressure to find yR = yCP as follows.
yCP 


I xc
72 ft 4
 yc 
 3 ft  4 ft
yc A
3 ft  24 ft 2


We next take a moment balance around the hinge as shown in the freebody diagram to the left. The resultant force is normal to the gate at a
location 2 ft above the hinge. The cable tension, T, acts 8 ft above the
hinge and the component normal to the gate is Tsin60o. The weight acts
4 ft above the hinge and has a component Wcos60o normal to the gate.
Using these results gives the following moment balance.
FR 2 ft   W cos 60 o 4 ft   T sin 60 o 8 ft 
Solving this equation for T and substituting the values for weight and
resultant force gives the desired numerical result.


3
o
FR 2 ft   W cos 60 o 4 ft  3.89 x10 lb f 2 ft   800 lb f cos 60 4 ft 
T

sin 60 o 8 ft 
sin 60 o 8 ft 
T =1.35x103 lbf
February 12 homework solutions
2.57
ME 390, L. S. Caretto, Spring 2008
Page 3
A 3-m wide, 8-m high rectangular gate is
located at the end of a rectangular
passage that is connected to a large
open tank filled with water as shown in
the figure at the right (copied from the
text). The gate is hinged at its bottom
and held closed by a horizontal force, FH
located at the center of the gate. The
maximum value for FH is 3500 kN. (a)
Determine the maximum water depth, h,
above the center of the gate that can
exist without the gate opening. (b) Is your answer the same if the gate is hinged at the
top? Explain your answer.
The resultant force on the gate, whose area is (3 m)(8 m) = 24 m 2, is given by the usual formula,
FR = ycAsin. Here yc must be computed from the top of the liquid to the center of the gate
giving yc = h. Since we are trying to find h, we have to carry h (or yc) as an unknown throughout
the calculation. Thus the resultant force becomes
FR  yc A sin  
9.80 kN
m3
 yc 24 m 2 sin 90o  232 yc
kN
m
The location of the resultant force requires the calculation of the moment of inertia about the x
axis through the centroid. See the solution to problem 49 for a discussion of this calculation. For
the rectangular gate here the moment of inertia, Ixc = ba3/12 = (3 m)(8 m)3/12 = 128 m4. We can
now obtain an equation for yCP from the usual equation.
yCP


I xc
128 m 4
5.333 m 2

 yc 
 yc 
 yc
yc A
yc
y c 24 m 2


The free-body diagram to the left shows that there are two forces on
the gate. The first is the resultant pressure force that is a distance ℓ
above the hinge and the opposing force, FH that is 4 m above the
hinge. The distance from the surface of the liquid to the bottom of
the liquid is yc + 4 m from the diagram. The distance, ℓ, is the
difference between this distance to the bottom and yCP so, ℓ = yc +
(4 m) – yCP. With these dimensions a momentum balance about the
hinge gives:
 5.333 m 2
 
5.333 m 2 
kN  
kN 



FH 4 m   FR    232 yc
y

4
m


y

232
y
4
m



 c


c
c


m  
yc
m 
yc


 

When FH has its maximum value of 3500 kN, the value of yc = h is given by the following
equation.
5.333 m 2
kN 

FH 4 m   3500 kN 4 m    232 y c
 4 m 
m 
yc


  918 y c kN  1237 kN  m


February 12 homework solutions
ME 390, L. S. Caretto, Spring 2008
y c  FH 4 m 
Page 4
3500 kN 4 m  1237 kN  m  16.6 m
918 kN
h = yc = 16.6 mf
The free-body diagram at the left shows the forces when the gate is
hinged at the top. Here the length, ℓ1 = yCP – (yc – 4) = (5.333 m2)/yc +
yc – (yc – 4 m) = (5.333 m2)/yc + 4m. Applying this length to the
momentum balance gives.
FH 4 m  3500 m4 kN   14000 kN  m  FR 

kN  5.333 m 2

  232 y c
 4 m   1237 kN  m  928 y c kN


m 
yc


Solving gives y c 
14000 kN  m  1237 kN  m
 13.8 m ; so h = yc = 13.8 m .
928 kN
The depth is less when the gate is hinged at the top because the pressure distribution creates
more momentum on the gate then. When the gate is hinged at the bottom, the center of
momentum is closer to the greater pressures. When the gate is hinged at the top it is further from
these greater pressures so more momentum is exerted by the pressure distribution.
2.58
A gate having the cross section shown in the figure at the
right (copied from the text) closes an opening 5 ft wide
and 5 ft high in a water reservoir. The gate weighs 500 lbf
and its center of gravity is 1 ft to the left of AC and 2 ft
above BC. Determine the horizontal reaction that is
developed on the gate at C.
The gate is at a different angle than the wall. We are told that
the gate is 5 ft high and the diagram shows that the projection
of the gate on a vertical surface is 4 ft high. We thus conclude
that the length AB is 5 ft; the length AC is 4 ft, and the length
BC =
5 ft 2  4 ft 2 = 3 ft.
If the gate were to continue along a wall that was at the same
angle as the gate, the length of the hypothetical wall could be
found by the relation for similar triangles. (See the diagram to
the left, below.)

The 8 ft height of the vertical wall would be equivalent to a
(5/4)(8 ft) = 10 ft slanted distance to the surface of the water.
The angle, , that the slanted length makes with the surface of
the water is sin-1(4 ft/5 ft) = 0.9273 radians. The slanted length
to the centroid of the gate would be yc = 10 ft + 2.5 ft = 12.5 ft.
The value of yc sin would be (12.5 ft)sin(0.9273) = 10 ft. We
could have also found this number by looking at the vertical
depth of the water at the centroid of the gate AB. From the
figure above we see that this is 10 ft below the surface of the
water. Regardless of how we find this length, we can now use
this hc= ycsin = 10 ft in our equation for the resultant force on
February 12 homework solutions
ME 390, L. S. Caretto, Spring 2008
Page 5
the side of the gate, Fside. We still use the actual area of the gate (5 ft)2 = 25 ft2.
Fside  hc A 
62.4 lb f
ft
3
10 ft 25

ft 2  1.56 x10 4 lb f
To compute the location of this resultant force we use our usual formula for yCP, and we use the
distance along the hypothetical slanted wall with yc = 12.5 ft. The moment of inertia about the
centroid, Ixc = ba3/12 = (5 ft)(5 ft)3/12 = 52.08 ft4.
yCP 


I xc
52.08 ft 4
 yc 
 12.5 ft  12.67 ft
yc A
12.5 ft  25 ft 2


To find the reaction force at C we have to use a moment balance of all the moments about the
hinge at A. The point, relative to A, where the resultant pressure force acts is found by
subtracting the 10 ft distance along the hypothetical slanted wall above the gate. This gives the
location of the center of pressure as 2.67 ft. below the start of the gate. There is also a pressure
force on the bottom of the triangular gate. The water at this point is 12 ft deep and the area of the
lower portion of the triangular gate is (3 ft)(5 ft) = 15 ft2. So the upward force that the water
pressure gives on this surface is
Fbottom  hbottom Abottom 
62.4 lb f
ft
3
12 ft 15

ft 2  1.123x10 4 lb f
This force acts at the centroid of the bottom plate which is 1.5 ft to the left of the line AC. We are
also told that the center of gravity for the 500-lbf weight of the gate is 1 ft to the left of the line AC.
With this information we can form the momentum balance around the hinge at A as follows.
Fside (2.67 ft)  Fbottom (1.5 ft)  W (1 ft)  FC (4 ft)  0
Substituting the values for the forces and weight allows us to solve for F C.
Fside (2.67 ft)  Fbottom (1.5 ft )  W (1 ft)

4 ft
ft (2.67 ft )  1.123x10 4 ft (1.5 ft )  500 ft (1 ft)
4 ft
FC 
1.56x10 
4


Fc = 6.33x103 lbf
2.68
Dams can vary from very large structures with
curved faces holding back water to great depths, as
shown in Video V2.3, to relative small structures
with plane faces as shown in the figure at the right
copied from the text. Assume that the concrete
dam shown in this figure weighs 23.6 kN/m3and
rests on a solid foundation. Determine the
minimum coefficient of friction between the dam
and the foundation required to keep the dam from
sliding at the water depth shown. You do not need
to consider the possible uplift along the base. Base
your analysis on a unit length of the dam.
February 12 homework solutions
ME 390, L. S. Caretto, Spring 2008
Page 6
The angular portion of the dam decreases by 4 m (from 6 m to
2 m) as the dam rises 5 m. Thus, the angle at the base of the
dam = tan-1(5 m/4 m) = 0.8961 radians = 51.34o. This is also
the angle, , that the water makes with the sloping surface of
the dam, which is used in our formula for the resultant force.
See the sketch at the left for the forces acting on the dam.
The slanted length of the dam, below the surface of the water,
is (4 m)/sin = (4 m)/sin(0.8961) = 5.122 m. The area of a unit
width of the dam is (5.122 m)(1 m)2 = 5.122 m2. The value of
yc = (5.122/2 m) = 2.5561 m. With these data we can find the
resultant force of the water on a unit length of the dam.
FR  yc A sin  
9.80 kN
m
3
2.5561 m5.122 m 2 sin 0.8931  100.4 kN
For force equilibrium in the x direction, the x component of the resultant force, F Rsin= (100.4
kN)sin(0.8931) = 78.40 kN, must equal the friction force, Ff. Force equilibrium in the y direction
requires the weight of the dam, which is the product of the specific weight times the volume of the
unit length of the dam. The cross sectional area has two parts: (1) a rectangle with an area of (2
m)(5 m) = 10 m2 and (2) a triangle with an area of (4 m)(5 m)/2 = 10 m 2. The volume of a unit
length of the dam is thus (10 m 2 + 10 m2)(1 m) = 20 m3. Thus the weight of concrete in the unit
length of he dam is (23.6 kN/m 3)(20 m3) = 472 kN. With this weight we have the following force
balance in the y direction.
N  FR cos   W  100.4 kN cos 0.8931  472 kN  534 kN
The friction coefficient, , is simply the friction force divided by the normal force.

2.83
Ff
N

78.40 kN
534 kN
 = 0.146
The homogenous wooden block A shown in the
figure at the right copied from the text is 0.7 m
by 0.7 m by 1.3 m and weighs 2.4 kN. The
concrete block, B, (specific weight = 23.6
kN/m3) is suspended from A by means of the
slender cable causing A to float in the position
indicated. Determine the volume of B.
We can consider two separate force balances
linked by the force in the cable. (We will assume
that the weight of the “slender” cable can be
ignored in this force balance.) In the first force
balance on the wood, we can compute the force in
the cable that causes the wood to float in the position shown where half of its volume of
(0.7 m)(0.7 m)(1.3 m) = 0.647 m 3 is floating. In the second force balance for the concrete, we will
use the force in the cable found in the first force balance to obtain
the volume of concrete (B).
The vertical force balance for the wood is shown in the figure on
the left. The cable force, T = FB – W , the buoyancy force minus
the weight. The weight of the wooden block is given as 2.4 kN.
The buoyancy force, FB, is the specific weight of water times the
volume of water displaced (which is half the volume of the wood =
0.3235 m3 from the computation in the previous paragraph). This
February 12 homework solutions
ME 390, L. S. Caretto, Spring 2008
Page 7
gives FB = (9.80 kN/m3)(0.3235 m3) = 3.121 kN. Thus the cable force, T = FB – W = 3.121 kN –
2.4 kN = 0.721 kN
The vertical force balance on the concrete is shown in the figure on the left. (Here
the subscript c stands for concrete.) Both the weight and the buoyancy force are
proportional to the volume of the concrete. Using this result we can write the
vertical force balance as follows.
0  T  FBc  W  0.721 kN   H2OVc   cVc
Solving for the unknown concrete volume and substituting the known specific weights gives the
desired answer.
Vc 
2.89
0.721 kN
 c   H 2O

0.721 kN
23.6 kN
m3

9.80 kN
 0.0523 m3
m3
When a hydrometer shown in the figure on the right copied
from the text (see video V2.6 to see the operation of a
hydrometer) having a stem diameter of 0.30 in is placed in
water the stem protrudes 3.15 in above the water surface. If
the water is replaced with a liquid having a specific gravity of
1,10, how much of the stem would protrude above the liquid
surface? The hydrometer weighs 0.042 lbf.
The hydrometer measures density (or specific weight) by floating
in the water. The weight of the hydrometer is balanced by the
buoyant force caused by the water it displaces. The hydrometer
contains air spaces to give it a specific weight that is lighter than
that of water. In actual operation the scale of the hydrometer is
calibrated directly in specific weight or specific gravity. This
problem illustrates how such a calibration would be done.
For both the initial water and the second fluid with a specific gravity of 1.10, the weight of the
hydrometer, W, is balanced by the buoyant force which is the specific gravity of the fluid times the
submerged volume of the hydrometer.
W  FB1   H 2OV1  V1 
0.042 lb f
W

 6.73x10 4 ft 3
 H 2O 62.4 lb f
ft 3
W  FB1   H 2OV1  FB 2  SG H 2OV2  V2 
V1 6.73x10 4 ft 3

 6.12 x10 4 ft 3
SG
1.1
The difference between these two volumes is 6.73x10-4 ft3 – 6.12x10-4 ft3 = 0.61x10-4 ft3. This
difference in volume displaced will elevate the hydrometer tip by exactly the same volume. Since
the hydrometer tip has a diameter of 0.30 in = 0.025 ft, giving it an area of (0.025 ft)2/2 =
0.000491 ft2, it will protrude an additional distance of (0.61x10-4 ft3) / (0.000491 ft2) = 0.125 ft =
1.5 in. Since it originally protruded 3.15 in with water, the new height will be 4.65 in .