148. Problem
A container holds two layers of different liquids, one fluid having a specific gravity of f.2 is 200 mm deep
and the other fluid having a specific gravity of 1.5 is 250 mm deep. A solid spherical metal having a diam
of 225 mm and a sp.gr. of 7.4 is submerged such a manner that half of the sphere is on the top layer and
the other half in the bottom layer of fluids.
1. Compute the weight of the spherical metal.
2. Compute the buoyant force acting on the object.
3. Compute the tension in the wire holding the sphere to maintain its position.
Solution:
1. Weight of the spherical metal.
W=
4
( 0.1125 )3 ( 7.4 )( 9.81 )
3
W = 0.433 kN
W = 433 N
2. Buoyant force acting on the sphere:
4
BF 1= π ¿ ¿
3
BF1=0.035 kN
4
BF 2= π ¿ ¿
3
BF 2=0.044 kN
Total buoyant force = 0.035 + 0.044
Total buoyant force = 79 N
3. Tension of wire holding the sphere : T + 79 = 433
T = 354 N
149. Problem:
Two spheres each 1.20 m. in diameter are connected by means of a short rope. One weighs 4 kN and
other weighs 12 kN. When placed in water.
1. Compute the tension in the rope.
2. Compute the depth of floatation of the 4 kN sphere.
3. Compute the volume of sphere exposed above the water surface.
Solution:
1. Tension in rope:
Vol. of sphere =
4π
¿
3
Vol. of sphere = 0.9048
Unit wt. of 4 kN sphere =
T = 12 -
4π
¿ (9.81)
3
T = 12 - 0.9048(9.81)
4
0.9048
T = 12 – 8.876
T = 3.124 kN
Unit wt. = 4.42 ¿ 9.81 kN /m3
it will float in water
Unit wt. of 12 kN sphere =
12
0.9048
Unit wt. = 13.26 ¿ 9.61
it will submerged in water
2. Depth of flotation of 4 kN sphere:
BF
2
1=9.81
( π3h )(3 r−h)
Try h = 0.86
π h2
0.693 = (0.86¿2 [ 1.8−0.86 ]
(3 r−h)
3
0.639 = h2 [ 3 ( 0.6 )−h ]
0.693 = almost equal
4 + 3.124 = 9.81
Use h = 0.86 m.
3. Vol. of sphere exposed above the water surface:
π h2
(3 r−h)
3
π (0.34)2
V=
[ 3 ( 0.6 )−0.34 ]
3
V = 0.177 m 3
V=
150. Problem:
A hollow cylinder 1.0 m. in diameter and 2 m. long weighs 3825 n.
1. How many kN of lead weighing 110 kN/m3 must be fastened to the outside bottom to make the
cylinder float vertically with 1.50 m. submerged in fresh water.
2. How many kN of lead weighing 110 kN/m3 must be placed inside the cylinder to make the
cylinder float vertically with 1.50 m. submerged in fresh water.
3. What additional load must be placed inside the cylinder to make the top of the cylinder flush
with the water surface.
Solution:
1. Load placed outside the cylinder:
W 1 + W 2 = BF 1 + BF 2
π
3.825 +W 2 = (1¿2(1.5)(9.81) + V(9.81)
4
w
V=
D
W2
V=
110
W 2 (9.81)
3.825 + W 2 = 11.56 +
110
W
0.91 2 = 7.735
W 2 = 8.5 kN
2. Load placed inside the cylinder:
W 1 + W 3 = BF
π
3.825 + W 3 = ¿
4
W 3 = 7.73 kN
3. Additional load to make the top flush with the water surface:
3.825 + 7.73 + W 4 =
W 4 = 3.85 kN
π 2
(1 (2) (9.81)
4 ¿
151. Problem:
A concrete block with a volume of 0.023 cu,m. is tied to one end of a wooden post having dimensions of
200 mm x 200 mm by 3m. long and placed in fresh water. Weight of wood is 6.4 kN/m3 and that
concrete is 23.5 kN/m3.
1. Det, the length of the wooden post above the water surface.
2. Det, the volume of additional concrete to be tied to the bottom of the post to make its top flush
with the water surface.
3. Det, the total weight of concrete to make its top flush with the water surface.
Solution:
1. Length above the water surface:
W + W 1 = BF 1 + BF 2
3(0.20)(0.20)(6.4) + 0.023(23.5)
= 0.2(0.2)h(9.81) + 0.023(9.61)
h = 2.76
Length above water surface = 3-2.76
Length above water surface = 0.24 m.
2. Additional vol. concrete:
W + W 1 = BF 1 + BF 2
0.20(0.20)(3)(6.4) + (0.023 + V) 23.5
= 0.2(0.2)(3)(9.81) + (0.023 + V)9.81
0.768 + 0.5405 + 23.5V = 1.772
+ 0.22563 + 9.81V
13.69V = 0.09433
V = 0.0069 m 2
3. Total weight of concrete:
W c = ( 0.023+0.0069 ) 23.5
W c = 0.70265 kN
W c = 702.65 N
152. Problem:
A rectangular barge weighing 200000 kg is 14 m long, 8 m. wide 4.5 m deep. It will transport to Manila
20 mm diameter, 6 m. long steel reinforcing bars having a density of 7850 kg/m3. Density of salt water is
1026 kg/m3.
1. Det, the drafty of the barge on sea water before the bar was loaded.
2. If a draft is to be maintained at 3 m., how many pieces of steel bars could it carry?
3. What is the draft of the barge when one half of its cargo is unloaded in fresh water?
Solution:
1. Draft of empty barge on sea water:
200000 = 14(8)(d)(1026)
d = 1.74 m.
2. No. of bars loaded:
8(14)(3)(1026) = 200000 + W b
W b = 144736 kg
Wb =
π
(0.02¿2 (6) N (7850)
4
144763 =
π
(0.02¿2 (6) (7850) N
4
N = 9782 bars
3. Draft of barge on fresh water when one half of its cargo is unloaded:
200000 +
1
(144736) = 8 (14)(d)(1000)
2
d = 2.43 m.
153. Problem:
The gate in the figure weighs 2340 N/m normal to the page. It is in equilibrium as shown. Neglect the
weight of the arm and brace supporting the counterweight.
1. Find the hydraulic force normal to the gate.
2. Find the submerged weight of concrete (W).
3. Find the weight of concrete in the air of it has a sp.gr.of 2.50
Solution:
1. Hydrostatic force normal to the gate:
P = ywh A
P = 9.81 (0.9)(1.8)(1)
P = 15.89 kN
2. Submerged weight of concrete:
∑ M0
=0
P(0.6) = W(1.5) Sin 30`
15.89(0.6) = W(1.5) Sin 30`
W = 12.712 kN
3. Weight of concrete in air:
∑ Fv = 0
9.81V(2.50) – 12.712 = 9.81V
14.715V = 12.712
V = 0.864 m 3
W c = 9.81V (2.5)
W c = 9.81 (0.864)(2.5)
W c = 21.19 kN (wt. of concrete in air)
154. Problem:
A board weighing 30 N/m has a cross sectional area of 0.005 m2 and a length of 3.4 m. placed in the
tank of oil having a sp.gr. of 0.85. assuming the hinged to be frictionless.
1. Compute the specific gravity of the board.
2. Compute the length of the board which is submerged in oil.
3. Compute the angle 0 for equilibrium conditions.
Solution:
1. A specific gravity of board:
W = 30(3.4)
W = 102 N
W = 102 N
0.102 = 9.81(3.4)(0.005.)(sp.gr.)
sp.gr = 0.61
2. Length of board which is submerged in oil:
BF = x(0.005)(9.81)(0.85)
BF = 0.042x
W = 30(3.4)
W = 102 N
W = 0.102 kN
∑ MA
=0
W(1.7) Sinθ = BF (3.4 -
x
) Sinθ
2
0.102 (1.7) = 0.042x(3..4 – 0.5x)
0.1734 = 0.1428 x – 0.021 x 2
x 2 6.8x + 8.257 = 0
x2 –
6.8 ±3.63
2
X = 1.58 m.
3. Angle θ :
Cos θ =
1.5
3.4−1.58
θ=34.5
155. Problem:
The timber beam shown has dimensions of 0.15 x 0.15 x 3 m. and weighs 560 N.
1. Compute the length of the beam submerged in water.
2. Compute the angle 0 when the water surface is 2.1m above the pivot.
3. At what depth of water will the timber stand vertically?
Solution:
1. Length of the beam:
BF = 0.15(0.15) x (9810)
BF = 220.725x
∑ M0
=0
x
2
560(1.5) Cos θ = 220.725x ( ) Cos θ
x 2 = 7.61
x = 2.76 m.
2. Angle θ :
Sin θ=
2.1
2.76
θ = 49.54`
3. Depth when the timber stands vertically:
560 = 0.15 (0.15) h (9810)
h = 2.54 m.
156. Problem:
A cubical block having an edge equal to 3 m. has its upper half of sp.gr.= 0.80 and its lower half of sp.gr.
= 1.4. it is resting on a two layer fluid with an upper sp.gr. of 0.90 and a lower sp.gr. of 1.2.
1. Compute the total weight of the cube in quintals.
2. Compute the height of the top of the cube above the interface of the two layer fluid.
3. If the depth of the upper layer of fluid having sp.gr. of 0.90 is equal to 1.2 m. how far below the
liquid surface is the top of the cube.
Solution:
1. Weight of block:
W = 9.81(3) (3) (1.5) (0.80) + 9.81(3)(3)(1.5)(1.4)
W = 291.36kN
W=
291360
9.81
W = 29700 kg
W = 297 quintals
Note 100 kg = 1 quintal
2. Height of cube above the interface of the two layer fluid:
9.81(1.2) (3) (3) (3-h) + 9.81(0.90) (3) (3)h = 291.36
105.95(3-h) + 79.46h = 291.36
26.49h = 26.29
h = 1.0 m.
3. Distance below the liquid surface from top of tue cube:
x = 1.2 – 1
x = 0.2 m.
157. Problem:
A concrete cube 0.5 m on each side is to be held in equilibrium under water by attaching a light foam
buoy to it. Specific weight of concrete and foam are 23.58 kN/m3 and 0.79 kN/m3 respectively. Assume
unit weight of water = 9.79 kN/m3.
1. What is the minimum volume of foam required?
2. What is the weight of the foam?
3. What is the total weight of the concrete and foam?
Solution:
1. Minimum volume of foam required:
W 1 + W 2 = BF 1 + BF 2
V (0.79) + 0.5(0.5)(0.5)(23.58)
= V (9.79) + 0.5(0.5)(0.5)(9.79)
V = 0.192 m 3
2. Weight of the foam:
W = 0.79 (0.192)
W = 0.152 kN
3. Total weight of the concrete and foam:
Total weight = 0.152 + 0.5 (0.5)(0.5)(0.5)(23.58)
Total weight = 3.0995 kN
158. Problem:
A piece of wood of sp.gr. 0.651 is 3 in. square and 5 ft. long.
1. What is the volume of the lead having a unit weight of 700 pcf that should be fastened of one
end of the stick so that it will float upright in 1 ft. out of water?
2. Determine the weight of the lead?
3. Determine the total weight of lead and the wood.
Solution:
1. Volume of lead:
W 1 + W 2 = BF 1 + BF 2
3 3
0.651 ( ) ( ) (5)(62.4) + 700 V
12 12
3 3
= ( ) ( ) (4) 62.4 + 62.4 V
12 12
= 0.00456 ft 3
2. Weight of lead:
w=VxD
w = 0.00456 (700)
w = 3.19 lb.
3. Total weight:
= 3.19 + 0.651 (
3 3
) ( ) (5)(62.4)
12 12
= 15.88 lb.
159. Problem:
A circular log having a diameter of 8 ft. has a length of 15 ft. it has a specific gravity of 0.425.
1. Det, the weight of the log.
2. To what depth will the log sink in fresh water.
3. What load must be applied in order to submerged the log fully?
Solution:
1. Weight of the log
W=
π
(8 2 (15) (0.425) (62.4)
4 ¿
W = 19,995.61 lb.
2. Depth of the log:
As
2θ π 2 2
= π ¿¿ =
=θ
2θ
2π
As = 16 θ
4 sin θ ( 2 ) ( 4 ) cos θ
2
A = 16 θ - 16 Sin θ Cos θ
A = 16 θ –
W = BF
19995.61 = 15 (16 θ−16 sin θ cos θ) (62.4)
Try θ=83.22
21..363 = 16 θ−16 sin θ cos θ
21.363 = 16
(83.22)π
180
- 16 Sin 83.22` Cos 83.22`
21.36 = 21.363 (almost equal)
h = 4-4 Cos 83.22`
h = 3.53 ft.
3. Load applied to submerged it fully:
A1 = 16 θ−16 sin θ cos θ
16(83.22) π
A1 =
- 16 Sin 83.22` Cos 83.22
180
A1 = 21.36 ft 2
V= Ah
π
(8 2 - 21.36
4 ¿
A = 28.91 ft 2
A=
Load applied = A h (62.4)
Load = 28.91 (15)(62.4)
P = 27,060 lb.
216-A Problem:
An open cylindrical tank having a radius of 0.30 m. and a height of 1.20 m. is filled with water at depth of
0.90 m.
1. How fast will it be rotated about its vertical axis if half of its volume is spilled out?
2. How fast will it be rotated about its vertical axis so that no water will be spilled out?
3. How fast will it be rotated about its vertical axis so as to produce a zero pressure with 0.20 m.
from the center of the tank.
Solution:
1. Speed of rotation so that one half of its volume is spilled out:
π x 2 ( y−1.2)
2
= 0.05 π (0.30¿2 (0.90)
π ¿¿ - π ¿¿ -
0.045y – 0.027 – 0.5(x¿2(y-1.2) = 0.0405
0.045y – 0.5 x 2 (y=1.2) = 0.0675
x 2 = ky
When x = 0.30
(0.30¿2 = ky
k=
0.09
y
when x = x
y = y – 1.20
x 2 = k(y – 1.20)
x2 =
0.09
(y – 1.20)
y
0.045y – 0.5 x 2 (y – 1.2) = 0.0675
0.045y – 0.5 ( 0.09 ) ¿ ¿ = 0.0675
0.090 y 2 - 0.09(y – 1.2¿2 = 0.135y
0.090 y 2 - 0.09 y 2 + 0.216y – 0.1296 = 0.135y
0.081y = 0.1296
Y = 1.6 m.
y=
∞2 r 2
2g
∞ 2 (0.3)2
2(9.81)
∞=18.68 rad /sec
18.68(60)
∞=
2π
∞=178.34 rpm
1.6 =
2. Speed of rotation so that no water will be spilled out:
y=
∞2 r 2
2g
∞ 2 (0.3)2
2(9.81)
∞=11.44 rad / sec
0.60 = =
11.44 (60)
2π
∞=109.21 rpm
∞=
3. Speed of rotation when x=0.15:
x = 0.20 m
x = 0.30
y=y
y = y – 1.20
¿¿ = ¿¿ =
0.09y – 0.108 = 0.04y
0.05y = 0.108
y = 2.16
y=
∞2 r 2
2g
∞ 2 (0.30)2
2(9.81)
∞=21.70rad /sec
21.70(60)
∞=
2π
∞=207.22rpm
(2.16) =
217. CE Board may 2004
An open cylindrical tank one meter in diam., and 2.5m high is 3/5 full of water. If the tank is rotated
about its vertical axix, what speed should it have in rpm so that.
1. The water could just reach the rim of the tank.
2. The depth of water at the center is zero.
3. There is no water at the bottom with in 20 cm. from the vertical axis.
Solution:
1. Water could just reach the rim of the tank.
∞2 r 2
2g
∞ 2 (0.5)2
2=
2(9.81)
∞=12.53 rad /sec
12.53(60)
∞=
2π
∞=119.64 rpm
y=
2. Depth of water at the center is zero.
∞2 r 2
2g
∞ 2 (0.5)2
2.5 =
2(9.81)
∞=14.01 rad / sec
14.01(50)
∞=
2π
∞=133.76 rpm
y=
3. No water at the bottom within 20 cm. from the vertical axis.
h + 2.5 =
∞2 r 2
2g
∞ 2 (0.5)2
h + 2.5 =
2g
2 2
∞ r
h=
2g
∞ 2 (0.20)2
h=
2g
2
∞ (0.20)2
∞ 2 (0.5)2
+ 2.5 =
2g
2g
0.21 ∞2
= 2.5
2g
∞=15.28 rad /sec
15.28(60)
∞=
2π
∞=145.94 rpm
218. Problem:
An open cylindrical tank 3 m. high and 1 meter diameter contains 2 m. of water. If the cylinder rotates
about its geometrical axis.
1. What constant angular velocity can be attained without spilling any water?
2. What constant angular velocity can be attained if the depth of water at the center is 0.5 m.
3. What constant angular velocity can be attained if the volume spilled out is equal to 0.196 m3
Solution:
1. Constant angular velocity that can be attained without spilling any water.
∞2 r 2
2g
∞ 2 (0.5)2
2=
2(9.81)
y=
w = 12.53 rad/sec.
2. Constant angular velocity that can be attained if the depth of water at the center is 0.5 m.
∞2 r 2
2g
∞ 2 (0.5)2
2.5 =
2(9.81)
∞ = 14.01 rad/sec.
y=
3. Constant angular velocity that can be attained if the vol. spilled out is 0.196 m3.
219. Problem:
An open cylindrical tank is 1.20 m. in diam. and 2.10 m. high is 2/3 full of water.
1. Compute the amount of water in liters that will be spilled out if the vessel is rotated about its
vertical axis at a constant speed of 90 rpm.
2. At what speed in rpm can the vessel be rotated without spilling any water.
3. At what speed in rpm can the vessel be rotated so that the pressure at the center of the bottom
of the tank will not exceed 6.867 kPa.
Solution:
1. Amount of water in liters
2. Speed in rpm can the vessel be rotated without spilling any water.
3. Speed in rpm can the vessel be rotated.
220.Problem:
An open cylindrical tank 30 cm. in diam. And 60 cm. high partially filled with water is rotated about its
vertical axis at a speed of 240 rpm.
1. What would be the depth of water in the tank if the cylindrical tank is brought to rest so that no
water will be spilled out.
2. At what speed would it be rotated so that 1.40 liters of water will be spilled out.
3. At what speed would it be rotated so that the pressure at the center of the bottom of tank is
zero?
Solution:
1. Depth of water in the tank of the cylindrical is brought to rest.
2. Speed would it be rotated so that 1.40 liters water sill be spilled out.
3. Speed would it be rotated so that the pressure at the center of the bottom of the tank is zero.
221. Problem:
An open cylindrical vessel, 3 m. high and 1.5 m in diameter is filled with water to a depth of 2.5 m. in its
side is a 50 mm circular orifice C = 0.60 placed 300 mm above the bottom of the vessel. The vessel is
rotated about the orifice for a period of 3 minutes.
1. Determine the volume of water allowed to escape through the orifice.
2. Determine the average rate of discharge flowing through the orifice in liters/sec.
3. How deep will the water stand in the vessel after it is brought to rest.
Solution:
1. Volume of water allowed to escape:
2. Average rate of discharge:
3. Depth of water if the vessel is brought to rest:
222. Problem:
An open cylindrical vessel 1 meter in diameter contains water at a depth of 3.56 m.
1. If it is rotated at a speed of 80 rpm about its vertical axis, determine the least depth of the
vessel so that no water will be spilled out.
2. If the vessel is rotated at 120 rpm, how much water will be spilled out?
3. If the vessel is rotated at 140 rpm, what would be the pressure at the center bottom of the tank.
Solution:
1. Least depth of vessel:
2. Water spilled out when α = 120 rpm
3. Pressure at the center bottom of the tank:
223. Problem:
A vertical cylindrical tank 6 m. high and 4 m. in diam. two thirds full of water, is rotated uniformly about
its axis until on the point of overflowing.
1. Compute the linear velocity at the circumference of the tank.
2. How fast will it have to be rotated in order that 6 cu.m of water will be spilled out.
3. If it is rotated at a speed of 48 rpm, what is the pressure at the center bottom of the tank.
Solution:
1. Linear velocity
2. Speed in order that 6 cu.m. will be spilled out.
3. Pressure at the center bottom of the tank: