CHAPTER 23 HYDROSTATICS
EXERCISE 126, Page 273
1. A force of 280 N is applied to a piston of a hydraulic system of cross-sectional area 0.010 m 2 .
Determine the pressure produced by the piston in the hydraulic fluid.
Pressure, p =
force
280 N
=
= 28000 Pa = 28 kPa
0.010 m 2
area
That is, the pressure produced by the piston is 28 kPa
2. Find the force on the piston of question 1 to produce a pressure of 450 kPa.
Pressure, p = 450 kPa = 450000 Pa
Pressure p =
force
hence, force = pressure × area
area
= 4540000 × 0.010 = 4500 N = 4.5 kN
3. If the area of the piston in question 1 is halved and the force applied is 280 N, determine the new
pressure in the hydraulic fluid.
New area =
0.010
= 0.005 m 2
2
New pressure, p =
force
280 N
=
= 56000 Pa = 56 kPa
area
0.005 m 2
299
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EXERCISE 127, Page 275
1. Determine the pressure acting at the base of a dam, when the surface of the water is 35 m above
base level. Take the density of water as 1000 kg/m 3 .
Pressure at base of dam, p = ρgh = 1000 kg/m 3 × 9.8 m / s 2 × 0.35 m = 343000 Pa = 343 kPa
2. An uncorked bottle is full of sea water of density 1030 kg/m 3 . Calculate, correct to 3 significant
figures, the pressures on the side wall of the bottle at depths of (a) 30 mm, and (b) 70 mm below
the top of the bottle.
Pressure on the side wall of the bottle, p = ρgh
(a) When depth, h = 30 mm = 30 ×10−3 m ,
pressure, p = 1030 kg/m 3 × 9.8 m / s 2 × 30 ×10−3 m = 303 Pa
(b) When depth, h = 70 mm = 70 ×10−3 m ,
pressure, p = 1030 kg/m 3 × 9.8 m / s 2 × 70 ×10−3 m = 707 Pa
3. A U-tube manometer is used to determine the pressure at a depth of 500 mm below the free
surface of a fluid. If the pressure at this depth is 6.86 kPa, calculate the density of the liquid used
in the manometer.
Pressure, p = ρgh
hence,
6.86 ×103 Pa = ρ × 9.8 m / s 2 × 500 ×10−3 m
from which, density of liquid, ρ =
6.86 ×103
= 1400 kg/m 3
−3
9.8 × 500 ×10
300
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4. A submarine pressure hull in the form of a circular cylinder is of external diameter 10 m and
length 200 m. It dives to the bottom of the Mariana Tench which is 11.52 km deep. What will be
the mass of water acting on the submarines’ circular surface in terms of the number of London
double-decker buses, given that the mass of a London double-decker bus is 7 tonnes. Assume
that the density of water, ρ = 1020 kg/m3 and gravitational acceleration, g = 9.81 m/s2.
Surface area of submarine pressure hull = π × d × L = π × 10 × 200 = 2000 π
= 6283.19 m2
Pressure = ρgh = 1020 × 9.81 × (11.52 ×103) = 115.27 ×106 Pa
Force = pressure × area = 115.27 ×106 × 6283.19 = 7.243 ×1011 N
Hence, weight = (7.243 ×1011 N)/(9.81 N/kg) = 7.383 ×1010 kg
= (7.383 ×1010 kg)/(1000 kg/t) = 73.833 ×106 tonne
Hence, number of buses = (73.833 ×106 tonne)/(7 tonnes/bus) = 10.55 ×106
i.e. the mass of water acting on the submarines’ circular surface is equivalent to 10.55 million
London double-decker buses.
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EXERCISE 128, Page 276
1. The height of a column of mercury in a barometer is 750 mm. Determine the atmospheric
pressure, correct to 3 significant figures. Take the gravitational acceleration as 9.8 m/s 2 and the
density of mercury as 13600 kg/m.
Atmospheric pressure, p = ρgh = 13600 kg/m 3 × 9.8 m / s 2 × 750 ×10−3 m
= 99960 Pa = 100 kPa
2. A U-tube manometer containing mercury gives a height reading of 250 mm of mercury when
connected to a gas cylinder. If the barometer reading at the same time is 756 mm of mercury,
calculate the absolute pressure of the gas in the cylinder, correct to 3 significant figures. Take the
gravitational acceleration as 9.8 m/s 2 and the density of mercury as 13600 kg/m.
Pressure, p1 = ρgh = 13600 kg/m 3 × 9.8 m / s 2 × 250 ×10−3 m
= 33320 Pa = 33.32 kPa
Pressure, p 2 = ρgh = 13600 kg/m 3 × 9.8 m / s 2 × 756 ×10−3 m
= 100760 Pa = 100.76 kPa
Absolute pressure = atmospheric pressure + gauge pressure
= p 2 + p1 = 100.76 + 33.32 = 134 kPa
3. A water manometer connected to a condenser shows that the pressure in the condenser is
350 mm below atmospheric pressure. If the barometer is reading 760 mm of mercury, determine
the absolute pressure in the condenser, correct to 3 significant figures.
Take the gravitational
acceleration as 9.8 m/s 2 and the density of water as 100 kg/m.
Pressure, p1 = - ρ 1 gh 1 = - 1000 kg/m 3 × 9.8 m / s 2 × 350 ×10−3 m
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© John Bird & Carl Ross Published by Taylor and Francis
= - 3430 Pa = - 3.43 kPa
Pressure, p 2 = ρ 2 gh 2 = 13600 kg/m 3 × 9.8 m / s 2 × 760 ×10−3 m
= 101293 Pa = 101.3 kPa
Absolute pressure = atmospheric pressure + gauge pressure
= p 2 + p1 = 101.3 - 3.43 = 97.9 kPa
4. A Bourdon pressure gauge shows a pressure of 1.151 MPa. If the absolute pressure is 1.25 MPa,
find the atmospheric pressure in millimetres of mercury. Take the gravitational acceleration as
9.8 m/s 2 and the density of mercury as 13600 kg/m.
Atmospheric pressure = absolute pressure - gauge pressure
= 1.25 MPa – 1.151 MPa = 0.099 MPa = 0.099 ×106 Pa
Atmospheric pressure, p = ρgh = 13600 kg/m 3 × 9.8 m / s 2 × h
i.e.
0.099 ×106 Pa = 13600 kg/m 3 × 9.8 m / s 2 × h
from which,
height, h =
0.099 ×106 Pa
= 0.743 m
13600kg / m3 × 9.8m / s 2
i.e. atmospheric pressure in millimetres of mercury = 0.743 m ×
1000 mm
= 743 mm
1m
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EXERCISE 129, Page 277
1. A body of volume 0.124 m 3 is completely immersed in water of density 1000 kg/m 3 . What is
the apparent loss of weight of the body? Take the gravitational acceleration as 9.8 m/s 2 .
Mass, m = density, ρ × volume, V = 1000 kg/m 3 × 0.124 m 3
= 124 kg
Apparent loss of weight of the body, W = ρ × V × g
= 124 kg × 9.8 m/s 2 = 1215 N = 1.215 kN
2. A body of weight 27.4 N and volume 1240 cm 3 is completely immersed in water of specific
weight 9.81 kN/m 3 . What is its apparent weight? Take the gravitational acceleration as
9.8 m/s 2 and the density of water as 1000 kg/m 3 .
Body weight, W 1 = 27.4 N
Apparent weight, W 2 = 27.4 - ρ × V × g
= 27.4 – (1000 kg/m 3 × 1240 ×10−6 m3 × 9.8 m/s 2 )
= 27.4 N – 12.152 N
= 15.25 N
3. A body weighs 512.6 N in air and 256.8 N when completely immersed in oil of density
810 kg/m 3 . What is the volume of the body? Take the gravitational acceleration as 9.8 m/s 2 .
W = ρ oil × V × g
i.e.
i.e.
(512.6 – 256.8) = ρ oil × V × g
255.8 = 810 × V × 9.8
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© John Bird & Carl Ross Published by Taylor and Francis
from which, volume, V =
255.8
= 0.03222 m 3 or 32.22 dm 3
810 × 9.8
4. A body weighs 243 N in air and 125 N when completely immersed in water. What will it weigh
when completely immersed in oil of relative density 0.8? Take the gravitational acceleration as
9.8 m/s 2 and the density of water as 1000 kg/m 3 .
W=ρ×V×g
i.e.
i.e.
(243 – 125) = ρ water × V × g
118 = 1000 × V × 9.8
from which, volume, V =
118
= 0.012041 m 3
1000 × 9.8
Weight in oil = 243 - ρ oil × V × g = 243 – (0.8 × 1000) × 0.012041 × 9.8
= 243 – 94.4 = 148.6 N
5. A watertight rectangular box, 1.2 m long and 0.75 m wide, floats with its sides and ends vertical
in water of density 1000 kg/m 3 . If the depth of the box in the water is 280 mm, what is its
weight? Take the gravitational acceleration as 9.8 m/s 2 .
Volume of box, V = 1.2 m × 0.75 m × 280 ×10−3 m = 0.252 m 3
Weight of box, W = ρ × V × g
= 1000 × 0.252 × 9.8
= 2469.6 N = 2.47 kN
6. A body weighs 18 N in air and 13.7 N when completely immersed in water of density
1000 kg/m 3 . What is the density and relative density of the body? Take the gravitational
acceleration as 9.8 m/s 2 .
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© John Bird & Carl Ross Published by Taylor and Francis
W=ρ×V×g
(18 – 13.7) = ρ water × V × g
i.e.
i.e.
4.3 = 1000 × V × 9.8
from which, volume, V =
4.3
= 4.388 ×10−4 m 3
1000 × 9.8
mass
18 N
1
Density of body,=
ρ=
×
−4
3
volume 4.388 ×10 m 9.8 N / kg
= 4186 kg/ m 3 or 4.186 tonnes/ m 3
Relative density =
density
4186 kg / m3
= 4.186
=
density of water 1000 kg / m3
7. A watertight rectangular box is 660 mm long and 320 mm wide. Its weight is 336 N. If it floats
with its sides and ends vertical in water of density 1020 kg/m 3 , what will be its depth in the
water? Take the gravitational acceleration as 9.8 m/s 2 .
Volume of box, V = 660 ×10−3 m × 320 ×10−3 m × D where D = depth of box
Weight, W = ρ × V × g
i.e.
336 = 1020 kg/m 3 ×( 660 ×10−3 m × 320 ×10−3 m × D) × 9.8
from which, depth of box, D =
336
= 0.159 m = 159 mm
1020 × 0.66 × 0.32 × 9.8
8. A watertight drum has a volume of 0.165 m 3 and a weight of 115 N. It is completely submerged
in water of density 1030 kg/m 3 , held in position by a single vertical chain attached to the
underside of the drum. What is the force in the chain? Take the gravitational acceleration as
9.8 m/s 2 .
Weight of drum , W = 115 N
Upthrust = ρ × V × g = 1030 kg/m 3 × 0.165 m 3 × 9.8 m/s 2 = 1665.5 N
306
© John Bird & Carl Ross Published by Taylor and Francis
Hence, the force in the chain = 1665.5 – 115 = 1551 N = 1.551 kN
307
© John Bird & Carl Ross Published by Taylor and Francis
EXERCISE 130, Page 283
1. Determine the gauge pressure acting on the surface of a submarine that dives to a depth of
500 m. Take water density as 1020 kg/m 3 and g as 9.81 m/s 2 .
Gauge pressure, p = ρgh
= 1020
kg
m
× 9.81 2 × 500 m
3
m
s
= 5003100 N/m 2 ×
i.e.
1bar
10 N / m 2
5
pressure, p = 50.03 bar
2. Solve Problem 1, when the submarine dives to a depth of 780 m.
Gauge pressure, p = ρgh
= 1020
kg
m
× 9.81 2 × 780 m
3
m
s
= 7804836 N/m 2 ×
i.e.
1bar
10 N / m 2
5
pressure, p = 78.05 bar
3. If the gauge pressure measured on the surface of the submarine of Problem 1 were 92 bar, at
what depth has the submarine dived to?
Pressure, p = 92 bar = 92 × 105 Pa
Gauge pressure, p = ρgh
i.e.
92 × 105 = 1020
kg
m
× 9.81 2 × h
3
m
s
92 ×105
from which, depth, h =
= 919.4 m
1020 × 9.81
308
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4. A tank has a flat rectangular end, which is of size 4 m depth by 3 m width. If the tank filled with
water to its brim and the flat end is vertical, determine the position of its centre of pressure and
the thrust on this end. Take water density as 1000 kg/m 3 and g as 9.81 m/s 2 .
Distance of centroid from free surface, h =
2
× 4 = 2.667 m
3
i.e. centre of pressure = 2.667 m
Pressure, p = ρgh
= 1000
kg
m
× 9.81 2 × 4 m
3
m
s
= 39240 Pa
Thrust = pressure × area
1

= 39240 ×  × 4 × 3 
2

= 235440 N = 0.235 MN
5. If another vertical flat rectangular end of the tank of Problem 4 is of size 6 m depth by 4 m
width, determine the position of the centre of pressure and the thrust on this end. The depth
of water at this end may be assumed to be 6 m.
Distance of centroid from free surface, h =
2
×6 = 4 m
3
i.e. centre of pressure = 4 m
Pressure, p = ρgh
= 1000
m
kg
× 9.81 2 × 6 m
3
s
m
= 58860 Pa
Thrust = pressure × area
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© John Bird & Carl Ross Published by Taylor and Francis
1

= 58860 ×  × 4 × 6 
2

=706320 N = 0.706 MN
6. A tank has a flat rectangular end, which is inclined to the horizontal surface, so that θ = 30°,
where θ is as defined in Figure 23.11, page 283. If this end is of size 6 m height and 4 m width,
determine the position of the centre of pressure from the top and thrust on this end. The tank
may be assumed to be just full.
Total thrust on plane surface = F = ρg sin θ ∫ y dA
_
= ρg sin θ × A h
_
where h = distance of the centroid of the plane from the free surface.
_
Hence, thrust, F = 1000 × 9.81 × sin 30º × 4 × 6 × h
_
= 117720 h
_
The position of the centre of pressure from the top, h =
2
× 6 sin 30º = 2 m
3
Hence, thrust, F = 117720 × 2 = 235440 N = 0.235 MN
310
© John Bird & Carl Ross Published by Taylor and Francis
EXERCISE 131, Page 287
1. A ship is of mass 10000 kg. If the ship floats in the water, what is the value of its buoyancy?
Take g = 9.81 m/s 2 and density of water ρ = 1020 kg/m 3 .
Buoyancy = 1020
m
kg
× 9.81 2 = 98100 N = 98.1 kN
3
s
m
2. A submarine may be assumed to be in the form of a circular cylinder of 10 m external diameter
and of length 100 m. If the submarine floats just below the surface of the water, what is the value
of its buoyancy? Take g = 9.81 m/s 2 and density of water ρ = 1020 kg/m 3 .
Volume = πr 2 h = π× 52 ×100 = 7854 m3
Buoyancy = ρVg = 1020 × 7854 × 9.81
= 78588500 N = 78.59 MN
3. A barge of length 20 m and of width 5 m floats on an even keel at a depth of 2 m. What is the
value of its buoyancy?
Volume below the water line = 20 × 5 × 2 =200 m3
Buoyancy = ρVg = 1020 × 200 × 9.81
= 2001240 N = 2 MN
4. An inclining experiment is carried out on the barge of Problem 3 where a mass of 20 kg is
moved transversely across the deck by a distance of 2.2 m. The resulting angle of keel is 0.8°.
Determine the metacentric height, GM.
w × d = W × GM × tan θ
i.e. 20 × 9.81 × 2.2 = 2 ×106 × GM × tan 0.8º
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from which, metacentric centre, GM =
20 × 9.81× 2.2
= 0.0155 m
2 ×106 × tan 0.8°
5. Determine the value of the radius of curvature of the centre of buoyancy, namely, BM, for the
barge of Problems 3 and 4, and hence the position of the centre of gravity above the keel, KG.
BD3 20 × 53
I
12
12 = 1.0417 m
BM == =
V
V
200
KM = KB + BM
=
2
m + 1.0417 m = 2.042 m
1
KG = KM – GM = 2.042 – 0.0155 = 2.026 m
6. If the submarine of Problem 2 floats so that its top is 2 m above the water, determine the centre
of curvature of the centre of buoyancy, BM.
From chord theory, c 2 = 2 × 8 from which, c = 4 m
B = 2c = 8 m
I=
100 × 83
= 4267 m 4
12
R2
Now, volume = 7854 ( α − sin α ) × l
2
α
c
α
R = 5 m;
α 106.26° = 1.85 rad
= sin
=
  0.8 ; = 53.13° = 0.96 rad;=
2
R
2
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© John Bird & Carl Ross Published by Taylor and Francis
Volume = 7854 -
BM =
52
(1.85 − 0.96 ) ×100 = 6741.5 m3
2
I
4267
= 0.633 m
=
V 6741.5
EXERCISE 132, Page 287
Answers found from within the text of the chapter, pages 272 to 286.
EXERCISE 133, Page 288
1. (b) 2. (d) 3. (a) 4. (a) 5. (c) 6. (d) 7. (b) 8. (c) 9. (c) 10. (d) 11. (d) 12. (d) 13. (c)
14. (b) 15. (c) 16. (a) 17. (b) 18. (f) 19. (a) 20. (b) 21. (c) 22. (c) 23. (a) 24. (c)
313
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