This document contains examples and solutions related to fluid statics concepts such as pressure, density, buoyancy, and Pascal's principle. It begins with examples calculating the mass, weight, density, and pressure using given values. Later examples apply concepts like buoyancy, pressure at depths, and pressure transmission using hydraulic jacks. Key formulas introduced include pressure (p=F/A), fluid pressure (p=hρg), and buoyancy (B=Vfluidρfluid). Overall, the document provides practice problems and solutions for understanding fundamental fluid statics principles.

1. Chapter 11
Fluid Statics
Younes Sina

3. Example :
The mass density of steel is 7.8 gr/cm3. A chunk of steel has a volume of
141cm3. Determine
(a) its mass in grams and
(b) its weight density in N/m3
Solution:
(a) ρ = M/V
M = ρV
M = (7.8 gr/cm3) (141 cm3)
M = 1100 grams
(b) ) ρ = 7.8 gr/cm3= 7800 kg/m3
D = ρg
D = (7800 kg /m3)(9.8 m/s2) = 76000 N/m3

4. Example : A piece of aluminum weighs 31.75 N. Determine
(a) its mass
(b) its volume if the mass density of aluminum is 2.7 gr/cm3.
Solution:
(a) w = Mg → M = w/g
M = 31.75 N /(9.8 m/s2)
M = 3.2 kg
M = 3200 gr
(b) ρ = M / V → V = M / ρ
V = 3200 gr/(2.7 gr/cm3) = 1200 cm3

5. Example : The mass densities of gold and copper are 19.3 gr/cm3 and 8.9 gr/cm3,
respectively. A piece of gold that is known to be an alloy of gold and copper has a
mass of 7.55 kg and a volume of 534 cm3. Calculate the mass percentage of gold in
the alloy assuming that the volume of the alloy is equal to the volume of copper
plus the volume of gold. In other words, no volume is lost or gained as a result of
the alloying process.

6. Solution:
Mgold = ρgoldVgold
Mcopper = ρcopperVcopper
Mg + Mc = 7550 gr
ρgVg + ρcVc = 7550
Vg + Vc = 534 cm3→ Vg = 534 - Vc
19.3 (534 - Vc) + 8.9 Vc = 7550
10306.2 -10.4Vc = 7550
2756.2 = 10.4Vc
Vc = 265 cm3
Vg = 534 - Vc → Vg = 534 - 265 = 269 cm3
Mg = ρgVg
Mg = (19.3 gr/cm3) ( 269 cm3 ) = 5190 gr
Mc = 2360 gr
Mgold / Malloy = (5190/7550) = 0.687 = 68.7 %
Karat means the number of portions out of 24 portions.
[68.7 /100] = [ x/24]
x = 16.5 karat

7. Pressure is defined as force per unit area.
p = F / A
The SI unit for pressure is N/m2 called " Pascal."
The American unit is lbf / ft2.
Two useful commercial units are:
kgf / cm2 and lbf / in2 (psi)
Pressure
Example : Calculate the average pressure that a 120-lbf table exerts on the floor
by each of its four legs if the cross-sectional area of each leg is 1.5 in2.

8. Solution:
p = F/A ; p = 120lbf /(4x 1.5 in2) = 20 lbf / in2 or 20 psi
Example : (a) Calculate the weight of a 102-gram mass piece of metal.
If this metal piece is rolled to a square sheet that is 1.0 m on each side,
and then spread over the same size (1.0m x 1.0m ) table,
(b) what pressure would it exert on the square table?

9. Solution:
(a) w = Mg
w = (0.102 kg)(9.8 m/s2)
w = 1.0 N
(b) p = F / A
p = 1.0N / (1.0m x 1.0m)
p = 1.0 N/m2
p = 1.0 Pascal (1.0 Pa)
1 Pa is a small amount of pressure.
The atmospheric pressure is 101,300 Pa.
We may say that the atmospheric pressure is
roughly 100,000 Pa, or 100kPa.

10. Fluid Statics:
The study of fluids at rest is called Fluid Statics.
If we assume that the density of a fluid remains fairly constant for relatively
small depths, the formula for fluid pressure my be written as:
p = hD
p = h ρg
mass density
weight density

11. Example :
Calculate
(a) the pressure due to just water at a depth of 15.0 m below lake surface.
(b) What is the total pressure at that depth if the atmospheric pressure is 101 kPa?
(c) Also find the total external force on a spherical research chamber which external
diameter is 5.0 m. Water has a mass density of ρ = 1000 kg/m3.
15 m

12. Solution:
(a) p = hD
p = h ρg
p = (15.0m)(1000 kg/m3)(9.8 m/s2) = 150,000 N /m2 or Pa
(b) [p total] external = p liquid + p atmosphere
[p total] external = 150,000Pa + 101,000Pa = 250,000Pa
(c) p = F / A → F = pA
Fexternal = (250,000 N/m2)(4π)(2.5m)2 = 20,000,000 N
F = 2.0x107N

13. Vacuum establishes the basis for absolute zero
pressure. Any gas pressure measured with
respect to vacuum is called " absolute pressure. "
Gravity pulls the air molecules around the Earth
toward the Earth's center.
As we move from outer space toward the Earth's
surface the air layers become denser and denser.
It is the weight of the atmosphere that causes
the atmospheric pressure.60 mi
vacuum

14. 76 cm
Torricelli
Torricelli’s Experiment at Ocean Level

15. Calculation of the Atmospheric Pressure:
p = F/A
patm = the mercury weight/the tube cross-sectional Area
VHg = (base area) (height) = (A)(hHg)
DHg =WHg/VHg→ WHg= (VHg)(DHg)
patm = (VHg)(DHg)/A = (A)(hHg)(DHg)/A = hHgDHg
patm = hHgDHg
(formula for pressure in a fluid)
In Torricelli's experiment:
hHg = 76.0 cm and DHg = 13.6 grf /cm3
patm = (76.0cm )(13.6 grf /cm3) = 1033.6 grf /cm2
patm = 1.0336 kgf / cm2
patm = 1.0 kgf /cm2

16. Atmosphere is pressing with a force of 1 kgf = 9.8 N on every cm2 of our bodies and we
are used to it
If you softly place a 2.2 lbf (or 1.0 kgf ) weight
over your finger nail ( A = 1 cm2 almost), you
will experience a pressure of 1.0 kgf /cm2
This pressure acts from all directions perpendicular to our bodies surfaces at any point.
An astronaut working outside a space station
must be in a very strong suit that can hold 1
atmosphere of pressure inside compared to
the zero pressure outside and not explode.

17. Example :
Convert the atmospheric pressure from 1.0336 kgf /cm2 to lbf / in2 or psi.
Solution:
1 kgf = 2.2 lbf
1 in = 2.54 cm
patm = 14.7 psi
Example:
Convert the atmospheric pressure from 1.0336 kgf /cm2 to N/m2 or Pa.
Solution:
1 kgf = 9.8N
1 m = 100 cm
patm = 101,300 Pa.

18. Solution:
p = F / A → F = pA
F = (101,300 N/m2 )(1 m2) =100,000 NF = (1.0336 kgf /cm2 )( 10,000 cm2 )
= 10,000 kgf =10 ton force
Example : The surface area of an average size person is almost 1m2. Calculate
the total force that the atmosphere exerts on such person.

19. Example : A submarine with a total outer area of 2200 m2 is at a depth of 65.0 m
below ocean surface. The density of ocean water is 1030 kg/m3. Calculate
(a) the pressure due to water at that depth
(b) the total external pressure at that depth
(c) the total external force on it. Let g = 9.81 m/s2.
Solution:
(a) p = hD
p = h ρg
p = (65.0m)(1030 kg/m3)(9.81 m/s2) = 657,000 N /m2 or Pa
(b) [p total] external = p liquid + p atmosphere
[ p total ]external = 657,000 Pa + 101,000 Pa = 758,000 Pa
(c) p = F/A → F= pA
F = (758,000 N/m2)(2200m) = 1.67x109 N

20. Buoyancy, Archimedes' Principle
When a non-dissolving object is submerged in a fluid (liquid or gas), the fluid
exerts an upward force onto the object that is called the buoyancy force (B).
B = Vobject Dfluid

21. Example : Calculate the downward force necessary to keep a 1.0-lbf basketball
submerged under water knowing that its diameter is 1.0 ft. The American unit
for the weight density of water is Dwater = 62.4 lbf /ft3.

22. Solution:
The volume of the basketball (sphere) is:
Vobject = (4/3) π R3 = (4/3)(3.14)(0.50 ft)3 = 0.523 ft3
The upward force (buoyancy) on the basketball is:
B = Vobject Dfluid = (0.523 ft3)(62.4 lbf / ft3) = 33 lbf
Water pushes the basketball up with a force of magnitude 33 lbf while gravity
pulls it down with a force of 1.0 lbf (its weight); therefore, a downward force
of 32 lbf is needed to keep the basketball fully under water.
The force diagram is shown below:

23. Example: Calculate the necessary upward force to keep a (5.0 cm)(4.0 cm)(2.0 cm)-
rectangular aluminum bar from sinking when submerged in water knowing that
Dwater= 1 grf /cm3 and DAl= 2.7 grf /cm3.

24. Solution:
Vobject = (5.0cm)(4.0cm)(2.0cm) = 40cm3
B = Vobject Dfluid = (40cm3)(1 grf/ cm3) = 40 grf
w = Vobject Dobject = (40cm3)(2.7 grf / cm3) = 110 grf
Water pushes the bar up with a force of magnitude 40 grf while gravity pulls
it down with 110 grf ; therefore, an upward force of 70 grf is needed to keep
the bar fully under water and to avoid it from sinking. The force diagram is
shown below:
The weight of the bar in air

25. Example: A boat has a volume of 40.0 m3 and a mass of 2.00 tons. What load will
push 75.0% of its volume into water?
Each metric ton is 1000 kg. Let g = 9.81 m/s2.

26. Solution:
Vobj = 0.750 x 40.0 m3 = 30.0 m3
B =Vobject Dfluid = (30.0m3)(1000 kg /m3)(9.81 m/s2) = 294,000 N
w = Mg = (2.00 x 103 kg)(9.81 m/s2) = 19600 N
F = B - w = 294,000N - 19600N = 274,000 N
F (load)w

27. Pascal's Principle
The pressure imposed at any point of a confined fluid transmits itself to all
points of that fluid without significant losses.
One application of the Pascal's principle is the mechanism
in hydraulic jacks. This pressure transmits throughout the
oil as well as onto the internal boundaries of the jack
specially under the big piston. On the big piston, the big
load F, pushes down over the big area A. This pressure
is F/A . The two pressures must be equal, according to
Pascal's principle. We may write:
f /a = F/A
Although, for balance, the force that pushes down on the
big piston is much greater in magnitude than the force that
pushes down on the small piston; however, the small piston
goes through a big displacement in order for the big piston
to go through a small displacement.

28. Example:
In a hydraulic jack the diameters of the small and big pistons are 2.00 cm and
26.00 cm respectively. A truck that weighs 33800 N is to be lifted by the big
piston. Find
(a) the force that has to push the smaller piston down
(b) the pressure under each piston
f /a = F/A

29. Solution:
(a) a = π r2 = π (1.00cm)2 = 3.14 cm2
A = π R2 = π (13.00cm)2 = 530.66 cm2
f / a = F / A
f / 3.14cm2 = 33800 N /530.66cm2
f = 200 N
(b) p = f /a = 63.7 N/cm2
p = F / A = 63.7 N/cm2

30. Homework :
Odd numbered problems
chapter 11