Chapter 11

Fluid Statics

Younes Sina

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/VM = ρVM = (7.8 gr/cm3) (141 cm3)M = 1100 grams(b) ) ρ = 7.8 gr/cm3= 7800 kg/m3

D = ρgD = (7800 kg /m3)(9.8 m/s2) = 76000 N/m3

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/gM = 31.75 N /(9.8 m/s2)M = 3.2 kgM = 3200 gr(b) ρ = M / V → V = M / ρV = 3200 gr/(2.7 gr/cm3) = 1200 cm3

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 ofthe alloying process.

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 = 755010306.2 -10.4Vc = 75502756.2 = 10.4Vc

Vc = 265 cm3

Vg = 534 - Vc → Vg = 534 - 265 = 269 cm3

Mg = ρgVg

Mg = (19.3 gr/cm3) ( 269 cm3 ) = 5190 grMc = 2360 grMgold / 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

Pressure is defined as force per unit area.

p = F / AThe 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 floorby each of its four legs if the cross-sectional area of each leg is 1.5 in2.

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?

Solution:(a) w = Mgw = (0.102 kg)(9.8 m/s2)w = 1.0 N(b) p = F / Ap = 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.

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 = hDp = h ρg

mass density

weight density

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 externaldiameter is 5.0 m. Water has a mass density of ρ = 1000 kg/m3.

15 m

Solution:(a) p = hDp = h ρgp = (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 = pAFexternal = (250,000 N/m2)(4π)(2.5m)2 = 20,000,000 NF = 2.0x107N

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

76 cm

Torricelli

Torricelli’s Experiment at Ocean Level

Calculation of the Atmospheric Pressure:

p = F/Apatm = 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

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.

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 cmpatm = 14.7 psi

Example:Convert the atmospheric pressure from 1.0336 kgf /cm2 to N/m2 or Pa.

Solution:1 kgf = 9.8N1 m = 100 cmpatm = 101,300 Pa.

Solution:p = F / A → F = pAF = (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.

Example : A submarine with a total outer area of 2200 m2 is at a depth of 65.0 mbelow 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 = hDp = h ρgp = (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= pAF = (758,000 N/m2)(2200m) = 1.67x109 N

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

Example : Calculate the downward force necessary to keep a 1.0-lbf basketballsubmerged 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.

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:

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 thatDwater= 1 grf /cm3 and DAl= 2.7 grf /cm3.

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

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.

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

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.

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

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 / Af / 3.14cm2 = 33800 N /530.66cm2

f = 200 N(b) p = f /a = 63.7 N/cm2

p = F / A = 63.7 N/cm2

Homework : Odd numbered problems

chapter 11