Chapter 11
Fluid Statics
In this chapter fluids at rest will be studied. Mass density, weight density, pressure, fluid pressure, buoyancy, and Pascal's principle will be discussed. In the following, the symbol ( ρ ) is pronounced " rho ."
Example 1 : The mass density of steel is 7.8 gr /cm^{3}. A chunk of steel has a volume of 141cm^{3}. Determine (a) its mass in grams and (b) its weight density in N/m^{3}. Solve before looking at the solution. Use horizontal fraction bars.
Solution: (a) Since ρ = M / V ; M = ρV ; M = (7.8 gr / cm^{3}) (141 cm^{3} ) ; M = 1100 grams.
Before going to Part (b), let's first convert (gr/cm^{3}) to its Metric version (kg/m^{3}). Use horizontal fraction bars.
7.8 gr / cm^{3 } = 7.8 (0.001kg) / (0.01m)^{3} = 7800 kg/m^{3}.
1 kg is equal to 1000gr. This means that 1 gr is 0.001kg as is used above.
Also, 1 m is 100cm. This means that 1cm is 0.01m. Cubing each, results in : 1cm^{3} = 0.000001m^{3} as is used above. Now, let's solve Part (b).
(b) D = ρg ; D = [7800 kg /m^{3}] [ 9.8 m/s^{2}] = 76000 N /m^{3}.
Not only you should write Part (b) with horizontal fraction bars, but also check the correctness of the units as well.
Example 2 : A piece of aluminum weighs 31.75N. Determine (a) its mass and (b) its volume if the mass density of aluminum is 2.7gr/cm^{3}.
Solution: (a) w = Mg ; M = w / g ; M = 31.75N / [9.8 m/s^{2}] ; M = 3.2 kg ; M = 3200 grams.
(b) ρ = M / V ; V = M / ρ ; V = 3200gr / [2.7 gr /cm^{3}] = 1200 cm^{3}.
Example 3 : The mass densities of gold and copper are 19.3 gr/cm^{3 }and 8.9 gr/cm^{3}, respectively. A piece of gold that is known to be an alloy of gold and copper has a mass of 7.55kg and a volume of 534 cm^{3}. 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. Do your best to solve it by yourself first.
Solution: Two equations can definitely be written down. The sum of masses as well as the sum of volumes are given. The formula M = ρV is applicable to both metals. M_{gold} = ρ_{gold}V_{gold} and M_{copper} = ρ_{copper}V_{copper} . Look at the following as a system of two equations in two unknowns:
M_{g} + M_{c } = 7550 gr  ρ_{g}V_{g} + ρ_{c}V_{c } = 7550  19.3 V_{g}+ 8.9V_{c } = 7550  19.3 V_{g} + 8.9V_{c } =7550 
V_{g} + V_{c } = 534 cm^{3}  V_{g} + V_{c } = 534  V_{g} + V_{c } = 534  Vg = 534  V_{c} 
Substituting for V_{g} in the first equation yields:
19.3 (534  V_{c}) + 8.9 V_{c} = 7550 ; 10306.2 10.4V_{c} = 7550 ; 2756.2 = 10.4V_{c} ; V_{c} = 265 cm^{3}.
Since Vg = 534  V_{c} ; therefore, Vg = 534  265 = 269 cm^{3}.
The masses are: M_{g} = ρ_{g}V_{g} ; M_{g} = (19.3 gr/cm^{3}) ( 269 cm^{3} ) = 5190 gr ; M_{c} = 2360 gr.
The mass percentage of gold in the alloy (7750gr) is M_{gold }/ M_{alloy} = (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:
Pressure is defined as force per unit area. Let's use lower case p for pressure; therefore, p = F / A. The SI unit for pressure is N/m^{2 } called " Pascal." The American unit is lb_{f} / ft^{2}. Two useful commercial units are: kgf / cm^{2} and lbf / in^{2} or psi.
Example 4: Calculate the average pressure that a 120lbf table exerts on the floor by each of its four legs if the crosssectional area of each leg is 1.5 in^{2}.
Solution: p = F / A ; p = 120lb_{f} / (4x 1.5 in^{2}) = 20 lbf / in^{2} or 20 psi.
Example 5: (a) Calculate the weight of a 102gram mass piece of metal. If this metal piece is rolled to a square sheet that is 1.0m 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 = Mg ; w = (0.102 kg)(9.8 m/s^{2}) ; w = 1.0 N
(b) p = F / A ; p = 1.0N / (1.0m x 1.0m) ; p = 1.0 N/m^{2} ; p = 1.0 Pascal (1.0Pa)
As you may have noticed, 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 We will calculate this later.
Fluid Statics:
Fluid Pressure: Both liquids and gases are considered fluids. The study of fluids at rest is called Fluid Statics. The pressure in stationary fluids depends on weight density, D, of the fluid and depth, h, at which pressure is to be calculated. Of course, as we go deeper in a fluid, its density increases slightly because at lower points, there are more layers of fluid pressing down causing the fluid to be denser. For liquids, the variation of density with depth is very small for relatively small depths and may be neglected. This is because of the fact that liquids are incompressible. For gases, the density increase with depth becomes significant and may not be neglected. Gases are called compressible fluids. 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 or p = h ρg
where ρ is the mass density and D is the weight density of the fluid.
Example 6: Calculate (a) the pressure due to just water at a depth of 15.0m below lake surface. (b) What is the total pressure at that depth if the atmospheric pressure is 101kPa? (c) Also find the total external force on a spherical research chamber which external diameter is 5.0m. Water has a mass density of ρ = 1000 kg/m^{3}.
Solution: (a) p = hD ; p = h ρg ; p = (15.0m)(1000 kg/m^{3})(9.8 m/s^{2}) = 150,000 N /m^{2} or Pa.
(b) [p _{total}]_{ external } = p _{liquid} + p _{atmosphere} ; [p _{total}]_{ external } = 150,000Pa + 101,000Pa = 250,000Pa.
(c) p = F / A ; solving for F, yields: F = pA ; F_{external } = (250,000 N/m2)(4π)(2.5m)^{2} = 20,000,000 N.
F = 2.0x10^{7}N ( How may millions?!)
Chapter 11 Test Yourself 1:
1) Average mass density, ρ is defined as (a) mass of unit volume (b) mass per unit volume (c) a &b. click here
2) Average weight density, D is defined as (a) weight per unit volume (b) mass of unit volume times g (c) both a & b.
3) D = ρg is correct because (a) w = Mg (b) D is weight density and ρ is mass density (c) both a & b.
4) 4.0cm^{3} of substance A has a mass of 33.0grams, and 8.0cm^{3} of substance B has a mass of 56.0 grams. (a) A is denser than B (b) B is denser than A (c) Both A and B have the same density. click here
Problem: 1gram was originally defined to be the mass of 1cm^{3} of pure water. Answer the following question by first doing the calculations. Make sure to write down neatly with horizontal fraction bars. click here
5) On this basis, one suitable unit for the mass density of water is (a) 1 cm^{3}/gr (b) 1 gr/cm^{3} (c) both a & b.
6) We know that 1kg = 1000gr. We may say that (a) 1gr = (1/1000)kg (b) 1gr = 0.001kg (c) both a & b.
7) We know that 1m = 100cm. We may say that (a) 1m^{3} = 100cm^{3} (b) 1m^{3} = 10000cm^{3} (c) 1m^{3} = 1000,000cm^{3}.
8) We know that 1cm = 0.01m. We may write (a) 1cm^{3} = 0.000001m^{3} (b) 1cm^{3} = 0.001m^{3} (c) 1cm^{3} = 0.01m^{3}.
9) Converting gr/cm^{3} to kg/m^{3} yields: (a)1gr/cm^{3} = 1000 kg/m^{3} (b)1gr/cm^{3} = 100 kg/m^{3} (c)1gr/cm^{3} = 10 kg/m^{3}.
10) From Q9, the mass density of water is also (a) 1000 kg/m^{3} (b) 1 ton/m^{3}, because 1ton=1000kg (c) both a & b.
11) Aluminum is 2.7 times denser than water. Since ρ_{water} = 1000kg/m^{3} ; therefore, ρ_{Alum.} = (a) 2700kg/m^{3} (b) 27kg/m^{3} (c) 27000kg/m^{3}. click here
12) Mercury has a mass density of 13.6 gr/cm^{3}. In Metric units (kg/m^{3}), its density is (a) 1360 kg/m^{3} (b) 13600 kg/m^{3} (c) 0.00136kg/m^{3}.
13) The weight density of water is (a) 9.8 kg/m^{3} (b) 9800kg/m^{3} (c) 9800N/m^{3}. click here
14) The volume of a piece of copper is 0.00247m^{3}. Knowing that copper is 8.9 times denser than water, first find the mass density of copper in Metric units and then find the mass of the copper piece. Ans. : (a) 44kg (b) 22kg (c) 16kg.
Problem: The weight of a gold sphere is 1.26N. The mass density of gold is ρ_{gold }=_{ }19300kg/m^{3}.
15) The weight density, D, of gold is (a) 1970 N/m^{3} (b) 189000 N/m^{3} (c) 100,000 N/m^{3}.
16) The volume of the gold sphere is (a) 6.66x10^{6}m^{3} (b) 6.66cm^{3} (c) both a & b. click here
17) The radius of the gold sphere is (a) 1.167cm (b) 0.9523cm (c) 2.209cm.
Pressure:
18) Pressure is defined as (a) force times area (b) force per unit area (c) force per length.
19) The Metric unit for pressure is (a) N/m^{3} (b) N/cm^{3} (c) N/m^{2}. click here
20) Pascal is the same thing as (a) lb_{f }/ ft^{2} (b) N/m^{2} (c) lb_{f }/ in^{2}.
21) psi is (a) lb_{f }/ ft^{2} (b) N/m^{2} (c) lb_{m }/ in^{2} (d) none of a, b, or c.
22) A solid brick may be placed on a flat surface on three different sides that have three different surface areas. To create the greatest pressure it must be placed on its (a) largest side (b) smallest side (c) middlesize side.
Problem: 113 grams is about 4.00 ounces. A 102 gram mass is 0.102kg. The weight of a 0.102kg mass is 1.00N. Verify this weight. If a 0.102gram piece of say copper that weighs 1.00N, is hammered or rolled to a flat sheet (1.00m by 1.00m), how thin would that be? May be one tenth of 1 mm? Note that a (1m) by (1m) rectangular sheet of metal may be viewed as a rectangular box which height or thickness is very small, like a sheet of paper. If you place your hand under such thin sheet of copper, do you hardly feel any pressure? Answer the following questions:
23) The weight density of copper that is 8.9 times denser than water is (a) 8900N/m^{2} (b) 1000N/m^{3} (c) 87220N/m^{3}.
24) The volume of a 0.102kg or 1.00N piece (sheet) of copper is (a) 1.15x10^{5}m^{3} (b) 1.15x10^{5}m^{3} (c) 8900m^{3}.
25) For a (1m)(1m) = 1m^{2 } base area of the sheet, its height or thickness is (a) 1.15x10^{5}m (b) 1.15x10^{5}m (c) 8900m.
26) The small height (thickness) in Question 25 is (a) 0.0115mm (b) 0.0115cm (c) 890cm.
27) The pressure (force / area) or (weight / area) that the above sheet generates is (a) 1N/1m^{2} (b) 1 Pascal (c) both a & b.
28) Compared to pressures in water pipes or car tires, 1 Pascal of pressure is (a) a great pressure (b) a medium pressure (c) a quite small pressure.
29) The atmospheric pressure is roughly (a) 100Pa (b) 100,000 Pa (c) 100kPa (d) both b & c.
30) The atmospheric pressure is (a) 14.7 psi (b) 1.0 kgf/m^{2} (c) 1.0 kgf/cm^{2} (d) a & c.
Atmospheric Pressure:
Gravity pulls the air molecules around the Earth toward the Earth's center. This makes the air layers denser and denser as we move from outer space toward the Earth's surface. It is the weight of the atmosphere that causes the atmospheric pressure. The depth of the atmosphere is about 60 miles. If we go 60 miles above the Earth surface, air molecules become very scarce to where we might travel one meter and not collide with even a single molecule (a good vacuum!). Vacuum establishes the basis for absolute zero pressure. Any gas pressure measured with respect to vacuum is called " absolute pressure. " 

Calculation of the Atmospheric Pressure:
The trick to calculate the atmospheric pressure is to place a 1m long test tube filled with mercury inverted over a pot of mercury such that air can not get in the tube. Torricelli ( Italian) was the first to try this. The figure is shown above. In doing this, we will see that the mercury level drops to 76.0cm or 30.0 inches if the experiment is performed at ocean level. The top of the tube lacks air and does not build up air pressure. This device acts as a balance. If it is taken to the top of a high mountain where there is a smaller number of air layers above one's head, the mercury level goes down. This device can even be calibrated to measure elevation for us based on air pressure.
The pressure that the 76.0cm column of mercury generates is equal the pressure that the same diameter column of air generates but with a length of 60 miles ( from the Earth's surface all the way up to the noair region).
Using the formula for pressure ( p = F / A ), the pressure of the mercury column or the atmospheric pressure can be calculated as follows:
p_{atm} = the mercury weight / the tube crosssectional Area. ( Write using horiz. fraction bars).
p_{atm} = (V_{Hg})(D_{Hg}) / A = (A)(h_{Hg})(D_{Hg}) / A = h_{Hg}D_{Hg } . Note that the tube's volume = V_{Hg}. = (base area) (height) = (A)(h_{Hg}.).
p_{atm} = h_{Hg}D_{Hg } ( This further verifies the formula for for pressure in a fluid).
In Torricelli's experiment, h_{Hg } = 76.0cm and D_{Hg }= 13.6 grf /cm^{3} ; therefore ,
p_{atm} = ( 76.0cm )( 13.6 grf /cm^{3} ) = 1033.6 grf / cm^{2}
Converting grf to kgf results in p_{atm} = 1.0336 kgf / cm^{2}
To 2 significant figures, this result is a coincidence: p_{atm} = 1.0 kgf /cm^{2}.
If you softly place a 2.2 lbf (or 1.0 kgf ) weight over your finger nail ( A = 1 cm^{2} almost), you will experience a pressure of 1.0 kgf / cm^{2} (somewhat painful) that is equivalent to the atmospheric pressure. The atmosphere is pressing with a force of 1 kgf = 9.8 N on every cm^{2} of our bodies and we are used to it. 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 7: Convert the atmospheric pressure from 1.0336 kgf / cm^{2} to lbf / in^{2} or psi.
Solution: 1 kgf = 2.2 lbf and 1 in. = 2.54 cm. Convert and show that p_{atm} = 14.7 psi.
Example 8: Convert the atmospheric pressure from 1.0336 kgf / cm^{2} to N / m^{2} or Pascals (Pa).
Solution: 1 kgf = 9.8N and 1 m = 100 cm. Convert and show that p_{atm} = 101,300 Pa.
Example 9: The surface area of an average size person is almost 1m^{2}. Calculate the total force that the atmosphere exerts on such person.
Solution: p = F / A ; F = pA ; F = ( 101,300 N/m^{2} )( 1 m^{2} ) = 100,000 N.
F = ( 1.0336 kgf / cm^{2} )( 10,000 cm^{2} ) = 10,000 kgf = 10 ton force.
Example 10: A submarine with a total outer area of 2200m^{2} is at a depth of 65.0m below ocean surface. The density of ocean water is 1030 kg/m^{3}. Calculate (a) the pressure due to water at that depth, (b) the total external pressure at that depth, and (c) the total external force on it. Let g = 9.81 m/s^{2}.
Solution: (a) p = hD ; p = h ρg ; p = (65.0m)(1030 kg/m^{3})(9.81 m/s^{2}) = 657,000 N /m^{2} or Pa.
(b) [p _{total}]_{ external }= p _{liquid} + p _{atmosphere} ; [ p _{total} ]_{external }= 657,000Pa + 101,000Pa = 758,000Pa.
(c) p = F / A ; solving for F, yields: F = pA ; F = (758,000 N/m^{2})(2200m) = 1.67x10^{9} N.
Buoyancy, Archimedes' Principle:
When a nondissolving 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). The magnitude of the buoyancy force is equal to the weight of displaced fluid. The formula for buoyancy is therefore,
B = V_{object} D_{fluid}
Example 11: Calculate the downward force necessary to keep a 1.0lb_{f} basketball submerged under water knowing that its diameter is 1.0ft. The American unit for the weight density of water is D_{water} = 62.4 lbf /ft^{3}.
Solution: The volume of the basketball (sphere) is: V_{object} = (4/3) π R^{3} = (4/3)(3.14)(0.50 ft)^{3} = 0.523 ft^{3}.
The upward force (buoyancy) on the basketball is: B = V_{object} D_{fluid } = (0.523 ft^{3})(62.4 lbf / ft^{3}) = 33 lb_{f }.
Water pushes the basketball up with a force of magnitude 33 lb_{f} while gravity pulls it down with a force of 1.0 lb_{f } (its weight); therefore, a downward force of 32 lb_{f }is needed to keep the basketball fully under water. The force diagram is shown below:
A Good Link to Try: http://www.mhhe.com/physsci/physical/giambattista/fluids/fluids.html .
Example12: Calculate the necessary upward force to keep a (5.0cm)(4.0cm)(2.0cm)rectangular aluminum bar from sinking when submerged in water knowing that D_{water} = 1 grf / cm^{3 } and D_{Al} = 2.7 grf / cm^{3}.
Solution: The volume of the bar is V_{object} = (5.0cm)(4.0cm)(2.0cm) = 40cm^{3}.
The buoyancy force is: B = V_{object} D_{fluid } = (40cm^{3})(1 grf / cm^{3}) = 40grf.
The weight of the bar in air is w = V_{object} D_{object} = (40cm^{3})(2.7 grf / cm^{3}) = 110grf.
Water pushes the bar up with a force of magnitude 40. grf while gravity pulls it down with 110grf ; 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:
Example13: A boat has a volume of 40.0m^{3} 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/s^{2}.
Solution: V_{obj} = 0.750 x 40.0m^{3} = 30.0m^{3}. B =V_{object} D_{fluid }= (30.0m^{3})(1000 kg /m^{3})(9.81 m/s^{2}) = 294,000N. w = Mg = (2.00 x 10^{3} kg)(9.81 m/s^{2}) = 19600N. F = B  w = 294,000N  19600N = 274,000N.

Pascal's Principle:
An important and useful principle in fluid statics is the " Pascal's Principle." Its statement is as follows: 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. As shown in the figure, a small force, f, applied to a small piston of area, a ,imposes a pressure onto the liquid (oil) equal to f/a. 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. 
Example14: In a hydraulic jack the diameters of the small and big pistons are 2.00cm and 26.00cm respectively. A truck that weighs 33800N is to be lifted by the big piston. Find (a) the force that has to push the smaller piston down, and (b) the pressure under each piston.
Solution: (a) a = π r^{2} = π (1.00cm)^{2} = 3.14 cm^{2} ; A = π R^{2} = π (13.00cm)^{2} = 530.66 cm^{2}
f / a = F / A ; f / 3.14cm^{2} = 33800N / 530.66cm^{2} ; f = 200N
(b) p = f /a = 63.7 N/cm^{2} ; p = F / A = 63.7 N/cm^{2}.
Chapter 11 Test Yourself 2:
1) In Torricelli's experiment of measuring the atmospheric pressure at the ocean level, the height of mercury in the tube is (a) 76.0cm (b) 7.6cm (c) 760mm (d) a & c. click here
2) The space above the tube in the Torricelli's experiment is (a) at regular pressure (b) almost vacuum (c) at a very small amount of mercury vapor pressure, because due to vacuum, a slight amount of mercury evaporates and creates minimal mercury vapor pressure (d) both b & c.
3) The pressure inside a stationary fluid (liquid or gas) depends on (a) mass density and depth (b) weight density and depth (c)depth only regardless of the fluid type. click here
4) A pressure gauge placed 50.0m below ocean surface measures (a) a higher (b) a lower (c) the same pressure compared to a gauge that is placed at the same depth in a lake.
5) The actual pressure at a certain depth in an ocean on a planet that has an atmosphere is equal to (a) just the liquid pressure (b) the liquid pressure + the atmospheric pressure (c) the atmospheric pressure only. click here
6) The formula that calculates the pressure at a certain depth in a fluid is (a) p = h ρ (b) p = hD (c) p = h ρg (d) both b & c.
Problem: Mercury is a liquid metal that is 13.6 times denser than water. Answer the following questions:
7) The mass density of mercury is (a) 13600 kg/m^{3} (b) 13.6 ton / m^{3} (c) both a & b. click here
8) The weight density of mercury is (a) 130,000N/m^{3} (b) 1400N/m^{3} (c) 20,100N/m^{3}.
9) In a mercury tank, the liquid pressure at a depth if 2.4m below mercury surface is (a) 213000N/m^{2} (b) 312000N/m^{3} (c) 312000N/m^{2}. click here
10) In the previous question, the total pressure at that depth is (a) 412000 N/m^{3} (b) 212000N/m^{2} (c) 412000 N/m^{2}.
Problem: This problem shows the effect of depth due to liquid pressure. A long vertical and narrow steel pipe of 1.0cm in diameter is connected to a spherical barrel of internal diameter of 1.00m. The barrel is also made of steel and can withstand an internal total force of 4,00,000N! The barrel is gradually filled with water through the thin and long pipe on its top while allowing the air out of the tank. When the spherical part of the tank is full of water, further filling makes the water level in the thin pipe to go up fast (Refer to Problem 7 at the very end of this chapter for a suitable figure). As the water level goes up, it quickly builds up pressure. p = hD. If the pipe is 15.5m long, for example, answer Questions 11 through 14.
11) The liquid pressure at the center of the barrel is (a) 151900N/m^{2} (b) 156800N/m^{2} (c) 16000kg/m^{2}.
12) The total internal area of the barrel is (a) 0.785m^{2}. (b) 3.141m^{2}. (a) 1.57m^{2}. click here
13) The total force on the internal surface of the sphere is (a) 246000N (b) 123000N (c) 492000N.
14) Based on the results in the previous question, the barrel (a) withstands the pressure (b) does not withstand the pressure. click here
15) The liquid pressure at a depth of 10m ( 33ft ) below water on this planet is roughly (a) 200,000Pa (b) 100,000Pa (c) 300,000Pa.
16) Since the atmospheric pressure is also roughly 100,000 Pa, we may say that every 10m of water depth or height is equivalent to (a) 1 atmosphere of pressure (b) 2 atmospheres of pressure (c) 3 atmospheres of pressure.
17) In the Torricelli's experiment, if the formula P = hD is used to calculate the pressure caused by 0.760m of mercury the value of atmospheric pressure becomes (a) 9800 N/m^{2} (b) 98000 N/m^{2} (c) 101,000 N/m^{2}. Perform the calculation. ρ_{mercury} = 13600kg/m^{3}. click here
18) To convert 101,000 N/m^{2} or the atmospheric pressure to lb_{f }/in^{2} or psi, one may replace (N) by 0.224 lb_{f} and (m) by 39.37 in. The result of the conversion is (a) 25.4ps (b) 14.7psi (c) 16.2psi. Perform the calculation.
19) To convert 101,000 N/m^{2} or the atmospheric pressure to kg_{f }/cm^{2}, one may replace (N) by 0.102 kg_{f} and (m) by 100cm. The result of the conversion is (a) 1.0 kg_{f }/cm^{2 } (b) 2.0 kg_{f } /cm^{2} (c) 3.0kg_{f } /cm^{2}. Perform the calculation.
20) Due to the atmospheric pressure, every cm^{2} of our bodies is under a force of (a) 1.0kgf (b) 9.8N (c) both a & b. click here
21) An example of an area approximately close to 1cm^{2} is the size of (a) a finger nail (b) a quarter (c) a dollar coin.
22) The formula that calculates the area of sphere is A_{sphere} = (a) πr^{2} (b) 2πr^{2} (c) 4πr^{2}.
23) The force due to liquid pressure on a 5.0m diameter spherical chamber that is at a depth of 40.0m below ocean surface is (a) 3.14x10^{6}N (b) 3.08x10^{7}N (c) 6.16x10^{6}N. click here
24) Buoyancy for a submerged object in a nondissolving liquid is (a) the upward force that the liquid exerts on that object (b) equal to the mass of the displaced fluid (c) equal to the weight of the displaced fluid (d) a & c.
25) The direction of the buoyancy force is (a) always downward (b) always upward (c) sometimes upward and sometimes downward. click here
26) The buoyancy on a cube 0.080m on each side and fully submerged in water is (a) 5.02N (b) 63N (c) 0.512N.
27) If the cube in the previous question is made of aluminum ( ρ = 2700 kg/m^{3}), it has a weight of (a) 13.5N (b) 170N (c) 0.189N. click here
28) The force necessary to keep the cube in the previous question from sinking in water is (a) 107N (b) 8.5N (c) 7.0N.
Problem: A (12.0m)(50.0m)(8.0mheight)barge has an empty mass of 1250 tons. For safety reasons and preventing it from sinking, only 6.0m of its height is allowed to go under water. Answer the following questions:
29) The total volume of the barge is (a) 480m^{3} (b) 60m^{3} (c) 4800m^{3}. click here
30) The effective (safe) volume of the barge that can be submerged in water is (a) 3600m^{3} (b) 50m^{3} (c) 360m^{3}.
31) The buoyancy force on the barge when submerged in water to its safe height is (a) 1.83x10^{6}N (b) 5.43x10^{8}N (c) 3.53x10^{7}N. click here
32) The safe load that the barge can carry is (a) Buoyancy + its empty weight (b) Buoyancy  its empty weight (c) Buoyancy  its volume.
33) The mass of the barge in kg is (a) 1.25x10^{3} ton (b) 1.25x10^{6} kg (c) a & b.
34) The weight of the barge in N is (a) 1.23x10^{7} N (b) 2.26x10^{7} N (c) neither a nor b. click here
35) The safe load in N that the barge can carry is (a) 3.41x10^{7}N (b) 2.3x10^{7}N (c) 2.53x10^{7}N.
36) The safe load in Metric tons is (a) 1370 ton (b) 2350 ton (a) 5000 ton.
37) According to Pascal's principle, a pressure imposed (a) on any fluid (b) on a confined fluid (c) on a monoatomic fluid, transmits itself to all points of that fluid without any significant loss. click here
Problem: In a hydraulic jack the diameter of the big cylinder is 10.0 times the diameter of the small cylinder. Answer the following questions:
38) The ratio of the areas (of the big piston to the small piston) is (a) 10.0 (b) 100 (c) 50.0. click here
39) The ratio of the applied forces (on the small piston to that of the big piston) is (a) 1/100 (b) 1/10 (c) 1/25. click here
40) If the applied force to the small piston is 147.0N, the mass of the car it can lift is (a) 1200kg (b) 3500kg (c) 1500kg.
Problems:
1) The mass density of mercury is 13.6 gr /cm^{3}. A cylindrical vessel that has a height of 8.00cm and a base radius of 4.00cm is filled with mercury. Find (a) the volume of the vessel. Calculate the mass of mercury in (b) grams, (c) kg, and (d) find its weight both in N and kgf. Note that 1kgf = 9.81N.
2) A piece of copper weighs 49N. Determine (a) its mass and (b) its volume. The mass density of copper is 8.9gr/cm^{3}.
3) The mass densities of gold and copper are 19.3 gr/cm^{3 }and 8.9 gr/cm^{3}, respectively. A piece of gold necklace has a mass of 51.0 grams and a volume of 3.50 cm^{3}. Calculate (a) the mass percentage and (b) the karat 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.
4) Calculate the average pressure that a 32ton tenwheeler truck exerts on the ground by each of its ten tires if the contact area of each tire with the ground is 750 cm^{2}. 1 ton = 1000kg. Express your answers in (a) Pascal, (b) kgf/cm^{2},^{ } and (c) psi.
5) Calculate (a) the water pressure at a depth of 22.0m below ocean surface. (b)What is the total pressure at that depth if the atmospheric pressure is 101,300Pa? (c) Find the total external force on a shark that has an external total surface area of 32.8 ft^{2}. Ocean water has a mass density of ρ = 1030 kg/m^{3}.
6) A submarine with a total outer area of 1720m^{2} is at a depth of 33.0m below ocean surface. The mass density of ocean water is 1025 kg/m^{3}. Calculate (a) the pressure due to water at that depth. (b) the total external pressure at that depth, and (c) the total external force on it. Let g = 9.81 m/s^{2}.
7) In the figure shown, calculate the liquid pressure at the center of the barrel if the narrow pipe is filled up to (a) Point A, (b) Point B, and (c) Point C. Using each pressure you find (in Parts a, b, and c) as the average pressure inside the barrel, calculate (d) the corresponding internal force on the barrel in each case. If it takes 4.00x10^{7}N for the barrel to rupture, (e) at what height of water in the pipe will that happen? A _{sphere} = 4πR^{2} and g = 9.81m/s^{2}. 

8) In problem 7, why is it not necessary to add the atmospheric pressure to the pressure you find for each case?
9) A volleyball has a diameter of 25.0cm and weighs 2.0N. Find (a) its volume. What downward force can keep it submerged in a type of alcohol that has a mass density of 834 kg/m^{3} (b) in Newtons and (c) lbforce? V_{sphere}=(4/3)πR^{3}.
10) What upward force is needed to keep a 560cm^{3} solid piece of aluminum completely under water avoiding it from sinking (a) in Newtons, and (b) in lbf.? The mass density of aluminum is 2700kg/m^{3}.
11) A boat has a volume of 127m^{3} and weighs 7.0 x10^{4}N. For safety, no more than 67.0% of its volume should be in water. What maximum load (a) in Newtons, (b) in kgf, (c) in tonforce, and (d) in lbf can be put in it?
Answers:
1) 402cm^{3}, 5470grams, 5.47kg, 53.7N & 5.47kgf 2) 5.0kg , 562 cm^{3}
3) 72.3% , 17.4 karat
4) 420 kPa, 4.3 kgf/cm^{2}, 61 psi
5) 222kPa, 323kPa, 969 kN
6) 332kPa, 433kPa, 7.45x10^{8}N
7) 176kPa, 225kPa, 255kPa, 2.00x10^{7}N, 2.55x10^{7}N, 2.88x10^{7}N,
36.1m above the barrel's center
8) For students to answer 9) 8.18x10^{3}m^{3}, 64.9N, 14.6 lbf
10) 9.3N, 2.1 lbf 11) 765000 N, 78000 kgf, 78 tonforce, 170,000 lbf