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Physics: Principles and Applications, 6e Giancoli Chapter 10 Fluids Conceptual Questions 1) The three common phases of matter are A) solid, liquid, and gas. B) solid, liquid, and vapor. C) solid, plasma, and gas. D) condensate, plasma, and gas. It is common knowledge that the three common phases of matter are solid, liquid and gas. You should be familiar with the concepts related to the models of these three phases of matter as outlined in the document “Models of the Phases of Matter.” Especially be familiar with how the properties of each phase are related to the strength of interparticle forces and kinetic energy of particles. Answer A 2) Density is A) proportional to both mass and volume. B) proportional to mass and inversely proportional to volume. C) inversely proportional to mass and proportional to volume. D) inversely proportional to both mass and volume.

The definition of density is ρ = m

V. From this formula you can see that if mass is increased, density

increases, so density is proportional to mass. You can also see that if V increases, density decreases, so , density is inversely proportional to volume. Answer B 3) Substance A has a density of 3.0 g/cm3 and substance B has a density of 4.0 g/cm3. In order to obtain equal masses of these two substances, the ratio of the volume of A to the volume of B will be equal to A) 1:3. B) 4:3. C) 3:4. D) 1:4. Manipulating ratios is formulas is not intuitive enough to most people that you can see the answer to this question right away. Because you will be able to see the manipulation visually, it is more likely that you will be successful answering this question quickly if you immediately write down the relationships involved, manipulate the formulas to get the correct ratio, and plug the values in:

ρA =mA

VA

ρB =mB

VB

mA = ρAVA mB = ρBVB → ρAVA = ρBVB

VA

VB

=ρB

ρA

= 43

Answer B

4) Pressure is A) proportional to both force and area. B) proportional to force and inversely proportional to area. C) inversely proportional to force and proportional to area. D) inversely proportional to both force and area.

The definition of pressure is P = F

A= N

m2 From this formula, you can see that if you increase force,

pressure increases, so pressure is proportional to force. You can also see that if you increase area, pressure decreases, so pressure is inversely proportional to A. Answer B

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5) Which of the following is not a unit of pressure? A) atmosphere B) N/m C) Pascal D) mm of mercury Units of pressure are either in units of force per units of area, relate to the height of a column of mercury supported by the weight of the atmosphere, or are given an arbitrary value of “1 atmophere.” (A) is a statement of the unit of an atmosphere; (C) is the name given for the unit representing one N per cubic meter; (D) is a statement of the height of a column of mercury supported by the weight of the atmosphere in mm. (C) is a unit that would represent a unit of force along a one dimensional length, not per unit of area. Answer B 6) Consider three drinking glasses. All three have the same area base, and all three are filled to the same depth with water. Glass A is cylindrical. Glass B is wider at the top than at the bottom, and so holds more water than A. Glass C is narrower at the top than at the bottom, and so holds less water than A. Which glass has the greatest liquid pressure at the bottom? A) Glass A B) Glass B C) Glass C D) All three have equal pressure. Always remember that the pressure in fluid only depends on the depth (the “height” in the formula), according to the formula, P =ρhg . This means that it does not matter what the shape of the volume of water is, because all three glasses have the same depth, all three will have the same pressure at the bottom. This is because pressure is exerted uniformly in all directions in a fluid. Remember the

sequence of logic we used to derive this formula P = F

A= mg

A= ρVg

A= ρ Ahg

A= ρhg Answer D

7) What is the difference between the pressures inside and outside a tire called? A) absolute pressure B) atmospheric pressure C) gauge pressure D) N/m2 Whenever there is a situation in which a valve separates a pressure inside a container and the atmospheric pressure, if the pressure is measured when the valve is open, the atmospheric pressure will be pushing against the pressure coming through the valve, and so, the measured pressure in this circumstance will actually be lessened by an amount equal to the atmospheric pressure. If this lessened pressure is measured by gauge, the measured pressure will be called the gauge pressure, and it is required to remember that the real pressure in the container will be the measured pressure plus the atmospheric pressure. The real pressure in the container is called the absolute pressure. Therefore, the difference between the pressure inside and outside the tire will be the gauge pressure. Answer C 8) When atmospheric pressure changes, what happens to the absolute pressure at the bottom of a pool? A) It does not change. B) It increases by a lesser amount. C) It increases by the same amount. D) It increases by a greater amount. As the pool is open to the atmospheric pressure, the additional pressure of the weight of the atmosphere will be transferred throughout the water, so the pressure at the bottom of the pool will be the pressure exerted by the water plus the pressure of the atmosphere. As the atmospheric pressure is uniformly transmitted throughout the water, however, the atmospheric pressure changes, that also will be the change in pressure at the bottom of the pool. Answer C

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9) You are originally 1.0 m beneath the surface of a pool. If you dive to 2.0 m beneath the surface, what happens to the absolute pressure on you? A) It quadruples. B) It more than doubles. C) It doubles. D) It less than doubles. If the depth of the water doubles, the pressure due to the weight of the water will double—however, the pressure due to the weight of the atmosphere overlying the pool will not have undergone a corresponding doubling. Therefore, the pressure in the pool will increase, but won’t undergo the full doubling. Answer D 10) State Pascal's principle. If an external pressure is applied to a confined fluid, the pressure at every point within the fluid increases by that amount. 11) State Archimedes' principle. The buoyant force on an object immersed in a fluid is equal to the weight of the fluid displaced by the object. 12) 50 cm3 of wood is floating on water, and 50 cm3 of iron is totally submerged. Which has the greater buoyant force on it? A) the wood B) the iron C) Both have the same buoyant force. D) cannot be determined without knowing their densities. Many will immediately jump to the conclusion that because the block of wood is floating, the wood must be under the influence of a greater buoyant force. But—the question is NOT asking, what is the likelihood that the buoyant force will be able to hold up either block. Remember that buoyant force is equal to the weight of the water displaced. Both objects have the same volume, so the maximum buoyant force would equal the weight of 50 cm3 of water which, BTW, is

However, only the cube of iron will feel this buoyant force because not all of the cube of wood is submerged. The amount of water displaced, and so, the buoyant force will be less for the wood. Answer B 13) As a rock sinks deeper and deeper into water of constant density, what happens to the buoyant force on it? A) It increases. B) It remains constant. C) It decreases. D) It may increase or decrease, depending on the shape of the rock. Many will think that as the pressure becomes greater and greater as the rock descends, this increase in pressure will result in an increase in buoyant force. However, remember that buoyant force depends only on the weight of the water displaced. Therefore, as the volume of the water displaced by the rock does not change during the rocks descent, the buoyant force will not change. Answer B 14) Salt water has greater density than fresh water. A boat floats in both fresh water and in salt water. Where is the buoyant force greater on the boat? A) salt water B) fresh water C) Buoyant force is the same in both. D) impossible to determine from the information given

ρwVdispg = (1000kg / m3) 50cm3

1,000,000 cm3

m3

⎛

⎝⎜

⎞

⎠⎟ (9.8m / s2 ) = .49N

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The maximum buoyant force will occur when an object is completely submerged because that is the point where the object will displace the most fluid. However, when the object is not completely submerged, the buoyant force will equal the weight of the object, no matter what the density of the fluid is. What the density of the fluid will affect is how much of the object is submerged so the boat will sink deeper in fresh water. But, the buoyant force will be the same in both. Answer C 15) Salt water is more dense than fresh water. A ship floats in both fresh water and salt water. Compared to the fresh water, the volume of water displaced in the salt water is A) more. B) less. C) the same. D) cannot be determined from the information given The maximum buoyant force will occur when an object is completely submerged because that is the point where the object will displace the most fluid. However, when the object is not completely submerged, the buoyant force will equal the weight of the object, no matter what the density of the fluid is. What the density of the fluid will affect is how much of the object is submerged—the greater the

density of the fluid, the less of the object submerged (remember, ). So, the volume displaced

in salt water is less than in fresh water. Answer B 16) A steel ball sinks in water but floats in a pool of mercury. Where is the buoyant force on the ball greater? A) floating on the mercury B) submerged in the water C) It is the same in both cases. D) cannot be determined from the information given Knowing that when an object is submerged the buoyant force equals the mass of the weight of the fluid displaced, we can reason that this will ultimately equal the weight of the object times the ratio of the density of the fluid to the density of the object. If the object is sinking this ratio must be less than one, so the buoyant force would be less than the objects weight for water. In contrast, if the object is floating on the mercury, the buoyant force will be equal to the weight of the object, which in this circumstance would be greater than the buoyant force when the object is submerged in water.

When the object is submerged:

When the object is floating: 17) A 10-kg piece of aluminum sits at the bottom of a lake, right next to a 10-kg piece of lead. Which has the greater buoyant force on it? A) the aluminum B) the lead C) Both have the same buoyant force. D) cannot be determined without knowing their volumes Remember that for submerged objects, the buoyant force will equal weight of the water displaced. As you are not asked for a quantitative answer you can simply reason this out. The object that displaces more water (the greater volume) will have the greater buoyant force. As both masses are equal, the mass with the greater volume will be the one with the lesser density, which will be the aluminum. This means that the aluminum will have the greater buoyant force. Answer A 18) A piece of iron rests on top of a piece of wood floating in a bathtub. If the iron is removed from the wood, what happens to the water level in the tub? A) It goes up.

Vdisp

Vo

=ρo

ρ fl

FB = mfl dispg = ρ flVog = ρ fl

mo

ρo

g

FB = mog

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B) It goes down. C) It does not change. D) impossible to determine from the information given The overall density of the block of wood plus the iron must not have been so great that the combined mass sank. But it should make sense that when you remove the iron from the block of wood that the block of wood will displace less water. Therefore, the level of the water must decrease. Answer B 19) A piece of wood is floating in a bathtub. A second piece of wood sits on top of the first piece, and does not touch the water. If the top piece is taken off and placed in the water, what happens to the water level in the tub? A) It goes up. B) It goes down. C) It does not change. D) cannot be determined from the information given As the amount of water displaced when one block is on top of the other will be equal to the weight of the blocks combined, when the top block is removed and placed in the water, the weight of water displaced will still be equal to the weight of both block, so the level of the water will not change. Answer C 20) Water flows through a pipe. The diameter of the pipe at point B is larger than at point A. Where is the speed of the water greater? A) point A B) point B C) same at both A and B D) cannot be determined from the information given Remember that according to the equation of continuity/volume flow equation ( or

), the greater the cross-sectional area of a tube, the slower the rate of

flow through the tube. Therefore, the speed of the water will be the greatest at point A. Answer A 21) An ideal fluid flows at 12 m/s in a horizontal pipe. If the pipe widens to twice its original radius, what is the flow speed in the wider section? A) 12 m/s B) 6.0 m/s C) 4.0 m/s D) 3.0 m/s (An ideal fluid would be a fluid that has no internal resistance-its viscosity would be stated to be 0—these do not actually exist in nature.) This would be a question where instead of trying to reason out the answer because you cannot remember how the actual relationship between velocity and diameter works without seeing it visually, you simply write out the relationship, showing what happens to the result when you change a variable as described in the problem. Make sure you remember in this case that if you double the radius you multiply the area by 4.

A1

A2

=v2

v1

so v2 =A1

A2

v1

A1

4 A2

v1 → 14

v2 So, when you double the radius you

divide the velocity by 4, which means 3.0 m/s. Answer D 22) An ideal fluid flows at 12 m/s in a horizontal pipe. If the pipe narrows to half its original radius, what is the flow speed in the narrower section? A) 12 m/s B) 24 m/s

ρ1A1v1 = ρ2 A2v2

A1v1 = A2v2 →

ΔV1

Δt=ΔV2

Δt

6

C) 36 m/s D) 48 m/s (An ideal fluid would be a fluid that has no internal resistance-its viscosity would be stated to be 0—these do not actually exist in nature.) This would be a question where instead of trying to reason out the answer because you cannot remember how the actual relationship between velocity and diameter works without seeing it visually, you simply write out the relationship, showing what happens to the result when you change a variable as described in the problem. Make sure you remember in this case that if you halv the radius you divide the area by 4.

A1

A2

=v2

v1

so v2 =A1

A2

v1

A1

14 A2

v1 → 4v2 So, when you halve the radius you

multiply the velocity by 4, which means 48 m/s. Answer D 23) State Bernoulli's principle. Where the velocity of fluid is high, the pressure is low; and where the velocity is low, the pressure is high. 24) Which one of the following is associated with the law of conservation of energy in fluids? A) Archimedes' principle B) Bernoulli's principle C) Pascal's principle D) equation of continuity Remember, Pascals’s principle states that if an external pressure is applied to a confined fluid, the pressure at every point within the fluid increases by that amount. Archimede’s principle states that the buoyant force on an object immersed in a fluid is equal to the weight of the fluid displaced by the object. Bernoulli’s principle states that where the velocity of fluid is high, the pressure is low; and where the velocity is low, the pressure is high. In analyzing Bernoulli’s principle to create Bernoulli’s equation, we find that

P2 +12 ρv2

2 + ρgy2 = P1 +12 ρv1

2 + ρgy1 . This essentially shows that the forces plus the mechanical energy present at one point in a fluid system equal the forces plus mechanical energy plus the mechanical energy at every other point—conservation of energy. Answer B 25) As the speed of a moving fluid increases, the pressure in the fluid A) increases. B) remains constant. C) decreases. D) may increase or decrease, depending on the viscosity. Remember, Bernoulli’s equation states P + 1

2 ρv2 + ρgy is constant—therefore, if velocity of the fluid increases, pressure must undergo a corresponding decrease. Answer C 26) Water flows through a pipe. The diameter of the pipe at point B is larger than at point A. Where is the water pressure greatest?

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A) point A B) point B C) same at both A and B D) cannot be determined from the information given Remember, Bernoulli’s equation states P + 1

2 ρv2 + ρgy is constant—therefore, the pressure will be greatest wherever velocity is the lowest. Remember also that according to the equation of continuity, velocity will be the lowest wherever the diameter is greatest. Therefore, the pressure will be greatest wherever the diameter is the greatest. Answer B 27) When you blow some air above a paper strip, the paper rises. This is because A) the air above the paper moves faster and the pressure is higher. B) the air above the paper moves faster and the pressure is lower. C) the air above the paper moves slower and the pressure is higher. D) the air above the paper moves slower and the pressure is lower. Remember that Bernoulli’s principle states, where velocity of fluid is high, pressure is low; where velocity of fluid is low, pressure is high. If you blow air above a paper strip, as the air above the strip will be moving faster, the pressure above the strip will be lower than underneath the strip—therefore, the paper rises. Answer B 28) A sky diver falls through the air at terminal velocity. The force of air resistance on him is A) half his weight. B) equal to his weight. C) twice his weight. D) cannot be determined from the information given When a sky diver falls through the air at terminal velocity, this means that he is no longer accelerating—the force due to gravity must be balanced by the upward force due to the buoyancy of the fluid atmosphere. (Keep in mind, however, that the atmosphere is compressible and its density increases the closer you get to the earth—in addition, as a skydiver goes faster they will compress the air particles beneath them at a greater and greater rate. So not only does the air density increase as one moves toward the surface of the earth, the faster the velocity, the greater this effect. Also, the more surface area exposed to the atmosphere—the greater this effect. The actual relationship that determines terminal velocity includes these and other factors and is quite complex.) Under the conditions described, even though we may not know all the factors that determine the force of air resistance, we know that to balance gravity, it must equal the weight of the skydiver. Answer B Questions 29 through 37 require information that is not part of the AP Curriculum Quantitative Problems 1) A plastic block of dimensions 2.00 cm × 3.00 cm × 4.00 cm has a mass of 30.0 g. What is its density?

ρ = m

V= .0300kg

(.0200m)(.0300m)(.0400m)= 1.25 x 103kg / m3

You are asked to find density so write out the formula for this. You have been given the mass so you can plug this in directly, remembering to convert to .0300 kg. You have not been given the volume but have the dimensions so you can plug these in the formula to find V, remembering to convert to meters. Check your answer by knowing that in general, densities range in the 1000’s of kg per cubic meter. 3SF 2) A liquid has a specific gravity of 0.357. What is its density? Remember that specific gravity is defined as the ratio of the density of the given substance to the density

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of water, and that as the density of water is “1” (1.00 x 103kg/m3), the value of the specific gravity is

3.57 x 102 kg / m3

3) A brick weighs 50.0 N, and measures 30.0 cm × 10.0 cm × 4.00 cm. What is the maximum pressure it can exert on a horizontal surface?

As P = F

A, pressure is inversely proportional to area, and the face of the brick through which the

greatest pressure will be exerted is the smallest face, 10.0 cm x 4.00 cm. Therefore, the greatest pressure

that can be exerted is:

P = F

A= 50.0N

(.10m)(.04m)= 1.25x104 Pa = 12.5kPa

A drawing may be useful. How does the pressure exerted by the brick compare to that of atmospheric pressure. 4) A person weighing 900 N is standing on snowshoes. Each snowshoe has area 2500 cm2. What is the pressure on the snow? You are asked for pressure so write down the formula for pressure. You have been given the force, and you have been given the area. Recognize that the force of the body weight is spread over the area of both shoes combined. Also, you can either convert 2500 cm2 + 2500 cm2 = 5000 cm2 in your mind to .50 m2, or you can convert in the calculation using a conversion factor—I have shown both ways.

P = FA= 900N

.50m2 = 1800Pa = 1.8x103 Pa = 1.8kPa

P = FA= 900N

5000cm2 x(100cm x 100cm)

1m2 = 1800Pa = 1.8x103 Pa = 1.8kPa

5) How much pressure (absolute) must a submarine withstand at a depth of 120.0 m in the ocean? The term absolute is a clue to remind you that in this calculation you need to include the atmospheric pressure in the calculation, so you will need to add in 101,325 Pa. Additionally, the density of seawater is greater than that of fresh water so the density of seawater will be required. Remember that the formula for determining the pressure at a given depth h is: P = ρhg . So, the calculation would be:

P = ρhg + 101,325 N

m2 = 1.025 x 103kgm3

⎛⎝⎜

⎞⎠⎟

(120.0m)(9.8m / s2 ) + 101,325 Nm2 = 1.307x106 Pa = 1.307x103kPa

6) A circular window of 30 cm diameter in a submarine can withstand a maximum force of 5.20 × 105 N. What is the maximum depth in a lake to which the submarine can go without damaging the window? The formula to find the pressure of water at a certain depth is P = ρhg . This problem involves a simple rearrangement of the pressure of water formula to isolate h. F/A will need to be substituted in for P, and A of a circle is πr 2 . Atmospheric pressure will also need to be added in.

P = ρhg + Patm

h =Pwater − Patm

ρg=

FA − Patm

ρg=

(5.20x105 N )(π )(.15m)2( )−101,325 N

m2

(1.000x103kg / m3)(9.8m / s2 )= 740. m

7) What is the gauge pressure if the absolute pressure is 300 kPa? Whenever there is a situation in which a valve separates a pressure inside a container and the atmospheric pressure, if the pressure is measured when the valve is open, the atmospheric pressure will be pushing against the pressure coming through the valve, and so, the measured pressure in this

9

circumstance will actually be lessened by an amount equal to the atmospheric pressure. If this lessened pressure is measured by gauge, the measured pressure will be called the gauge pressure, and it is required to remember that the real pressure in the container will be the measured pressure plus the atmospheric pressure. The real pressure in the container is called the absolute pressure. Therefore, the difference between the pressure inside and outside the tire will be the gauge pressure. This means that to find the gauge pressure we need to subtract Patm from Pabs:

Pgauge = Pabs − Patm = 300kPa −101kPa = 199kPa

8) What is the gauge pressure at a location 15.0 m below the surface of sea? (The density of seawater is 1.03 × 103 kg/m3.) Because the ocean is under the influence of the atmosphere, the pressure at a certain depth in the ocean will be the pressure of the weight of the water at that depth, plus the atmospheric pressure. The gauge pressure is considered to be the pressure at that depth without including the pressure due to the weight of the atmosphere. Therefore, the gauge pressure at 15.0 m can be found by simply applying the formula to determine water pressure:

Pgauge = ρhg = 1.03x103kg

m3

⎛⎝⎜

⎞⎠⎟

(15.0m)(9.8m / s2 ) = 151,410 Pa = 151kPa

9) What is the absolute pressure at a location 15.0 m below the surface of sea? (The density of seawater is 1.03 × 103 kg/m3.) In conjunction with the explanation for question 8, the absolute pressure would be considered the sum of the gauge pressure and the atmospheric pressure:

Pabs = ρhg + Patm = 1.03x103kg

m3

⎛⎝⎜

⎞⎠⎟

(15.0m)(9.8m / s2 ) + 101,325Pa = 252735 Pa = 253kPa

10) A 500-N weight sits on the small piston of a hydraulic machine. The small piston has area 2.0 cm2. If the large piston has area 40 cm2, how much weight can the large piston support? Remember that for hydraulic systems, the pressure exerted by the input system is distributed and felt uniformly throughout the contained fluid, including at the output piston. Therefore,

Pi =

Fi

Ai

= Po =Fo

Ao

→Fo

Fi

=Ao

Ai

You will not be given this formula on the test, you will have to derive

it if you need it. For this problem we are being asked for the weight that can be supported by the output piston, which is the output force, so isolate Fo, plug in the values and solve:

Fo = Fi

Ao

Ai

= (500N ) 402

= 10,000N

If the units of area are the same for both pistons, it does not matter if the units are not m2, it is the ratio of the output piston to the input piston that matters. For multiple choice questions it is probably easy enough to remember that the hydraulic advantage is simply this ratio times the input force. However, for free response questions you would still have to show all the work. 11) In a hydraulic garage lift, the small piston has a radius of 5.0 cm and the large piston has a radius of 15 cm. What force must be applied on the small piston in order to lift a car weighing 20,000 N on the large piston? Remember that for hydraulic systems, the pressure exerted by the input system is distributed and felt uniformly throughout the contained fluid, including at the output piston. Therefore,

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Pi =

Fi

Ai

= Po =Fo

Ao

→Fo

Fi

=Ao

Ai

You will not be given this formula on the test, you will have to derive

it if you need it. For this problem we are being asked for the input force, so isolate this, plug in values and solve.

Fi = Fo

Ai

Ao

= Fo

π ri2

π ro2

= (20000N ) 52

152

⎛⎝⎜

⎞⎠⎟= 2222N = 2200N

Make sure you are paying attention to recognize that you are given the radius, not the area. 12) A 13,000 N vehicle is to be lifted by a 25 cm diameter hydraulic piston. What force needs to be applied to a 5.0 cm diameter piston to accomplish this? Remember that for hydraulic systems, the pressure exerted by the input system is distributed and felt uniformly throughout the contained fluid, including at the output piston. Therefore,

Pi =

Fi

Ai

= Po =Fo

Ao

→Fo

Fi

=Ao

Ai

You will not be given this formula on the test, you will have to derive

it if you need it. For this problem we are being asked for the input force, so isolate this, plug in values and solve. As you are only concerned about the ratio of areas, as long as the units are the same you do not need to convert to m. You have been given diameters, however, so make sure these are converted to radii.

Vdisp =Vo

Fi = Fo

Ai

Ao

= Fo

π ri2

π ro2= (13000N ) 2.52

12.52

⎛⎝⎜

⎞⎠⎟= 520N

FB = w− ′w = ρ flVdispg

ρo Vo g

ρ fl Vdisp g=ρo

ρ fl

ww− ′w

=ρo Vo g

ρ fl Vdisp g=ρo

ρ fl

13) The small piston of a hydraulic lift has a diameter of 8.0 cm, and its large piston has a diameter of 40 cm. The lift raises a load of 15,000 N. (a) Determine the force that must be applied to the small piston. Remember that for hydraulic systems, the pressure exerted by the input system is distributed and felt uniformly throughout the contained fluid, including at the output piston. Therefore,

Pi =

Fi

Ai

= Po =Fo

Ao

→Fo

Fi

=Ao

Ai

You will not be given this formula on the test, you will have to derive

it if you need it. For this problem we are being asked for the input force, so isolate this, plug in values and solve. As you are only concerned about the ratio of areas, as long as the units are the same you do not need to convert to m. You have been given diameters, however, so make sure these are converted to radii.

11

Fi = Fo

Ai

Ao

= Fo

π ri2

π ro2= (15000N ) 42

202

⎛⎝⎜

⎞⎠⎟= 600N

(b) Determine the pressure applied to the fluid in the lift. Pressure equals force divided by area. In this case we are talking about the input force and the input area. We know the input force is 600 N as we just calculated it. You have been told that the input area

is a piston with a radius of .04m, so calculate the input area using πr 2 . Plug the values in and solve:

P = F

A= Fπr 2 = 600N

π (.040m)2 = 1.2x105 Pa

14) A block of metal weighs 40 N in air and 30 N in water. What is the buoyant force of the water? You are aware that the weight in water is also known as the apparent weight. You know a simple formula to determine the apparent weight—apparent weight is the difference between the actual weight and the buoyant force. Rearranging you can isolate FB. You have been given both the actual and apparent weight. Plug in the values and solve.

15) An object has a volume of 4.0 m3 and weighs 40,000 N. What will its weight be in water? You are aware that the weight in water is also known as the apparent weight. You know a simple formula to determine the apparent weight—apparent weight is the difference between the actual weight and the buoyant force. You have been given the actual weight and you know the buoyant force is equal

to the weight of the displaced water, which is also equal to . You have been given the volume

of the object and this will be the volume of the displaced water. You know the density of water. Plug the values in and solve:

16) A 4.00-kg cylinder of solid iron is supported by a string while submerged in water. What is the tension in the string? (The specific gravity of iron is 7.86.) From reading the problem you know that as there is no acceleration, you will have the force due to gravity balanced by the sum of the buoyant force and tension in the string. Drawing a free body diagram leads you to a net force equation, from which you can isolate the force of tension, which will equal the force due to gravity minus the buoyant force. You know how to find the force due to gravity.

The buoyant force will be . However, you do not have the volume of water displaced. But, you

have been given the mass of the iron cylinder and its density so you can find the volume of the cylinder and so that of the water displaced. 17) If the density of gold is 19.3 × 103 kg/m3, what buoyant force does a 0.60-kg gold crown experience when it is immersed in water?

′w = w− FB → FB = w− ′w = 40N − 30N = 10N

ρwVdispg

′w = w− FB = w− ρwVdispg = (40000N )− (1000kg / m3)(4.0m3)(9.8m / s2 ) = 800N

ρwVdispg

Fn = FT + FB − FG = 0N∑FT = FG − FB = mFeg − ρWVdispg = mFeg − ρW

mFe

ρFe

g

(4.0kg)(9.8m / s2 )− (1000kg / m3) 4.0kg7860kg / m3

⎛⎝⎜

⎞⎠⎟

(9.8m / s2 ) = 34.2N

12

Remember that the buoyant force equals the weight of the displaced fluid which also equals ρ flVdispg .

You know the density of water. As the volume displaced will equal the volume of the crown, you can find this by knowing that the density of the gold will equal m/V, and so, V = m/d. We have been given the both the mass and the density of the crown so we can determine V. Plug the values in and solve:

ρ flVdispg = ρ fl

mcrown

ρcrown

g = (1000kg / m3) .60kg19300kg / m3

⎛⎝⎜

⎞⎠⎟

(9.8m / s2 ) = .30N

18) A crane lifts a steel submarine (density = 7.8 × 103 kg/m3) of mass 20,000 kg. What is the tension in the lifting cable (1) when the submarine is submerged, and (2) when it is entirely out of the water? From reading the problem you know that as there is no acceleration, you will have the force due to gravity balanced by the sum of the buoyant force and tension in the string. Drawing a free body diagram leads you to a net force equation, from which you can isolate the force of tension, which will equal the force due to gravity minus the buoyant force. You know how to find the force due to gravity.

The buoyant force will be . However, you do not have the volume of water displaced. But, you

have been given the mass of the iron cylinder and its density so you can find the volume of the cylinder and so that of the water displaced.

Fnet = FT + FB − FG = 0N∑FT = FG − FB = msubg − ρ flVdispg = msubg − ρ fl

msub

ρsub

g

= (20,000kg)(9.8m / s2 )− (1000kg / m3) 20000kg7800kg / m3

⎛⎝⎜

⎞⎠⎟

(9.8m / s2 )

1.7x105 N

When the sub is out of water the weight will only be balanced by the tension so the tension will simply

equal the force due to gravity: FT = mg = (20000kg)(9.8) = 2.0x105 N

19) An object weighs 7.84 N when it is in air and 6.86 N when it is immersed in water. What is the specific gravity of the object? Remember that if we know the weight and apparent weight we can easily determine the specific gravity

using the following relationship:

ww− ′w

=ρo Vo g

ρ fl Vdisp g=ρo

ρ fl

We have been given both the weight

and the apparent weigh—plug the values in and solve:

ρo

ρ fl

= ww− ′w

= 7.84N7.84− 6.86

= 8.00

20) A container of water is placed on a scale, and the scale reads 120 g. Now a 20-g piece of copper (specific gravity = 8.9) is suspended from a thread and lowered into the water, not touching the bottom of the container. What will the scale now read? We cannot assume that any water is lost from the pail as we are not given this information—we must assume all the water is still in the pail. Recognize that the water will generate a buoyant force on the cube of copper. Therefore, the cube will exert an equal and opposite force on the water, so this must be added to the weight the pail already has. This extra force will equal the weight of the water displaced by the cube, and we will want to divide by g to get the actual mass. We can find the volume

ρwVdispg

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displaced because we know the mass of copper and its density. Therefore, the mass added to the existing mass of the pail will equal density of the water displaced times the mass of the copper divided by its density:

21) A piece of aluminum with a mass of 1.0 kg and density of 2700 kg/m3 is suspended from a string and then completely immersed in a container of water. The density of water is 1000 kg/m3. (a) Determine the volume of the piece of aluminum. (b) Determine the tension in the string after the metal is immersed in the container of water. 22) A cylindrical rod of length 12 cm and diameter 2.0 cm will just barely float in water. What is its mass? 23) A rectangular box of negligible mass measures 5.0 m long, 1.0 m wide, and 0.50 m high. How many kilograms of mass can be loaded onto the box before it sinks in a lake? 24) A 1.0-m3 object floats in water with 20% of it above the waterline. What does the object weigh out of the water? 25) An object floats with half its volume beneath the surface of the water. The weight of the displaced water is 2000 N. What is the weight of the object?

final mass = m init + m

disp= m init + ρWVdisp = minit + ρW

mCu

ρCu

= 120g + (1.00g / cm3) 20g8.9g / cm3

⎛⎝⎜

⎞⎠⎟= 122g

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26) A solid object floats in water with three-fourths of its volume beneath the surface. What is the object's density? 27) A 200-N object floats with three-fourths of its volume beneath the surface of the water. What is the buoyant force on the object? 28) A polar bear of mass 200 kg stands on an ice floe 100 cm thick. What is the minimum area of the floe that will just support the bear in saltwater of specific gravity 1.03? The specific gravity of ice is 0.98. 29) Liquid flows through a pipe of diameter 3.0 cm at 2.0 m/s. Find the flow rate. 30) Liquid flows through a 4.0 cm diameter pipe at 1.0 m/s. There is a 2.0 cm diameter restriction in the line. What is the velocity in this restriction?

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31) Water flows at 12 m/s in a horizontal pipe with a pressure of 3.0 × 104 N/m2. If the pipe widens to twice its original radius, what is the pressure in the wider section? 32) How much pressure does it take for a pump to supply a drinking fountain with 300 kPa, if the fountain is 30.0 m above the pump? 33) A hole of radius 1.0 mm occurs in the bottom of a water storage tank that holds water at a depth of 15 m. At what rate will water flow out of the hole? 34) Water flows through a horizontal pipe of cross-sectional area 10.0 cm2 at a pressure of 0.250 atm. The flow rate is 1.00 × 10-3 m3/s. At a valve, the effective cross-sectional area of the pipe is reduced to 5.00 cm2. What is the pressure at the valve? 35) SAE No. 10 oil has a viscosity of 0.20 Pa-s. How long would it take to pour 4.0 L of oil through a funnel with a neck 15 cm long and 2.0 cm in diameter? Assume the surface of the oil is kept 6 cm above the top of the neck, and neglect any drag effects due to the upper part of the funnel. 36) Suppose that the build-up of fatty tissue on the wall of an artery decreased the radius by 10%. By how much would the pressure provided by the heart have to be increased to maintain a constant blood flow?

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37) Two narrow tubes are placed in a pan of water. Tube A has twice the diameter of tube B. If water rises 24 cm in tube A, how high will it rise in tube B? 38) The surface tension of water is 0.073 N/m. How high will water rise in a capillary tube of diameter 1.2 mm? 39) To what height would water at 0 degree Celsius rise in a glass capillary tube with a diameter of 0.50 mm? 40) A narrow tube is placed vertically in a pan of water, and the water rises in the tube to 4.0 cm above the level of the pan. The surface tension in the liquid is lowered to one-half its original value by the addition of some soap. What happens to the height of the liquid column in the tube? 41) Consider a rectangular frame of length 0.120 m and width 0.600 m with a soap film formed within its confined area. If the surface tension of soapy water is 0.0260 N/m, how much force does the soap film exert on the 0.600 m side?

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42) An adjustable rectangular frame has length 12 cm and width 5 cm, and is built in such a way that one of the longer sides can be moved without bursting the soap film contained in the frame. Soapy water has surface tension 0.026 N/m. How much work is required to move the adjustable long side so that the frame is 12 cm square?