AP PHYSICS B

AP* PHYSICS B

Fluids

Teacher Packet

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Fluids

Objective To review the student on the concepts, processes and problem solving strategies necessary to successfully answer questions on fluid mechanics. Standards Fluids is addressed in the topic outline of the College Board AP* Physics Course Description Guide as described below. AP Physics Exam Connections Topics relating to fluids are tested every year on the multiple choice and in most years on the free response portion of the exam. The list below identifies free response questions that have been previously asked over fluids. These questions are available from the College Board and can be downloaded free of charge from AP Central. http://apcentral.collegeboard.com. Free Response Questions

2008 Question 4 2008 Form B Question 4 2007 Question 4 2007 Form B Question 4 2005 Question 5 2005 Form B Question 5 2004 Question 2 2004 Form B Question 2 2003 Question 6 2003 Form B Question 6 2002 Question 6

II. Fluid Mechanics and Thermal Physics A. Fluid Mechanics

1. Hydrostatic Pressure 2. Buoyancy 3. Fluid Flow Continuity 4. Bernoulli’s equation


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Fluids

What I Absolutely Have to Know to Survive the AP* Exam

The study of fluids is separated into two distinct parts: hydrostatics (fluids at rest) and hydrodynamics (fluids in motion). The static pressure due to an incompressible fluid depends primarily upon the depth of the fluid. Pascal’s principle states that the pressure exerted at one location in a confined fluid is transmitted undiminished throughout the fluid. Archimedes principle states that the buoyant force acting on an object in a fluid is equal to the weight of the fluid displaced by the object. As a fluid flows through a pipe, the flow rate through the cross sectional area is the same at any point in the pipe. Bernoulli’s equation relates static pressure of a fluid to its dynamic (moving) pressure.

Key Formulas and Relationships

Volume: V = (length * width * height) lwh= Unit: m3

Density: mV

ρ =

Unit: kg/m3

Note: Density of water is 1000 kg/m3

Pressure: FPA

=

Unit: N/m2 = 1 Pascal = 1 Pa Note: 1 atmosphere of pressure is approximately equal to Pa 51 10× Gauge Pressure: gaugeP ghρ= Absolute Pressure: 0 0absolute gaugeP P P P ghρ= + = +

Pascal’s Principal: 1 2

1 2

F FA A

=

Buoyant Force: BuoyF Vgρ= = Weight of Displaced Fluid Unit: N Volume Flow Rate: Avυ = Unit: m3/s Continuity Equation: 1 1 2 2v A v A=

Bernoulli’s Equation: 21 constant2

P gy vρ ρ+ + =


Fluids

Important Concepts

• Pressure: o Pressure is not a force. o Pressures from fluids always generate forces that are perpendicular to the surface of

whatever the fluid is in contact with.

• Static Pressure in a Fluid: Remember from our study of forces and Newton’s Laws that an object at rest must be in equilibrium. Consider the dot pictured in the container of fluid to the right. For the dot to be in equilibrium all the forces on it must cancel out. The gravitational force pulls the dot down and must be canceled by a force pushing upward on the dot.

upward on the dot downward on the dot

and =

and V=Area HeightPA= gAhP= gh

F mg

mF PAv

PA gV

ρ

ρρ

ρ

=

=

= ×

The static pressure due to the fluid depends upon the depth of the fluid, the density of the fluid, and the acceleration due to gravity. Note that the area does not make any difference! The pressure is the same at the bottom of both of the containers pictured above because they have the same depth of fluid. This means that at the same vertical location in a continuous fluid, the pressure is always the same.

mg

F

h


Fluids

Consider the “U” shaped tube filled with a fluid and fitted with a piston pictured below. Along the horizontal line the pressure in the fluid is the same: 1P ghρ= Now the piston is pushed downward with a force toward the horizontal line thus moving the fluid higher in the right hand side of the “U” shaped tube. This force increases the pressure in the fluid at the horizontal line to a new value: 2P ghρ=

horizontal line

fluid piston

original height h1

new height h2fluid

piston

horizontal line

Force


Fluids

o Absolute Pressure and Gauge Pressure If there is anything on top of the fluid itself, the pressure in the fluid would need to be greater to maintain equilibrium.

atmosphere

Mass and atmosphere

Example 1) The atmosphere might be pushing down on the fluid from above. In this case the pressure in the fluid would be:

atmosphereP P ghρ= + Example 2) The atmosphere pushes down from above and a mass is positioned on top of a piston above the fluid. In this case the pressure in the fluid would be:

atmosphere due to massP P P ghρ= + +

atmospherepiston

mgP P ghA

ρ= + +

To avoid as much confusion as possible, two terms have been developed to describe the static pressure in a fluid: Absolute Pressure and Gauge Pressure.

Gauge Pressure is the pressure caused only by the fluid itself:

gaugeP ghρ= Absolute Pressure is the pressure caused by the fluid (ρgh) and anything on top of the fluid, such as the atmosphere and any mass or piston (P0):

0 0absolute gaugeP P P P ghρ= + = +


Fluids

• In a confined fluid, a pressure exerted at one location is transmitted to all parts of the fluid. This is called Pascal’s Principle.

1 2P P=

1 2

1 2

F FA A

=

Pascal’s Principle is utilized in both hydraulic (liquid) and pneumatic (gas) devices where a small force is used to push the fluid through a smaller cylinder into a larger area cylinder where the force

is multiplied by a factor, the ratio of the cross sectional areas 2

1

AA

⎛ ⎞⎜ ⎟⎝ ⎠

.

Note that the force is amplified, but the distances through which the cylinders move are not the same. The large cylinder moves only a short distance, whereas the small cylinder moves a greater distance. The work output of a Pascal’s Principle device will equal the work input if there is no friction.

F2

A2

F1

A1


Fluids

• Objects submerged in a fluid will receive a greater fluid pressure on the bottom than on the top. This occurs because the bottom of the object is deeper in the fluid.

• Objects submerged in a fluid will receive a greater fluid pressure on the bottom than on the top. This occurs because the bottom of the object is deeper in the fluid.

FBuoy

mg

bottom topP P>

bottom topgh ghρ ρ>

bottom toph h> This difference in pressure produces a buoyant force that is directed This difference in pressure produces a buoyant force that is directed upward and is opposite the force of gravitational attraction. upward and is opposite the force of gravitational attraction.

The equation for buoyant force is: Buoy fluid displaced fluidF V gρ= The density in the equation is for the fluid and the volume is for the amount of fluid displaced by the object in the fluid. Note that fluid displaced fluidVρ is the same as the mass ( m Vρ= ) of the displaced fluid. The buoyant force does not depend upon the weight or the shape of the submerged object, but the weight of the displaced fluid. This was first discovered by Archimedes. Archimedes Principle: The buoyant force will equal the weight of the fluid that is displaced by the object. Archimedes Principle: The buoyant force will equal the weight of the fluid that is displaced by the object.

Buoy fluid displaced fluidF V gρ= g displaced fluidm= = Weight of Displaced Fluid = Weight of Displaced Fluid

When an object floats, the weight of the object is equal to the buoyant force (weight of displaced fluid). The volume of the object is not necessarily equal to the volume of the displaced fluid, however. See the figure above.

When an object floats, the weight of the object is equal to the buoyant force (weight of displaced fluid). The volume of the object is not necessarily equal to the volume of the displaced fluid, however. See the figure above.

Downward Pressure

Fluids


FBuoy

mg

bottom topP P>

bottom topgh ghρ ρ>

bottom toph h>

Buoy fluid displaced fluidF V gThe equation for buoyant force is: ρ=

The density in the equation is for the fluid and the volume is for the amount of fluid displaced by the object in the fluid. Note that fluid displaced fluidVρ is the same as the mass ( m Vρ=

Buoy fluid displaced fluidF V g

) of the displaced fluid. The buoyant force does not depend upon the weight or the shape of the submerged object, but the weight of the displaced fluid. This was first discovered by Archimedes.

ρ= displaced fluidm g=

Downward Pressure

Upward Pressure


Fluids

FBuoy

mg

When an object sinks, the weight of the object is greater than the buoyant force (weight of displaced fluid). Hence, the volume of the object will equal the volume of the displaced fluid as per the figure above.

• The continuity equation states that the volume of an incompressible fluid entering one part of a flow tube must be equal to the volume leaving downstream.

1 1 2 2v A v A=

(Note that the product of velocity and the cross sectional area of the flow tube is the volume flow rate of the fluid in units of m3/s. The volume flow rate remains constant throughout all parts of the flow tube.) The important consequence of the continuity equation is that the velocity of the fluid is slower in wider parts of the flow tube and faster in narrower parts of the flow tube.

A1

v1

A2

v2


Fluids

• Bernoulli’s principle applies the ideas of work and energy to dynamic fluid flow to derive what is now called Bernoulli’s equation:

2 2

1 1 1 2 2

1 2 1 2

2 21 1 2 2

1 12 2

thus 1 12 2

P gy v P gy v

y y gy gy

P v P v

2ρ ρ ρ

ρ ρ

ρ ρ

+ + = + +

= =

+ = +

ρ

(Using Bernoulli’s equation is analogous to using the law of conservation of energy. Simply identify two points on a streamline and equate them using Bernoulli’s equation.) Example 1) Consider the flow tube above. Points 1 and 2 are both centered in the middle of the flow tube. Bernoulli’s equation tells us:

2 21 1 1 2 2

1 12 2

P gy v P gy v2ρ ρ ρ+ + = + + ρ

The vertical location, densities of the fluid, and “g” at points 1 and 2 are the same. Thus, they cancel each other out on both sides of the equation.

2 21 1 1 2 2

1 2 1 2

2 21 1 2 2

1 12 2

thus 1 12 2

P gy v P gy v

y y gy gy

P v P v

2ρ ρ ρ ρ

ρ ρ

ρ ρ

+ + = + +

= =

+ = +

The Continuity equation tells us that the velocity of the fluid at point 2 must be greater than the velocity at point 1. This implies that the pressure in the fluid decreases as the speed of the fluid increases.

1 2


Fluids


Free Response Question 1 (15 pts)

18 m

1 m 2 m

4 m

1. A ship carrying cargo into a port is hit by a sudden storm that causes one of the cargo containers to fall overboard and sink to the bottom of the ocean. The ocean depth where the container sank is 18 m. The container itself has a mass of 2400 kg. 9000 kg of cargo was loaded inside the container. The density of ocean water is 1025 kg/m3. A. Calculate the gauge pressure on the top of the container.

1 point for the correct use of gauge pressure.

(3 points max)

Gauge pressure does not include the 1 atm of pressure above the ocean. 1 point for correct height of 17 m The top of the container is only 17 meters below the surface of the ocean. 1 point for the correct answer including

correct units and reasonable number of significant digits

Δh = 18 m – 1 m = 17 m Pgauge = ρgΔh = (1025 kg/m3)(9.8 m/s2)(18 m – 1 m) Pgauge = 1.7 ×105 Pa

Fluids

B. Calculate the total downward force exerted on the top of the container due to the water.

(3 points max) Total downward force implies that we need to use the absolute pressure to determine the force. F = PabsoluteA F = Pabsolute(length * width)

1 point for using absolute pressure 1 point for the correct force equation 1 point for the correct answer including correct units and reasonable number of significant digits F = (1 atm + Pgauge)A

F = (100,000 Pa + 1.7 E5 Pa)(4 m)(2 m) F = 2.2×106 N

C. A rescue ship with a crane is sent to retrieve the cargo. Using a cable attached to the top of the container, the ship hoists the container up at a slow constant rate. Assuming no resistance due to the water, determine the tension in the cable.

(5 points max) Since the velocity is constant, the container is in dynamic equilibrium and ΣF = 0. There are three forces acting on the container. Tension and Buoyant force act upward and Weight acts downward. ΣF = 0 FBuoy + FT = mg FT = mg - FBuoy

FT = (mcontainer + mcargo)g - ρVg FT = (mcontainer + mcargo)g - ρ(length * width * height)g FT = (2400 kg + 9000kg) (9.8 m/s2) - 1025 kg/m3)(4 m)(2 m)(1 m)(9.8 m/s2) = 31,000 N

1 point for a statement that ΣF = 0 1 point for the correct expression FBuoy + FT = mg 1 point for adding the masses 1 point for correct calculation of the volume 1 point for the correct answer including correct units and reasonable number of significant digits


Fluids

D. On the graph below sketch the tension in the cable as the container is lifted out of the water at a steady rate.

Top of container at the surface

of the ocean

Bottom of container at

the surface of the ocean

Container still completely

submerged in the ocean

Container lifted completely out

of the ocean

Container partially

submerged

Tension in the cable

Location of the container

1 point for correctly drawing the constant tension regions while the container is still submerged and after it has been lifted from the ocean

2 points max)

The tension is a constant 31,000 N while the container is still in the water. While the container is being lifted out of the water, the tension increases at a constant rate as the buoyant force decreases. Out of the water, tension equals the weight of the container and cargo (110,000 N).

1 point for correctly drawing a linear transition between the two constant regions


Fluids

1 m 2 m

4 m

d

E. All of the cargo is removed and the container is placed back into the ocean where it now floats. Calculate d the portion of the container below the waterline.

1 point for the correct expressions relating weight and buoyancy force

(2 points max)

For floating objects, buoyant force equals the weight.

( )

( )( )3

2400

1025 2 4

0.29 m

g BuoyF F

mg gVm lwd

m kgd kglw m mm

d

ρρ

ρ

=

=

=

= =

=

1 point for the correct answer including correct units and reasonable number of significant digits


Fluids

Question 2 (15 pts) A large container of fluid sits on a table. It has an opening at the bottom that is sealed by a plug. The opening has a cross sectional area of . The density of the fluid is 900 kg/m

31.8 10 m−× 2

3. The figure is not drawn to scale.

0.65 m 0.4 m

0.25 m

h

A. What is the net force on the plug?

1 point for finding the net force as the difference between the force from the fluid and the air

(3 points max) The net force on the plug will be the difference in forces caused by the external air pressure and the internal fluid pressure. Thus the net force is caused by the gauge pressure.

1 point for finding force from the pressure: F = PA Fnet = Ffluid – Fair Fnet = PfluidA – PairA 1 point for the correct answer including correct units and reasonable number of significant digits

Fnet = (Pfluid – Pair)A Fnet = (Pgauge)A Fnet = ρghA Fnet = (900 kg/m3)(9.8 m/s2)(0.4 m)(1.8 ×10-3 m2) Fnet = 6.35 N


Fluids

B. The plug is removed. Calculate the velocity of the fluid as it flows out of the opening.

(3 points max) This is a dynamic fluids problem that can be solved using Bernoulli’s Equation. (P + ρgy + ½ ρv2)at the top of the fluid = (P + ρgy + ½ ρv2)at the openingSince the external pressure is the same at the top of the fluid and at the opening (1 atm), the pressure can be canceled out. The velocity of the fluid at the top is very small and is assumed to be zero, Also, y = 0 at the opening. This leaves us with this equation: (p + ρgy + ½ ρv2)at the top of the fluid = (p + ρgy + ½ ρv2)at the opening ρgyat the top of the fluid = ½ ρv2

at the openinggyat the top of the fluid = ½ v2

at the opening

2 at the top ofopeningthe fluid

v gy=

1 point for the correct application of Bernoulli’s Equation 1 point for using Bernoulli’s Equation to find:

2 at the top ofopening

the fluidv gy=


( )22 9.8 0.4

m2.8 s

openingmv ms

v

⎛ ⎞= ⎜ ⎟⎝ ⎠

=

C. Calculate the volume flow rate of the fluid out of the opening.

1 point for determining volume flow rate using Avυ =

2 points max) Volume flow rate is equal to the velocity of the fluid times the area of the opening:

1 point for the correct answer including correct units and reasonable number of significant digits ( )3 2

3

1.8 10 2.8

m0.005 s

Avmms

υ

υ

υ

−

=

⎛ ⎞= × ⎜ ⎟⎝ ⎠

=


Fluids

D. The fluid lands on the floor 0.25 m away from the container. Determine how high h the container is above the floor.

1 point for the correct application of kinematics in the x-direction: x = vxt

(4 points max)

This is a 2-D trajectory problem from kinematics. In the x-direction: 1 point for calculating the correct time

0.25 0.09 s2.8

x

x

x v tx mt mv

s

=

= = =

1 point for the correct application of kinematics in the y-direction: y = y0 + vy0t + ½ at2


In the y-direction: y = y0 + vy0t + ½ at2

The initial y-position and initial y-velocity are both zero.

( )

2

22

121 9.8 0.0920.04 m

y gt

my ss

y

=

⎛ ⎞= ⎜ ⎟⎝ ⎠

=

E. Suppose that a fluid twice as dense is placed into the container to the same original height of 0.4 m and the plug is again removed. Will the new fluid land farther away from the container, the same 0.25 m distance from the container, or closer to the container. _____ Farther from the container _____ Same 0.25 m from the container _____ Closer to the container Justify your answer.

(3 points max) The new fluid lands the same 0.25 m away from the container. From part b) we derived that

2 at the top ofopeningthe fluid

v gy=

This equation is independent of the density of the fluid. Therefore, the 2-D trajectory of the new fluid will be identical to that of the original fluid.

1 point for stating that the fluid lands at the same distance of 0.25 m 2 points for a complete explanation that includes: The exit velocity of the fluid is independent of the density of the fluid. The trajectory motion will be identical because the launch velocity, acceleration due to gravity, and the time of flight have not changed.


Fluids

Multiple Choice

1. Shown below are five beakers that are filled with the same volume of mercury. Blocks of different masses and sizes are floating in the liquid mercury. The mass and volume of each block are given below. Rank in order from greatest to least the Buoyant Force exerted on each block. If there are any situations that have the same Buoyant Force list them as being equal. 1_____ 2_____ 3_____ 4_____ 5_____

Greatest Least

A B C D E Mass of block 100 g 200 g 200 g 400 g 100 g

25 cm3 50 cm3Volume of block

25 cm3 50 cm3 75 cm3

1 D 2 B&C 3 A&E

For floating objects the buoyant force up must equal the weight down.


Fluids

2. Shown below are five beakers that are filled with the same volume of water. Blocks of different masses and sizes are completely immersed in the water and have sunk to the bottom of each beaker. The mass and volume of each block are given below. Rank in order from greatest to least the Buoyant Force exerted on each block. If there are any situations that have the same Buoyant Force list them as being equal. 1_____ 2_____ 3_____ 4_____ 5_____

Greatest Least



25 cm3 50 cm3 75 cm3

1 E 2 B&D 3 A&C

FBuoy = ρVg The density of the fluid and g are the same for each case. Thus the objects with the greatest submerged volume will have the largest FBuoy.


Fluids

Questions 3-4 Shown below are five beakers that are filled with the same volume of water. Blocks of different masses and sizes are completely immersed in the water and are suspended from a string. All of the blocks would sink if not for the string holding them up. The mass and volume of each block are given below.



25 cm3 50 cm3 75 cm3

3. Rank in order from greatest to least the Buoyant Force exerted on each block. If there are any situations that have the same Buoyant Force list them as being equal. 1_____ 2_____ 3_____ 4_____ 5_____

Greatest Least

1 E 2 B&D

3 A&C

FBuoy = ρVg The density of the fluid and g are the same for each case. Thus the objects with the greatest submerged volume will have the largest FBuoy.

4. Rank in order from greatest to least the Tension in the supporting string. 1_____ 2_____ 3_____ 4_____ 5_____

Greatest Least

1 D 2 C 3 B 4 A 5 E

The suspended block has three forces acting on it (FBuoy , weight of the block, and tension) These three forces must cancel because the block is in equilibrium.

FBuoy + T = mg ρVg + T = mg T = mblockg - ρwaterVg = g(mblock - ρwaterV)


Fluids

Questions 5-6) Shown below are six graduated cylinders filled with different amounts of the same fluid. The radius of each cylinder and the height of fluid in each is given in the table below.

A B C D E F Radius of Cylinder 0.2 m 0.2 m 0.3 m 0.1 m 0.4 m 0.4 m

Depth of Fluid 0.4 m 0.25 m 0.6 m 1.0 m 0.4 m 0.1 m

5. Rank in order from greatest to least the pressure from the fluid on the bottom of each cylinder. 1 2 3 4 5 6

Greatest Least

1 D 2 C 3 A & E 4 B 5 F

Fluid static pressure: P = ρgh. The density and g are the same in each case. Only the depth of the fluid makes a difference.


Fluids

6. Rank in order from greatest to least the force from the fluid on the bottom of each cylinder. 1 2 3 4 5 6

Greatest Least

Force = (Pressure)(Area): F = PA = (ρgh)(πr2). The density, g, and π are the same in each case. The cylinders with the greatest h r2 have the greatest forces exerted on the bottom of the cylinder.

1 E 2 C 3 A & F

4 B & D Questions 7-8)

A 40 cm tall glass is filled with water to a depth of 30 cm.

7. What is the gauge pressure at the bottom of the glass?

A) 1000 Pa B) 3000 Pa

30 cm

C) 4000 Pa D) 103,000 Pa E) 104,000 Pa

Pgauge = ρgh = (1000 kg/m3)(10 m/s2)(0.3 m) = 3000 Pa B


Fluids

8. What is the absolute pressure at the bottom of the glass? A) 1000 Pa B) 3000 Pa C) 4000 Pa D) 103,000 Pa E) 104,000 Pa

D

Pabsolute = P0 + Pgauge = P0 + ρgh = 100,000 Pa + (1000 kg/m3)(10 m/s2)(0.3 m) = 103,000 Pa

9. Two cylinders filled with oil and fitted with pistons are connected by a tube that allows the oil to flow freely between them. The radius and heights of the cylinders are given. A 10 kg mass is placed on the right cylinder. What force must be applied to the left cylinder to keep the 10 kg mass at the same height?

F

r = 10 cm2

h = 70 cm

r = 50 cm2

h = 70 cm

A) 0.4 N B) 4.0 N C) 100 N D) 250 N E) 2500 N

Pascals Principle: An external pressure applied to a confined fluid is distributed throughout the fluid.

FL / AL = FR / AR

FL / πrL2 = mg / πrR

2B FL = mgrL

2 / rR2

FL = (10 kg)(10 m/s2)(10 cm2)2 / (50 cm2)2 = 4 N


Fluids

10. A 20 kg cube is placed on the bottom of a large container of water and then released. The cube has a height of 20 cm. What normal force will the bottom of the container exert on the cube?

A) The will be no normal force because the cube will float to the surface B) 80 N C) 120 N D) 200 N E) 280 N

C

The Buoyant Force on the cube is: FBuoy = ρVg = (1000 kg/m3)[(0.2 m)3](10 m/s2) = 80 N The weight of the cube is: mg = (20 kg)(10 m/s2) = 200 N The Buoyant Force is not great enough to make the cube float. Therefore,

FBuoy + FN = mg FN = mg - FBuoy = 120 N

Questions 11-13) A fluid flows through a pipe of varying diameters as shown in the figure at the right. 11. Which of the following is true of the fluid velocity as it passed through the pipe?

A) A cv v=B) B cv v>

B

A C

C) C Av v>D) A Bv v<E) C Av v<

The Continuity Equation tell us that inside of a pipe or stream tube the volume flow rate is a constant: A1v1 = A2v2 Where the pipe is narrowest the velocity will be the greatest: vA > vC > vB. E


Fluids

12. Which of the following is true of the pressure in the fluid as it passes through the pipe?

A) A BP P P= = C

C

B

A

B) A BP P P> >C) A CP P P> >D) B CP P P> >E) C B AP P P> >

Bernoulli’s Equation tells us: P + ρgy + ½ ρv2 = constant. As the velocity along a streamline increases the pressure must decrease to maintain the same constant. Wherever the velocity is high the pressure is low and when the velocity is low the pressure must increase:

D

PB > PC > PA

13. Which of the following is true of the volume flow rate (define the symbol, υ , for the volume flow rate) as the fluid flows through the pipe?

A) A B Cυ υ υ= = B) A B Cυ υ υ> > C) A C Bυ υ υ> > D) B C Aυ υ υ> > E) C B Aυ υ υ> >

A The Continuity Equation tell us that inside of a pipe or stream tube the volume flow rate is a constant: A1v1 = A2v2.


Fluids

14. A hurricane blows air at a steady 45 m/s across the top of a flat topped warehouse as shown in the figure above. The roof has an area of 300 m2. What is net force on the roof due to the wind? (Assume the air inside the house is stationary and the density of air is 1.29 kg/m3.)

A) The air blows parallel to the roof and does not cause a net force on the roof. B) 31.3 10 N×C) 38.7 10 N×D) 53.9 10 N×

E) 57.8 10 N×

Bernoulli’s Equation tells us: P + ρgy + ½ ρv2 = constant. (P + ρgy + ½ ρv2)above the roof = (P + ρgy + ½ ρv2)below the roofNote: the difference in heights between “above” and “below” the roof is small and so these two values cancel each other out. In addition, the velocity of the air inside the warehouse is zero. This leaves the following equation: D (P + ½ ρv2)above the roof = Pbelow the roof

½ ρv2above the roof = Pbelow the roof - Pabove the roof = ΔP

Fon the roof = (ΔP)A = (½ ρv2above the roof)A

F= ½ (1.29 kg/m3)(45 m/s)2 (300 m2) = 53.9 10 N×


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