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A two-dimensional view of a dam with dimensions L and h is shown. Force F at h is shown by a horizontal arrow. The force F exerted by water on the dam is F equals average pressure p bar into area A and pressure in turn is average height h bar into density rho into acceleration due to gravity g.
The dam must withstand the force exerted against it by the water it retains. This force is small compared with the weight of the water behind the dam.

Atmospheric pressure is another example of pressure due to the weight of a fluid, in this case due to the weight of air above a given height. The atmospheric pressure at the Earth’s surface varies a little due to the large-scale flow of the atmosphere induced by the Earth’s rotation (this creates weather “highs” and “lows”). However, the average pressure at sea level is given by the standard atmospheric pressure P atm size 12{P rSub { size 8{"atm"} } } {} , measured to be

1 atmosphere (atm) = P atm = 1.01 × 10 5 N/m 2 = 101 kPa . size 12{1`"atmosphere"` \( "atm" \) =P rSub { size 8{"atm"} } =1 "." "01" times "10" rSup { size 8{5} } `"N/m" rSup { size 8{2} } ="101"`"kPa"} {}

This relationship means that, on average, at sea level, a column of air above 1.00 m 2 of the Earth’s surface has a weight of 1.01 × 10 5 N size 12{1 "." "01" times "10" rSup { size 8{5} } `N} {} , equivalent to 1 atm . (See [link] .)

Figure shows a column of air exerting a weight of one point zero one times ten to the power five newtons on a rectangular patch of ground of one square meter cross section.
Atmospheric pressure at sea level averages 1 . 01 × 10 5 Pa size 12{1 "." "01" times "10" rSup { size 8{5} } `"Pa"} {} (equivalent to 1 atm), since the column of air over this 1 m 2 size 12{1`m rSup { size 8{2} } } {} , extending to the top of the atmosphere, weighs 1 . 01 × 10 5 N size 12{1 "." "01" times "10" rSup { size 8{5} } " N"} {} .

Calculating average density: how dense is the air?

Calculate the average density of the atmosphere, given that it extends to an altitude of 120 km. Compare this density with that of air listed in [link] .

Strategy

If we solve P = hρg size 12{P=hρg} {} for density, we see that

ρ ¯ = P hg . size 12{ { bar {ρ}}= { {P} over { ital "hg"} } } {}

We then take P size 12{P} {} to be atmospheric pressure, h size 12{h} {} is given, and g size 12{g} {} is known, and so we can use this to calculate ρ ¯ size 12{ { bar {ρ}}} {} .

Solution

Entering known values into the expression for ρ ¯ size 12{ { bar {ρ}}} {} yields

ρ ¯ = 1 . 01 × 10 5 N/m 2 ( 120 × 10 3 m ) ( 9 . 80 m/s 2 ) = 8 . 59 × 10 2 kg/m 3 . size 12{ { bar {ρ}}= { {1 "." "01" times "10" rSup { size 8{5} } `"N/m" rSup { size 8{2} } } over { \( "120" times "10" rSup { size 8{3} } `m \) \( 9 "." "80"`"m/s" rSup { size 8{2} } \) } } =8 "." "59" times "10" rSup { size 8{ - 2} } `"kg/m" rSup { size 8{3} } } {}

Discussion

This result is the average density of air between the Earth’s surface and the top of the Earth’s atmosphere, which essentially ends at 120 km. The density of air at sea level is given in [link] as 1 . 29 kg/m 3 size 12{1 "." "29"`"kg/m" rSup { size 8{3} } } {} —about 15 times its average value. Because air is so compressible, its density has its highest value near the Earth’s surface and declines rapidly with altitude.

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Calculating depth below the surface of water: what depth of water creates the same pressure as the entire atmosphere?

Calculate the depth below the surface of water at which the pressure due to the weight of the water equals 1.00 atm.

Strategy

We begin by solving the equation P = hρg size 12{P=hρg} {} for depth h size 12{h} {} :

h = P ρg . size 12{h= { {P} over {ρg} } } {}

Then we take P size 12{P} {} to be 1.00 atm and ρ size 12{ρ} {} to be the density of the water that creates the pressure.

Solution

Entering the known values into the expression for h size 12{h} {} gives

h = 1 . 01 × 10 5 N/m 2 ( 1 . 00 × 10 3 kg/m 3 ) ( 9 . 80 m/s 2 ) = 10 . 3 m . size 12{h= { {1 "." "01" times "10" rSup { size 8{5} } `"N/m" rSup { size 8{2} } } over { \( 1 "." "00" times "10" rSup { size 8{3} } `"kg/m" rSup { size 8{3} } \) \( 9 "." "80"`"m/s" rSup { size 8{2} } \) } } ="10" "." 3`m} {}

Discussion

Just 10.3 m of water creates the same pressure as 120 km of air. Since water is nearly incompressible, we can neglect any change in its density over this depth.

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What do you suppose is the total pressure at a depth of 10.3 m in a swimming pool? Does the atmospheric pressure on the water’s surface affect the pressure below? The answer is yes. This seems only logical, since both the water’s weight and the atmosphere’s weight must be supported. So the total pressure at a depth of 10.3 m is 2 atm—half from the water above and half from the air above. We shall see in Pascal’s Principle that fluid pressures always add in this way.

Section summary

  • Pressure is the weight of the fluid mg size 12{ ital "mg"} {} divided by the area A size 12{A} {} supporting it (the area of the bottom of the container):
    P = mg A . size 12{P= { { ital "mg"} over {A} } } {}
  • Pressure due to the weight of a liquid is given by
    P = hρg , size 12{P=hρg} {}

    where P size 12{P} {} is the pressure, h size 12{h} {} is the height of the liquid, ρ size 12{ρ} {} is the density of the liquid, and g size 12{g} {} is the acceleration due to gravity.

Questions & Answers

a 50kg mass is placed on a frictionless piston fitted to a gas cylinder .If 149 kelvin of heat is supplied to the cylinder increasing the internal energy by 100 joules,determine the height through which the mass of the piston raise
Lawal Reply
what is thermodynamics
wana Reply
thermodynamic is a branch of physics that teaches on the relationship about heat and anyother form of energy
Emmanuel
could you please help solve question on thermodynamics
Lawal
if a mass of 149 of heat is supplied and there's an increase in internal energy of 100jouls,find the raise in height
Lawal
if l cary box and stop is ther any work
Tamirat Reply
no that because u have moved no distance. for work to be performed a force needs to be applied and a distance needs to be moved
Emmanuel
Different between fundamental unit and derived unit
Alimi Reply
fundamental unit are independent quantities that do not depend on any other unit while derived unit are quantities that depend on two or more units for it definition
Emmanuel
what is nuclear fission
Sadik Reply
hello
Shawty
are you there
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miss your absence here...
Shawty
what is a vector
Temitayo Reply
vectors are quantities that have numerical value or magnitude and direction.
Muhammad
what is regelation
Oladele
vector is any quantity that has magnitude and direction
Emmanuel
Physics is a physical science that deals with the study of matter in relation to energy
Divine Reply
Hi
Jimoh
hello
Salaudeen
hello
Sadik
Yes
Maxamuud
hi everyone
Muhammad
what is physics
Rhema Reply
physics is a physical science that deals with the study of matter in relation to energy
Osayuwa
a15kg powerexerted by the foresafter 3second
Firdos Reply
what is displacement
Xolani Reply
movement in a direction
Jason
hello
Hosea
Hey
Smart
haider
Explain why magnetic damping might not be effective on an object made of several thin conducting layers separated by insulation? can someone please explain this i need it for my final exam
anas Reply
Hi
saeid
hi
Yimam
Hi
Jimoh
An object made of several thin conducting layers separated by insulation may not be affected by magnetic damping because the eddy current produced in each layer due to induction will be very small and the opposing magnetic flux produced by the eddy currents will be very small
Muhammad
What is thê principle behind movement of thê taps control
Oluwakayode Reply
while
Hosea
what is atomic mass
thomas Reply
this is the mass of an atom of an element in ratio with the mass of carbon-atom
Chukwuka
show me how to get the accuracies of the values of the resistors for the two circuits i.e for series and parallel sides
Jesuovie Reply
Explain why it is difficult to have an ideal machine in real life situations.
Isaac Reply
tell me
Promise
what's the s . i unit for couple?
Promise
its s.i unit is Nm
Covenant
Force×perpendicular distance N×m=Nm
Oluwakayode
İt iş diffucult to have idêal machine because of FRİCTİON definitely reduce thê efficiency
Oluwakayode
It is difficult to have an ideal machine in real life situation because in ideal machines all the input energy should be converted to output energy . But , some part of energy is always lost in overcoming friction and input energy is always greater than output energy . Hence , no machine is ideal.
Muhammad
Practice Key Terms 1

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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