Archimedes' Principle

16 Pages·2017·1.4 MB·English

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Archimedes’ Principle*
OpenStax
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Abstract
• De(cid:28)ne buoyant force.
• State Archimedes’ principle.
• Understand why objects (cid:29)oat or sink.
• Understand the relationship between density and Archimedes’ principle.
When you rise from lounging in a warm bath, your arms feel strangely heavy. This is because you no
longer have the buoyant support of the water. Where does this buoyant force come from? Why is it that
some things (cid:29)oat and others do not? Do objects that sink get any support at all from the (cid:29)uid? Is your
body buoyed by the atmosphere, or are only helium balloons a(cid:27)ected? (See Figure 1.)
Figure 1: (a) Even objects that sink, like this anchor, are partly supported by water when submerged.
(b)Submarineshaveadjustabledensity(ballasttanks)sothattheymay(cid:29)oatorsinkasdesired. (credit:
Allied Navy) (c) Helium-(cid:28)lled balloons tug upward on their strings, demonstrating air’s buoyant e(cid:27)ect.
(credit: Crystl)
*Version1.8: Jul16,20129:24pm+0000
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Answerstoallthesequestions,andmanyothers,arebasedonthefactthatpressureincreaseswithdepth
in a (cid:29)uid. This means that the upward force on the bottom of an object in a (cid:29)uid is greater than the
downward force on the top of the object. There is a net upward, or buoyant force on any object in any
(cid:29)uid. (See Figure 2.) If the buoyant force is greater than the object’s weight, the object will rise to the
surface and (cid:29)oat. If the buoyant force is less than the object’s weight, the object will sink. If the buoyant
forceequalstheobject’sweight,theobjectwillremainsuspendedatthatdepth. Thebuoyantforceisalways
present whether the object (cid:29)oats, sinks, or is suspended in a (cid:29)uid.
: The buoyant force is the net upward force on any object in any (cid:29)uid.
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Figure 2: Pressure due to the weight of a (cid:29)uid increases with depth since P =hρg. This pressure and
associated upward force on the bottom of the cylinder are greater than the downward force on the top
of the cylinder. Their di(cid:27)erence is the buoyant force FB. (Horizontal forces cancel.)
Justhowgreatisthisbuoyantforce? Toanswerthisquestion,thinkaboutwhathappenswhenasubmerged
object is removed from a (cid:29)uid, as in Figure 3.
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Figure 3: (a)Anobjectsubmergedina(cid:29)uidexperiencesabuoyantforceFB. IfFB isgreaterthanthe
weightoftheobject,theobjectwillrise. IfFB islessthantheweightoftheobject,theobjectwillsink.
(b) If the object is removed, it is replaced by (cid:29)uid having weight w(cid:29). Since this weight is supported by
surrounding (cid:29)uid, the buoyant force must equal the weight of the (cid:29)uid displaced. That is, FB = w(cid:29),a
statement of Archimedes’ principle.
The space it occupied is (cid:28)lled by (cid:29)uid having a weight w . This weight is supported by the surrounding
(cid:29)
(cid:29)uid, and so the buoyant force must equal w , the weight of the (cid:29)uid displaced by the object. It is a tribute
(cid:29)
to the genius of the Greek mathematician and inventor Archimedes (ca. 287(cid:21)212 B.C.) that he stated this
principle long before concepts of force were well established. Stated in words, Archimedes’ principle is
as follows: The buoyant force on an object equals the weight of the (cid:29)uid it displaces. In equation form,
Archimedes’ principle is
F =w , (3)
B (cid:29)
where F is the buoyant force and w is the weight of the (cid:29)uid displaced by the object. Archimedes’
B (cid:29)
principle is valid in general, for any object in any (cid:29)uid, whether partially or totally submerged.
: According to this principle the buoyant force on an object equals the weight of the (cid:29)uid it
displaces. In equation form, Archimedes’ principle is
F =w , (3)
B (cid:29)
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where F is the buoyant force and w is the weight of the (cid:29)uid displaced by the object.
B (cid:29)
Humm ... High-tech body swimsuits were introduced in 2008 in preparation for the Beijing Olympics. One
concern (and international rule) was that these suits should not provide any buoyancy advantage. How do
you think that this rule could be veri(cid:28)ed?
: The density of aluminum foil is 2.7 times the density of water. Take a piece of foil, roll it up
into a ball and drop it into water. Does it sink? Why or why not? Can you make it sink?
1 Floating and Sinking
Drop a lump of clay in water. It will sink. Then mold the lump of clay into the shape of a boat, and it will
(cid:29)oat. Because of its shape, the boat displaces more water than the lump and experiences a greater buoyant
force. The same is true of steel ships.
Example 1: Calculating buoyant force: dependency on shape
(a) Calculate the buoyant force on 10,000 metric tons (cid:0)1.00×107kg(cid:1) of solid steel completely
submerged in water, and compare this with the steel’s weight. (b) What is the maximum buoyant
force that water could exert on this same steel if it were shaped into a boat that could displace
1.00×105m3 of water?
Strategy for (a)
To (cid:28)nd the buoyant force, we must (cid:28)nd the weight of water displaced. We can do this by using
the densities of water and steel given in . We note that, since the steel is completely submerged,
its volume and the water’s volume are the same. Once we know the volume of water, we can (cid:28)nd
its mass and weight.
Solution for (a)
First, we use the de(cid:28)nition of density ρ= m to (cid:28)nd the steel’s volume, and then we substitute
V
values for mass and density. This gives
m 1.00×107kg
V = st = =1.28×103m3. (3)
st ρst 7.8×103kg/m3
Because the steel is completely submerged, this is also the volume of water displaced, V . We can
w
now (cid:28)nd themass ofwaterdisplaced fromthe relationshipbetween its volumeand density, bothof
which are known. This gives
m = ρ V =(cid:16)1.000×103kg/m3(cid:17)(cid:0)1.28×103m3(cid:1)
w w w
(3)
= 1.28×106kg.
By Archimedes’ principle, the weight of water displaced is m g, so the buoyant force is
w
F = w =m g =(cid:0)1.28×106kg(cid:1)(cid:16)9.80m/s2(cid:17)
B w w
(3)
= 1.3×107N.
The steel’s weight is m g = 9.80×107N, which is much greater than the buoyant force, so the
w
steel will remain submerged. Note that the buoyant force is rounded to two digits because the
density of steel is given to only two digits.
Strategy for (b)
Herewearegiventhemaximumvolumeofwaterthesteelboatcandisplace. Thebuoyantforce
is the weight of this volume of water.
Solution for (b)
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Themassofwaterdisplacedisfoundfromitsrelationshiptodensityandvolume,bothofwhich
are known. That is,
m = ρ V =(cid:16)1.000×103kg/m3(cid:17)(cid:0)1.00×105m3(cid:1)
w w w
(3)
= 1.00×108kg.
The maximum buoyant force is the weight of this much water, or
F = w =m g =(cid:0)1.00×108kg(cid:1)(cid:16)9.80m/s2(cid:17)
B w w
(3)
= 9.80×108N.
Discussion
The maximum buoyant force is ten times the weight of the steel, meaning the ship can carry a
load nine times its own weight without sinking.
: A piece of household aluminum foil is 0.016 mm thick. Use a piece of foil that measures 10 cm
by 15 cm. (a) What is the mass of this amount of foil? (b) If the foil is folded to give it four sides,
and paper clips or washers are added to this (cid:16)boat,(cid:17) what shape of the boat would allow it to hold
the most (cid:16)cargo(cid:17) when placed in water? Test your prediction.
2 Density and Archimedes’ Principle
Density plays a crucial role in Archimedes’ principle. The average density of an object is what ultimately
determines whether it (cid:29)oats. If its average density is less than that of the surrounding (cid:29)uid, it will (cid:29)oat.
This is because the (cid:29)uid, having a higher density, contains more mass and hence more weight in the same
volume. The buoyant force, which equals the weight of the (cid:29)uid displaced, is thus greater than the weight
of the object. Likewise, an object denser than the (cid:29)uid will sink.
The extent to which a (cid:29)oating object is submerged depends on how the object’s density is related to
that of the (cid:29)uid. In Figure 4, for example, the unloaded ship has a lower density and less of it is submerged
compared with the same ship loaded. We can derive a quantitative expression for the fraction submerged
by considering density. The fraction submerged is the ratio of the volume submerged to the volume of the
object, or
V V
fraction submerged = sub = (cid:29) . (3)
V V
obj obj
The volume submerged equals the volume of (cid:29)uid displaced, which we call V . Now we can obtain the
(cid:29)
relationship between the densities by substituting ρ= m into the expression. This gives
V
V m /ρ
(cid:29) = (cid:29) (cid:29) , (3)
V m /ρ
obj obj obj
where ρ is the average density of the object and ρ is the density of the (cid:29)uid. Since the object (cid:29)oats, its
obj (cid:29)
mass and that of the displaced (cid:29)uid are equal, and so they cancel from the equation, leaving
ρ
fraction submerged= obj. (3)
ρ
(cid:29)
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Figure 4: An unloaded ship (a) (cid:29)oats higher in the water than a loaded ship (b).
We use this last relationship to measure densities. This is done by measuring the fraction of a (cid:29)oating
object that is submerged(cid:22)for example, with a hydrometer. It is useful to de(cid:28)ne the ratio of the density of
an object to a (cid:29)uid (usually water) as speci(cid:28)c gravity:
ρ
speci(cid:28)c gravity= , (4)
ρ
w
where ρ is the average density of the object or substance and ρ is the density of water at 4.00◦C. Speci(cid:28)c
w
gravityisdimensionless, independentofwhateverunitsareusedforρ. Ifanobject(cid:29)oats, itsspeci(cid:28)cgravity
islessthanone. Ifitsinks,itsspeci(cid:28)cgravityisgreaterthanone. Moreover,thefractionofa(cid:29)oatingobject
that is submerged equals its speci(cid:28)c gravity. If an object’s speci(cid:28)c gravity is exactly 1, then it will remain
suspended in the (cid:29)uid, neither sinking nor (cid:29)oating. Scuba divers try to obtain this state so that they can
hover in the water. We measure the speci(cid:28)c gravity of (cid:29)uids, such as battery acid, radiator (cid:29)uid, and urine,
as an indicator of their condition. One device for measuring speci(cid:28)c gravity is shown in Figure 5.
: Speci(cid:28)c gravity is the ratio of the density of an object to a (cid:29)uid (usually water).
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Figure 5: This hydrometer is (cid:29)oating in a (cid:29)uid of speci(cid:28)c gravity 0.87. The glass hydrometer is (cid:28)lled
with air and weighted with lead at the bottom. It (cid:29)oats highest in the densest (cid:29)uids and has been
calibrated and labeled so that speci(cid:28)c gravity can be read from it directly.
Example 2: Calculating Average Density: Floating Woman
Supposea60.0-kgwoman(cid:29)oatsinfreshwaterwith97.0%ofhervolumesubmergedwhenherlungs
are full of air. What is her average density?
Strategy
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We can (cid:28)nd the woman’s density by solving the equation
ρ
fraction submerged= obj (5)
ρ
(cid:29)
for the density of the object. This yields
ρ =ρ =(fraction submerged)·ρ . (5)
obj person (cid:29)
We know both the fraction submerged and the density of water, and so we can calculate the
woman’s density.
Solution
Entering the known values into the expression for her density, we obtain
(cid:18) (cid:19)
kg kg
ρ =0.970· 103 =970 . (5)
person m3 m3
Discussion
Her density is less than the (cid:29)uid density. We expect this because she (cid:29)oats. Body density is
oneindicatorofaperson’spercentbodyfat,ofinterestinmedicaldiagnosticsandathletictraining.
(See Figure 6.)
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Figure 6: Subject in a (cid:16)fat tank,(cid:17) where he is weighed while completely submerged as part of a body
densitydetermination. Thesubjectmustcompletelyemptyhislungsandholdametalweightinorderto
sink. Correctionsaremadefortheresidualairinhislungs(measuredseparately)andthemetalweight.
His corrected submerged weight, his weight in air, and pinch tests of strategic fatty areas are used to
calculate his percent body fat.