PHY–302 K.
Solutions for Problem set # 11.
Textbook problem 10.16:
At the boundary between oil and water, the pressures in both liquids must be the same,
Poil = Pwater
at the boundary.
(1)
The oil column in the left side of the U-tube is open to the air, hence at depth Doil = 27.2 cm
below the oil surface, the oil pressure is
Poil = Patm + ρoil × g × Doil .
(2)
The water pressure at the boundary is equal to the water pressure in the right side of the
U-tube at the same level, i.e. at depth
Dwater = 27.2 cm − 9.41 cm = 17.8 cm
(3)
below the water surface. And since the water column on the right side of the U-tube is open
to the air, the water pressure at that depth is
Pwater = Patm + ρwater × g × Dwater .
(4)
Thus, at the water-oil boundary
Poil = Patm + ρoil gDoil
=
Pwater = Patm + ρwater gDwater ,
(5)
and hence
ρoil × Doil = ρwater × Dwater .
(6)
Solving this equation for the specific gravity of the oil — i.e., the ratio of its density to the
density of water, — we get
ρoil
ρwater
=
Dwater
17.8 cm
=
= 0.654.
Doil
27.2 cm
(7)
Since the water density is ρwater = 1.000 g/cm3 = 1000 kg/m3 , the oil density is ρoil =
0.654 g/cm3 = 654 kg/m3 .
1
Textbook problem 10.22:
The apparent weight of the rock submerged in water is the difference between the gravity
force and the buoyant force of the water,
Wapp = M g − FB .
(8)
By Archimedes’s Law, the buoyant force equal to the weight of the displaced water. For a
submerged rock, this is the weight of the same volume of water as the rock, thus
FB = g × ρwater × Vrock .
(9)
Wapp = gMrock − gρwater Vrock ,
(10)
Consequently,
and if the scales automatically convert this apparent weight into mass by dividing by g, then
they show the apparent mass of
Mapp =
Wapp
= Mrock − ρwater × Vrock .
g
(11)
Dividing this apparent mass of the rock by its true mass, we get
Mapp
ρwater
ρwater × Vrock
= 1 −
= 1 −
.
Mtrue
Mrock
ρrock
(12)
For the Moon rock in question, this formula gives us
Mapp
ρwater
6.18 kg
= 1 −
= 1 −
= 0.334.
ρrock
Mtrue
9.28 kg
(13)
Consequently, the Moon rock’s density is
ρrock =
ρwater = 1000 kg/m3
= 2990 kg/m3 .
0.334
(14)
This is the density of basalt — which is rather common on the Moon’s surface — so the Moon
rock in question is probably a piece of basalt.
2
Textbook problem 10.72:
The Earth’s structure has many layers. The top layer of rock under the atmosphere or ocean
is called the crust, while the next layer down is called the mantle. The mantle is very thick —
about 2900 km (1800 miles) — while the crust is much thinner: from 30 to 70 km (20 to 45
miles) under continents, and only 7 to 10 km (4.5 to 6 miles) under deep oceans. Although the
mantle is made out of solid rock, on the time scale of millions of years it can flow like a liquid.
Consequently, the crust essentially floats on top of the mantle, and its isostatic equilibrium
(balance of vertical forces) requires the buoyant force FB from the mantle to cancel the weight
M g of the crust.
For simplicity, let’s ignore the relatively thin oceanic crust (and the water on top of it)
and treat the deep ocean basins as an open surface of the mantle. The continents then are
thick slabs of less dense rock floating on that mantle surface like soap bars floating on the
surface of water in a bathtub or icebergs floating on the surface of the ocean.
By Archimedes’s Law, the buoyant force on a floating body — soap bar, iceberg, or a
continent — is the weight of the liquid it displaces, thus
FB = g × ρliquid × Vsubm
(15)
where Vsubm is the body’s volume that happens to be submerged — i.e., lies below the surface
of the liquid — and hence displaces the liquid; the volume that stays above the surface does
not affect the buoyant force. In isostatic equilibrium FB = M g and hence
ρliquid × Vsubm = Mbody = ρbody × Vtotal .
(16)
where the second equality applies to bodies of uniform density. For such bodies,
ρbody
Vsubmerged
=
.
Vtotal
ρliquid
(17)
For example, an iceberg is 90% submerged in water because the ice density is 90% of the
density of seawater.
3
For the problem at hand, the body in question is the continent and the “liquid” in which
it floats is the mantle rock. Thus,
Vsubmerged (continent)
ρ(continental rock)
2800 kg/m3
=
=
= 0.85 :
Vtotal (continent)
ρ(mantle rock
3300 kg/m3
(18)
85% of the contient’s volume lies below the mantle’s surface and only 15% sticks out. Approximating the continent as a slab of constant thickness D, we have
Dsubmerged
Vsubmerged
=
= 0.85
Dtotal
Vtotal
(19)
Vabove
Dabove
=
= 1 − 0.85 = 0.15.
Dtotal
Vtotal
(20)
and hence
In other words, only 15% of the continent total thickness of 35 km lies above the mantle
surface, i.e. Dabove = 0.15 × 35 km = 5.3 km.
PS: This is not a required part of the homework, but I would like to put this problem in the
context of the real Earth.
In real life, the mantle does not have an open surface — outside the continents, the mantle
is covered by the thin oceanic crust, which is in turn covered by ocean water. The isostatic
equilibrium of the oceanic crust requires
gMcrust + gMwater = FB (from mantle) = gρmantle Vsubm ,
(21)
hence in the slab approximation
gA × Dcrust ρcrust + Dwater ρwater
= gA × ρmantle Dsubmerged crust
(22)
and therefore
Dsubmerged crust =
Dcrust ρcrust + Dwater ρwater
≈ 0.85Dcrust + 0.30Dwater .
ρmantle
(23)
Consequently, the would-be open mantle surface would be lower than the ocean water surface
4
by
−Ymantle = Dwater + Dcrust − Dsubmerged crust = 0.70Dwater + 0.15Dcrust .
(24)
For the 8 km thick oceanic crust under 4.5 km of water, this evaluates to the would-be-mantle
surface being 4.3 km below the sea level.
We saw that the continent’s top should lie 5.3 km above that would-be mantle surface,
which puts it 1.0 km above the sea level. And indeed, the average altitude of dry land on
Earth is about 1 km above the sea level or 5.5 km above the average level of the ocean bottom.
The reason for this height different is the different thickness of crust that floats on the
mantle. The continental crust is much thicker, so its bottom lies much lower than the bottom
of the oceanic crust, but since the crust is less dense than the mantle, the top surface of the
thicker continental crust lies 5.5 km higher than the ocean bottom.
Textbook problem 10.80:
Consider a volume of water surrounded by other water in an accelerated bucket. The buoyant
force from the surrounding water should not only compensate the gravity force mg on the
selected volume but also give it the same acceleration as a the bucket, thus
FB − mg =
X
Fy = may ≡ ma
(25)
and hence
FB = mwater × (g + a).
(26)
For the same volume occupied by some other body — like a piece of granite rock — instead
of the water, we would get the same buoyant force
FB = mwater × (g + a) = ρwater × Vbody × (g + a).
(27)
Another way to derive this formula is to work in the non-inertial frame of the accelerating
bucket. The Newtonian mechanics — and hence hydrostatics and hydrodynamics — in such
5
frames can be described in terms of effective gravity which combine the true gravity force m~g
with the inertial force −m~aframe , hence
~ eff = m~g − m~aframe = m~geff
W
where ~geff = ~g − ~aframe .
(28)
For the bucket accelerating up, the effective gravity is directed down and has magnitude
geff = g + abucket ≈ 12.2 m/s2 ,
(29)
and as long as we do not look outside the bucket, we cannot distinguish this effective gravity
from the real gravity on some other planet where g = 12.2 m/s2 . Consequently, the buoyant
force of water in the accelerating bucket is the weight of displaced water in the effective
gravitational field,
FB = ρwater × Vbody × geff = ρwater × Vbody × (g + a).
(30)
The rock in question has specific gravity 2.7, so its volume is
Vrock =
ρrock
Mrock
= 2.7 × ρwater
=⇒
ρwater × Vrock =
Mrock
3.0 kg
=
= 1.1 kg. (31)
2.7
2.7
In other words, the rock displaces 1.1 kg of water, and the buoyant force on the rock is the
weight of that displaced water in the effective gravitational field inside the accelerating bucket,
FB = 1.1 kg × 12.2 m/s2 = 13.3 N = ≈ 3.0 lb.
6
(32)
Textbook problem 10.34:
In equilibrium, the net buoyant force on the floating piece of wood and on the chunk of lead
hanging from it must balance the net weight of the lead and the wood,
FB (on wood) + (on lead) = gMlead + gMwood .
(33)
Note that the lead is hanging on a sting below the floating piece of wood, so it is completely
immersed in water. Consequently, it is subject to the buoyant force
FB (on lead) = gρwater Vlead = gρwater ×
Mlead
gMlead
=
ρlead
SGlead
(34)
where SGlead = ρlead /ρwater is the specific gravity of lead; according to table 10–1 on page 256
of the textbook, SGlead = 11.3.
The piece of wood is also subject to the buoyant force which depends on how much wood
is submerged under water,
FB (on wood) = gρwater × Vsubmerged wood .
(35)
When the wood is completely sunk by the lead’s weight,
Vsubmerged wood → Vwood, total =
Mwood
ρwood
(36)
the buoyant force on the sunk wood becomes
FB (on wood) = gρwater ×
gMwood
Mwood
=
.
ρwood
SGwood
(37)
To sink the wood, the net weight of the lead and the wood must exceed the net buoyant
force on the wood and the lead, thus
gMlead + gMwood >
gMwood
gMlead
+
.
SGwood
SGlead
(38)
To solve this inequality, let’s move the terms involving lead to the left hand side and the terms
7
involving wood to the right hand side. This gives us
1
gMlead × 1 −
SGlead
> gMwood ×
1
SGwood
−1
(39)
and consequently
1
1
−1
−1
Mlead
= 0.50 1 ≈ 1.1.
> SGwood 1
Mwood
1 − SGlead
1 − 11.3
(40)
The minimal mass of lead needed to sink the piece of wood is
min
Mlead
= 1.1 × Mwood = 1.1 × 5.25 kg ≈ 5.8 kg.
(41)
Textbook question 10.7:
(a) The ice cube floats on the surface of water because the ice is less dense than the liquid
water.
(b) As the ice cube melts, it adds more liquid water to the glass. On the other hand, a smaller
ice cube displaces less water. The two effects precisely cancel each other, and the water level
in the glass remains exactly the same. If the water initially fills the glass to the brim, it would
continue to fill the glass to the brim, but it will not overflow.
To see this, consider the volume V glass of the glass up to the water level. Initially, some of
this volume is occupied by by the submerged part of the ice cube (the part below the water
level), and the rest is filled with liquid water, thus
ice
V glass = Vsubmerged
+ V water .
(42)
By Archimedes’s Law, the volume of the submerged part of the ice cube controls the buoyant
force on the ice cube,
ice
FB = g × ρwater × Vsubmerged
,
(43)
and for the ice cube floating in equilibrium, this force balances the whole cube’s weight M ice g.
8
Therefore,
ice
g × ρwater × Vsubmerged
= g × M ice
=⇒
ice
Vsubmerged
=
M ice
.
ρwater
(44)
Also, the liquid water in the glass fills up volume
V water =
M water
.
ρwater
(45)
Combining eqs. (44) and (45) according to eq. (42), we see that the volume of the glass up to
the water level is
V glass =
M water
M ice + M water
M ice
+
=
.
ρwater
ρwater
ρwater
(46)
As the ice cube melts, it becomes liquid water, indistinguishable from the rest of the liquid
water in the glass, and for every gram of ice which melts, we get exactly one extra gram of
liquid water. Thus, while the ice melts, M ice decreases, M water increases, but the total mass
of ice and water stays exactly the same,
M tot = M ice + M water = const.
(47)
Consequently, eq. (42) tells us that
V glass =
M tot
= const,
ρwater
(48)
the volume of the glass below the water level remains constant as the ice cube melts. And for
any fixed shape of the glass, this immediately tells us that the water level in the glass remains
?
constant.
? For a cylindrical glass, V glass = πR2 × H where R is the fixed radius of the glass and H is the water
level above the bottom. This makes it obvious that constant V glass implies constant water level H.
For glasses of other shapes, V glass is a more complicated function of the water level H, but it always
increases with H. (For small ∆H, ∆V glass = ∆H × area of the water level inside the glass.) Therefore,
constant V glass implies constant water level H.
9
Textbook problem 10.39:
The hose has inside diameter 2R =
A = πR
2
5
8
inch = 1.59 cm, so its cross-sectional area is
= π
1.59 cm
2
2
= 1.98 cm2 .
(49)
The water flow through this hose at speed v = 0.40 m/s, so the volume flow rate is
F = A × v = 1.98 cm2 × 40 cm/s ≈ 79 cm3 /s.
(50)
The volume of water needed to fill the pool is
V = h × Apool =
2
h × πRpool
= 1.2 m × π
6.1 m
2
2
= 35 m3 .
(51)
The time needed to supply this volume of water at the rate (50) is
t =
35 m3 = 35 · 106 cm3
V
=
= 440, 000 s = 123 hours = 5.1 days.
F
79 cm3 /s
(52)
PS: To speed up this process, we would need a much faster speed of water in the hose. And a
wide hose would help too. For example, doubling the hose’s diameter to
5
4
inch and increasing
the flow of water to 10 m/s would increase the flow rate by a factor of 100 and hence shorten
the pool-filling time from 5+ days to an hour and a quarter.
Textbook problem 10.38:
According to the Bernoulli equation,
P1 + ρgy1 +
1 2
2 ρv1
= P2 + ρgy2 +
1 2
2 ρv2
(53)
for any two points (1) and (2) of a flow. Let (1) be the surface of water in the tank while (2)
is the water jet just outside the hole. Both points (1) and (2) are exposed to the outside air,
10
hence
P1 = P2 = Patm
(54)
and consequently
ρgy1 +
1 2
2 ρv1
= ρgy2 +
1 2
2 ρv2 .
(55)
With a little algebra, we may rewrite this equation as
v22 − v12 = 2g(y1 − y2 ).
(56)
Moreover, the velocities v1 and v2 are related by the continuity equation
v1 × A1 = F = v2 × A2
(57)
where A1 is the cross-sectional area of the tank and A2 is the cross-sectional area of the
jet coming out of the hole. Obviously A2 A1 , which immediately gives us v1 v2 : the
velocity of water inside the tank is much smaller than the velocity of the jet outside the hole.
Consequently, on the left hand side of eq. (56) we may neglect the v12 term compared to the
v22 , which gives us the Torricelli equation
v2 =
p
2g(y1 − y2 ).
(58)
In words, the speed of the jet coming out of the tank is equal to the speed of body free-falling
from height y1 = y2 .
The tank in question has a hole at depth y1 − y2 = 4.6 m below the water surface in the
tank, hence the water jets out from the hole at speed
v2
q
2(9.8 m/s2 )(4.6 m) = 9.5 m/s ≈ 21 MPH.
=
11
(59)
Textbook problem 10.41:
For a horizontal pipe (y = const), the Bernoulli equation (53) simplifies to
P1 +
1 2
2 ρv1
= P2 +
1 2
2 ρv2 .
(60)
Using the continuity equation (57), we may relate the flow speeds v1 and v2 to the pipe
diameters d1 = 2R2 and d2 = 2R2 at the corresponding points of the pipe and to the volume
flow rate F (which is constant throughout the pipe):
v1 =
F
4F
F
=
=
A1
πR12
πd21
(61)
4F
F
=
.
2
πR2
πd22
(62)
and likewise
v1 =
Plugging these velocities into the Bernoulli equation (60), we get
ρ
P1 + ×
2
4F
πd21
2
ρ
= P2 + ×
2
4F
πd22
2
.
(63)
To solve this equation for the flow rate F, we rewrite it as
P1 +
8ρF 2
8ρF 2
2
=
P
+
2
π 2 d41
π 2 d42
(64)
and re-group the terms to get
P1 − P2
8ρF 2
×
=
π2
1
1
− 4
4
d2
d1
.
(65)
Consequently,
F
2
π 2 (P1 − P2 )
=
×
8ρ
12
1
1
−
d42
d41
−1
.
(66)
For the pipe in question d1 = 6.0 cm, d2 = 4.0 cm, hence
1
1
− 4
4
d2
d1
−1
3.91 · 10−3 cm−4 − 0.77 · 10−3 cm−4
=
−1
= 319 cm4 .
(67)
Also, we are given P1 = 32.0 kPa and P2 = 24.0 kPa, thus
π 2 (P1 − P2 )
π 2 × 8.0 · 103 Pa
=
= 9.86 m2 /s2 .
8ρ
8 × 1.00 kg/m3
(68)
Plugging these numbers into eq. FF gives us
F 2 = 9.86 m2 /s2 × 319 cm4 = 3150 m2 cm4 /s2 = 3.15 · 107 cm6 /s2
(69)
and hence
F =
q
3.15 · 107 cm6 /s2 ≈ 5600 cm3 /s = 7.8 L/s.
(70)
Alternative Solution:
Using the continuity equation (57), we can obtain the ratio of flow speeds in terms of the pipe
diameters,
A1 v1 = A2 v2
=⇒
A1
v2
=
=
v1
A2
d1
d2
2
=
6.0 cm
4.0 cm
2
= 2.25.
(71)
Plugging this ratio in the Bernoulli equation (60), we have
P1 +
1 2
2 ρv1
= P2 +
1
2ρ
2
2.25 v1 .
(72)
We can solve this equations for the v1 by re-writing it as
1 2
2 ρv1
× 2.252 − 1 = P1 − P2 ,
(73)
hence
s
v1 =
2(P1 − P2 )
=
ρ(2.252 − 1)
s
2(8 · 103 Pa)
≈ 2.0 m/s.
(1000 kg/m3 )(2.252 − 1)
(74)
Given this flow speed in the wider part of the pipe where the diameter is 6.0 cm, the flow rate
13
is
F = v1 ×A1 = v1 ×π(d1 /2)2 = 2.0 m/s×π×(3.0 cm)2 = 56 m cm2 /s = 5600 cm3 /s (75)
or in other units, 5.6 liters per second or 5.6 · 10−3 m3 /s.
Non-textbook problem #1:
To lift force on the plane results from the air pressure on the bottom of the wings being larger
than the pressure on the top of the wings. Consequently, the air pressure pushed the plane
up harder than it pushed it down, which creates a net upward force
F lift = F up − F down = A × P bot − A × P top = A × P bot − P top
(76)
where A is the net area of both wings of the plane.
To understand the pressure difference across the wings, let’s consider how the air moves
relative to the plane. If the plane flies with velocity ~vp relative to the air, then in the reference
frame of the plane, the air far ahead of the plane moves with velocity −~vp towards the plane.
Closer to the plane, the air stream splits into two — one goes above the the wings and the
other below the wings — and each stream satisfies its own Bernoulli equation (53). Thus,
P wing top + ρgy wing top +
wing top 2
1
)
2 ρ(v
= P ahead,top + ρgy ahead,top +
ahead,top 2
1
) ,
2 ρ(v
P wing bot + ρgy wing bot +
wing bot 2
1
)
2 ρ(v
= P ahead,bot + ρgy ahead,bot +
ahead,bot 2
1
) .
2 ρ(v
(77)
Neglecting the vertical motion within each airstream, we approximate
y ahead,top = y wing top
and y ahead,bot = y wing bot ,
(78)
which simplifies eqs. (77) to
P wing top +
wing top 2
1
)
2 ρ(v
= P ahead,top +
ahead,top 2
1
) ,
2 ρ(v
P wing bot +
wing bot 2
1
)
2 ρ(v
= P ahead,bot +
ahead,bot 2
1
) .
2 ρ(v
(79)
Far ahead of the plane, both air-streams flow with exactly the same speed relative to the
14
plane
v ahead,top = v ahead,bot = vp ,
(80)
and the pressure in both air-streams is simply the atmospheric pressure at the plane’s altitude,
P ahead,top = P ahead,bot = P atm .
(81)?
Thus, the right hand sides of the two equations (79) are equal to each other, so the left hand
sides must also be equal to each other,
P wing top +
wing top 2
1
)
2 ρ(v
= P wing bot +
wing bot 2
1
) .
2 ρ(v
(83)
The wings of an airplane are designed to speed up the air flowing above the wings and
slow down the air flowing below the wings,
v wing top > vp > v wing bot .
(84)
According to eq. (83), this creates pressure difference across the wing,
P bot − P top =
ρ wing top 2
× (v
) − (v wing bot )2 ,
2
(85)
which results in the lift force on the plane
F lift = A ×
ρ wing top 2
× (v
) − (v wing bot )2
2
(86)
For the plane in question, the lift force is
F lift = 12 m2 ×
2
1.00 kg/m3
× (55 m/s)2 − (45 m/s)2 = 6.0 · 103 N.
2
(87)
In a level flight, this lift force balances the plane’s weight M g. So the plane in question must
weigh 6000 Newtons, i.e. 1350 pounds.
? To be precise, there is a small hydrostatic pressure difference between the two air-streams ahead of the
plane,
P ahead,bot − P ahead,top = ρg × y ahead,top − y ahead,bot .
(82)
But this difference is so small compared to the other pressure differences in this problem that we may
safely ignore it.
15
Non-textbook problem #2:
The soccer ball on the video veers to the side because of the Magnus effect which creates a
sideways force similar to the lift force on a wing of an airplane, cf. the previous problem.
Let’s go in the reference frame of the flying ball and consider how the air flows around
the ball. Far ahead of the ball, the air flows back — towards the ball — with speed equal to
v ball . Closer to the ball, the air speeds up or slows down due to friction against the spinning
surface of the ball. Since the air flows backwards (with respect to the ball’s direction of flight),
it speeds up over the side of the ball which spins back and slows down over the side of the
ball which spins forward. This speed-up / slow-down of the air stream lowers / raises the
local air pressure according to the Bernoulli’s equation, exactly as for the airplane’s wings in
the previous problem. And this difference in pressure creates a net Magnus force on the ball,
similar to the lift force on the plane.
The side of the ball which spins forward slows down the air flow, so it acts as a bottom of
a wing. The opposite side — which spins back — speeds up the air flow, so it acts as a top of
the wing. So by analogy with the lift force, the Magnus force pushes the ball in the direction
of the side which spins back.
In baseball, a batter wishing for a home run hits the ball below the center. This gives the
ball a back spin about a horizontal axis — the top of the ball spins back while the bottom spins
forward. Consequently, the Magnus force is directed up and creates a lift, which drastically
increases the horizontal range of the ball.
In soccer, to make the ball veer left or right, one needs a horizontal Magnus force. This
calls for a ball spinning around a vertical axis. If the spin is clockwise (as viewed from above),
the left side of the ball spins forward while the right side spins back; this creates for a Magnus
force directed to the right (relative to the ball’s direction), and the ball veers right, as in the
video. If the ball spins counterclockwise, it veers left.
To make the ball veer right, the player should kick it left from center. The force of such a
kick creates a clockwise torque relative to the ball’s center of mass, which makes the ball spin
clockwise as it flies forward. Consequently, the Magnus force will make the ball veer to the
right.
16