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PHY–302 K. Solutions for problem set #12.
Textbook conceptual question 9.12: 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.
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
V glass^ = V (^) submergedice + V water. (S.1)
By Archimedes’s Law, the volume of the submerged part of the ice cube controls the buoyant force on the ice cube,
FB = g × ρwater^ × V (^) submergedice , (S.2)
and for the ice cube floating in equilibrium, this force balances the whole cube’s weight M iceg. Therefore,
g × ρwater^ × V (^) submergedice = g × M ice^ =⇒ V (^) submergedice = M^
ice ρwater^.^ (S.3)
Also, the liquid water in the glass fills up volume
V water^ = M^
water ρwater^.^ (S.4)
Combining eqs. (S.3) and (S.4) according to eq. (S.1), we see that the volume of the glass up to the water level is
V glass^ = M^
ice ρwater^ +^
M water ρwater^ =^
M ice^ + M water ρwater^.^ (S.5)
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. (S.6)
Consequently, eq. (S.1) tells us that
V glass^ = M^
tot ρwater^ =^ const,^ (S.7)
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.?
Textbook problem 9.32: As the ship floats in (vertical) equilibrium, its gross weight M g is balanced by the buoyant force
FB = g × ρwater^ × V subm^ (S.8)
where V subm^ is the volume of the submerged part of the ship’s hull (everything enclosed by the hull, but only below the waterline). Therefore,
g × M = g × ρsea water^ × V subm^ =⇒ V subm^ = (^) ρsea waterM. (S.9)
The density of sea water (at the surface pressure) varies between 1020 and 1030 kg/m^3 de- pending on the temperature and the salinity. For the problem at hand, I assume 3.5% salinity — typical for the open ocean where Nimitz carries her planes — and 20◦^ C temperature, which gives me ρsea water^ = 1025 kg/m^3.
? For a cylindrical glass, V glass^ = πR^2 × 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.
which gives the buoyant force capable of supporting gross mass
M max^ = F^ Bmax g =^ ρ
water (^) × V (^) maxsubm = 1000 .0 kg/m (^3) × 0 .08 m (^2) = 80 kg. (S.15)
(Assuming fresh water of density 1000.0 kg/m^3 .)
Note that eq. (S.15) gives the maximal gross mass supported by the air mattress. The net mass obtains by subtracting the 2.0 kg mass of the mattress itself. Thus, the maximal net mass this mattress can support without sinking is 80 kg − 2 kg = 78 kg.
Textbook problem 9.40: Water flows into the tank at rate
F = Vt = 20 .0 gallons 30 .0 s = 7530 ..7 L0 s = 2 .52 L/s = 2. 52 · 10 −^3 m^3 /s. (S.16)
In a pipe, the flow rate is related to the pipe’s cross-section A and speed v of the water in the pipe as
F = A × v. (S.17)
Given the flow rate (S.16) and the pipe’s cross-section A = 1.00inch^2 = 6.45 cm^2 = 6. 45 · 10 −^4 m^2 , we may find the water speed as
v = F A = 2.^52 ·^10
− (^3) m (^3) /s
- 45 · 10 −^4 m^2 =^3 .91 m/s,^ (S.18)
or about 8.75 miles/hour.
Textbook problem 9.42: (a) The flow rate
F = A × v (S.19)
is constant along the pipe. If one section of the pipe is narrower than the other, the liquid’s
speed in that section must increase so that the flow rate (S.19) remains the same. Specifically,
A 2 × v 2 = F = A 1 × v 1 =⇒ v 2 = A A^1 2 × v 1 = 10 .0 cm
2 2 .50 cm^2 ×^ 275 cm/s^ =^ 1100 cm/s. (S.20)
(b) According to the Bernoulli’s equation
P + ρgy + 12 ρv^2 = const (S.21)
along the pipe, so the speeds and pressures of the liquid in the two sections of the pipe are related to each other as
P 2 + ρgy 2 + 12 ρv^22 = P 1 + ρgy 1 + 12 ρv^21. (S.22)
The pipe in question is horizontal, so y 2 = y 1 and the ρgy terms on the two sides of this equation cancel each other. Consequently
P 2 + 12 ρv^22 = P 1 + 12 ρv^21 ,
and the pressure in the second section is
P 2 = P 1 + ρ 2 ×
v^21 − v 22
= 1. 20 · 105 Pa + 1650 kg/m
3 2 ×^
(2.75 m/s)^2 − (11.0 m/s)^2
= 26 , 400 Pa.
(S.23)
Textbook problem 9.46: (a) Consider a continuous stream of water which begins at the surface inside the tank (point 1) and ends as a little jet outside the tank (point 2). Applying Bernoulli’s equation (S.21) to this stream, we have
P 2 + ρgy 2 + 12 ρv^22 = P 1 + ρgy 1 + 12 ρv^21. (S.24)
Since the tank is open to the atmosphere, the pressure at the surface equals to the air pressure, P 1 = P atm. Likewise, the water jetting from the tank is open to the air, so P 2 = P atm.
Textbook problem 9.48: Consider a continuous stream of water beginning at the surface of the Colorado river (point 1), going through the inlet and the pump, up through the pipe, and finally emerging from the top of the pipe in the village (point 2). Without the pump, the pressure in this stream would be governed by the Bernoulli’s equation (S.21), but the pump increases the pressure by P pump without affecting the speed or altitude of the water. Consequently, the combination
P = P + ρgy + 12 ρv^2 (S.31)
stays constant until the water reaches the pump, abruptly increases by P pump^ at the pump, and then again stays constant until the water reaches the tank. Consequently, comparing points 1 and 2 of the stream, we have
P 2 + ρgy 2 + 12 ρv^22 = P 1 + ρgy 1 + 12 ρv 12 + P pump. (S.32)
Now consider the pressures P 1 and P 2 in this equation. P 1 is the pressure of water at the surface of the Colorado river, so it is equal to the air pressure at the bottom of the Grand Canyon, P 1 = P (^) GCair. P 2 is the water pressure at the outlet of the pipe; since this outlet is open to the air, P 2 is the air pressure in the village, P 2 = P (^) villageair. Because of the large difference in altitude, the two air pressures are not equal; instead,
P (^) GCair − P (^) villageair = ρair^ ×g ×(y 2 −y 1 ) = 1.2 kg/m^3 × 9 .8 N/kg×1548 m ≈ 18 · 103 Pa. (S.33)
and consequently
P 1 − P 2 ≈ 18 kPa. (S.34)
However, this pressure difference is much smaller than the other terms in eq. (S.32), so we are going to neglect it. Thus, we simplify eq. (S.32) to
ρgy 2 + 12 ρv 22 ρgy 1 + 12 ρv 12 + ∆P pump^ (S.35)
and hence
P pump^ = ρg × (y 2 − y 1 ) + 12 ρ × (v^22 − v 12 ). (S.36)
(a) According to eq. (S.36), the pump pressure required to get water to the village depends
on the speed of the water. At the minimal pressure, the water would barely get up tho the village, but its speed would be very slow, so
P (^) minpump = ρg × (y 2 − y 1 ) = 1000.0 kg/m^3 × 9 .8 N/kg × 1548 m = 15. 2 · 106 Pa ≈ 150 atm. (S.37)
(b) The flow rate in the pipe depends on the water speed v and the pipe’s cross-sectional area A as
F = A × v. (S.38)
For the pipe in question the area is
A = πR^2 = π(d/2)^2 = π(15.0 cm/2)^2 = 177 cm^2 = 1. 77 · 10 −^2 m^2 , (S.39)
so given the flow rate
F = 4500 m^3 /day = 5. 2 · 10 −^2 m^3 /s, (S.40)
we may find the water’s speed as
v = F A = 5.^2 ·^10
− (^2) m (^3) /s
- 77 · 10 −^2 m^2 =^2 .95 m/s.^ (S.41)
(c) According to eq. (S.36), besides the hydrostatic pressure (a), the pump needs to provide extra pressure
∆P pump^ = 12 ρ × (v 22 − v 12 ). (S.42)
Here v 2 is the speed of water at the outlet of the pipe, and v 1 is the speed at point 1, down from the surface of the river. By continuity equation,
v 2 = (^) AF 2 = 2 .95 m/s as found in part(b),
v 1 = (^) AF 1 v 1 because A 1 A 2 ,
(S.43)
hence
∆P pump^ ≈ 12 ρv^22 = 12 × 1000 .0 kg/m^3 × (2.95 m/s)^2 ≈ 4300 Pa. (S.44)
Although such an extra pressure would be quite noticeable in most other water pumps, but it
and the water in the tube. Once we have this difference, eq. (S.49) will tell us the air speed v above the left opening.
The liquids in the tube are not moving, so the pressure in the tube varies with depth according to the hydrostatic equation,
P = P surface^ + ρg × depth (S.50)
The water surface in the left half of the U-tube is open to air at pressure PL, so the water pressure in the tube at level y is
P water(y) = P (^) Lair + ρwater^ × g × (yL − y) (S.51)
here yL is the water surface level on the left side. Likewise, the oil surface in the right half is open to air at pressure PR, so the oil pressure at level y is
P oil(y) = P (^) Rair + ρoil^ × g × (YR − y). (S.52)
At the water-oil interface, both pressures must be equal,
P water(yI ) P oil(yI ) (S.53)
so
P (^) Lair + ρwater^ × g × (yL − yI ) = P (^) Rair + ρoil^ × g × (yR − yI ). (S.54)
Figure P9.80 (c) shows left and right surfaces being at the same level, yL = yR, while the water-oil interface is at depth L. Consequently,
YL − yI = yR − yI = L, (S.55)
and eq. (S.54) becomes
P (^) Lair + ρwater^ × gL = P (^) Rair + ρoil^ × gL. (S.56)
Using this equation, we obtain the difference between air pressures in the two sides of the
U-tube as
P (^) Rair − P (^) Lair =
ρwater^ − ρoil
× gL = (1000 kg/m^3 − 750 kg/m^3 ) × 9 .8 N/kg × 0 .0500 m = 123 Pa.
(S.57)
At this point, we go back to eq. (S.49) which relates the pressure difference to the speed of the air,
(^12) ρair (^) × v (^2) = P (^) Rair − P (^) Lair = 123 Pa. (S.58)
Consequently,
v^2 = 2 × P^ Rair −^ P^ Lair ρair^ =^2 ×^
123 Pa 1 .29 kg/m^3 =^ 190 m
(^2) /s (^2) (S.59)
and the air speed is v = 13.8 m/s or about 31 miles per hour.
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
(S.60)
where A is the combined area of the plane’s wings.
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
slow down the air flowing below the wings,
vwing top^ > vp > vwing bot. (S.68)
According to eq. (S.67), this creates pressure difference across the wing,
P bot^ − P top^ = ρ 2 ×
(vwing top)^2 − (vwing bot)^2
, (S.69)
which results in the lift force on the plane
F lift^ = A × ρ 2 ×
(vwing top)^2 − (vwing bot)^2
(S.70)
For the plane in question, the lift force is
F lift^ = 12 m^2 × 1 .00 kg/m
3 2 ×^
((55 m/s) (^2) − (45 m/s) 2 ) (^2) = 6. 0 · 103 N. (S.71)
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.
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 vball. 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 give 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.
Textbook problem 9.58: A thin layer of glycerin immediately adjacent to a glass slide does not like to move relative to that slide. When the two parallel slides move relative to each other, the bottom layer of glycerin moves with the bottom slide, the top layer moves with the top slide, and between the slides there is a velocity sheer where different layers of the glycerin move with different velocities. (See figure 9.48 on page 303 of the textbook for the diagram of the velocities in such a sheer.) For the plates in question, the velocity sheer is
∆vx ∆y =^
0 .30 m/s 1 .5 mm =^ 200 s
− 1. (S.72)
In a viscous fluid such as glycerin, velocity sheer creates a friction force between adjacent
For the hypodermic needle in question, the length is L = 3.0 cm, the radius is R = 0 .30 mm/2 = 0.15 mm, and we want the flow rate of 1.0 g/s — or in volume terms, F = 1 .0 cm^3 /s (note ρ(water) = 1.00 g/cm^3 ). By Poiseuille’s Law (S.78), this requires pressure differential
∆P ≡ Psyringe − Psharp end = 0 .030 m × 8 ×^1.^0 ·^10
− (^3) Pa · s × 1. 0 · 10 − (^6) m (^3) /s π(1. 5 · 10 −^4 m)^4 = 1. 5 · 105 Pa ≈ 1 .5 atm.
(S.79)
