6 b assignment 1

Physics 6B, Assignment 1

• 18-Karat Gold vs. 14-Karat Gold (Density)

• Fluid Pressure in a U-Tube (Force and Pressure)

• Problem 15.27 (Density and Pressure)

• A Water Tank on Mars (Force and Pressure)

• Problem 15.47 (Force and Pressure)

• A Submerged Ball (Archimedes’ Principle)

• Problem 15.50 (Archimedes’ Principle)

• Submerged Sphere in a Beaker (Archimedes’ Principle)

• Flow Velocity of Blood (Bernoulli’s Equation)

• Water Flowing from a Tank (Bernoulli’s Equation)

• Problem 15.56 (Bernoulli’s Equation)

1 18-Karat Gold vs. 14-Karat Gold

The composition of a gold alloy is calculated by mass, using the karat (kt) asa unit of measure. A karat represents a proportion by mass of one part intwenty-four. The higher the karat value, the higher the proportion of gold inrelation to the total metal content. Pure gold is therefore 24 karat, while an18-karat gold alloy contains 18 parts (by mass) of gold out of 24 parts total.

1.1 18-karat Gold

In a sample of 18-karat gold, 75% of the total mass is pure gold, while therest is typically 16% silver and 9% copper. If the density of pure gold isρgold = 19.3 g/cm3, the density of silver is ρsilver = 10.5 g/cm3 and the density ofcopper is ρcopper = 8.90 g/cm3, what is the overall density ρalloy of this alloy of18-karat gold?

Density is defined to be mass per unit volume,

1

ρ =m

V

So the density ρalloy of a volume Valloy of 18-karat gold is

ρalloy =malloy

Valloy

The volume of the 18-karat gold is the sum of the volumes of the pure metals,

Valloy = Vgold + Vsilver + Vcopper

So our equation for the density of the 18-karat gold alloy becomes

ρalloy =malloy

Valloy

=malloy

(Vgold + Vsilver + Vcopper)

We need to calculate the volume of each pure metal used in the alloy.

The density of a metal is

ρ =m

V

The mass of a given volume of metal is therefore

m = ρ V

The mass of each of the pure metals in terms of the density and the volume is

mgold = ρgold Vgold

msilver = ρsilver Vsilver

mcopper = ρcopper Vcopper

In a sample of 18-karat gold, 75% of the total mass is pure gold, while therest of the mass is typically 16% silver and 9% copper.

The mass of the pure gold in the alloy is 75% of the total mass of the alloy,

mgold = 0.75 malloy

The mass of the silver is 16% of the total mass of the alloy,

msilver = 0.16 malloy

and the mass of the copper is 9% of the mass of the alloy,

2

mcopper = 0.09 malloy

We now have two equations for the mass of each metal,

mgold = ρgold Vgold = 0.75 malloy

msilver = ρsilver Vsilver = 0.16 malloy

mcopper = ρcopper Vcopper = 0.09 malloy

We can use these two equations to obtain the volume of each metal. The mass of thepure gold in the alloy is


so the volume of the pure gold is

Vgold =

(0.75

ρgold

)malloy

Vgold =

(0.75

19.3 g/cm3

)malloy = 0.03886 malloy

[cm3

g

]

The mass of the silver is


so the volume of the silver is

Vsilver =

(0.16

ρsilver

)malloy =

(0.16

10.5 g/cm3


[cm3

g

]

The mass of the copper is


so the volume of the copper is

Vcopper =

(0.09

ρcopper

)malloy =

(0.09

8.90 g/cm3


[cm3

g

]

Now that we know the volumes of the pure metals, the equation for the density of the18-karat gold alloy becomes

ρalloy =malloy

(Vgold + Vsilver + Vcopper)

3

ρalloy =malloy

(0.03886 malloy + 0.01524 malloy + 0.01011 malloy)[cm3

g

]The total mass of the gold alloy cancels out. (The density of the alloy should not

depend on how many grams of the material you have.)

ρalloy =1(

0.03886 + 0.01524 + 0.01011)[cm3

g

])

ρalloy =1

0.06421

[g

cm3

]So the density of the 18-karat gold alloy is

ρalloy = 15.6g

cm3

1.2 14-karat Gold

The percentage composition by mass of the less expensive 14-karat gold istypically 58.5% gold, 4.0% silver, 31.2% copper, and 6.3% zinc. The densityof zinc is ρzinc = 7.14 g/cm3. Find the percentage composition by volume of14-karat gold.

The total volume of the alloy is the sum of the volumes of the pure metals,

Valloy = Vgold + Vsilver + Vcopper + Vzinc

We need to determine what fraction F by volume is attributed to each of the fourpure metals,

Fgold =Vgold

Valloy

Fsilver =Vsilver

Valloy

Fcopper =Vcopper

Valloy

Fzinc =Vzinc

Valloy

The percentage composition by mass of 14-karat gold is 58.5% gold, 4.0%silver, 31.2% copper, and 6.3% zinc.

4

58.5% of the total mass of the alloy is pure gold, so the mass of the pure gold is 58.5% ofthe total mass of the alloy.

mgold = 0.585 malloy

The mass of the silver is 4.0% of the total mass of the alloy,

msilver = 0.040 malloy

The mass of the copper is 31.2% of the mass of the alloy,

mcopper = 0.312 malloy

and the mass of the zinc is 6.3% of the mass of the alloy,

mzinc = 0.063 malloy

Density is mass per unit volume

ρ =m

V

so the masses of the metals are




mzinc = ρzinc Vzinc = 0.063 malloy

The volumes of the metals are thus

Vgold =

(0.585

ρgold

)malloy =

(0.585

19.3 g/cm3


[cm3

g

]

Vsilver =

(0.040

ρsilver

)malloy =

(0.040

10.5 g/cm3


[cm3

g

]

Vcopper =

(0.312

ρcopper

)malloy =

(0.312

8.90 g/cm3


[cm3

g

]

Vzinc =

(0.063

ρzinc

)malloy =

(0.063

7.14 g/cm3


[cm3

g

]

5

The total volume of the alloy is

Valloy = Vgold + Vsilver + Vcopper + Vzinc

Valloy = (0.03031 + 0.00381 + 0.03506 + 0.00882) malloy

[cm3

g

]

Valloy = 0.078 malloy

[cm3

g

]

(Since the mass of the alloy will be expressed in grams, the volume will be in cubiccentimeters, cm3.)

We can now calculate what fraction by volume of the alloy is attributed to each ofthe four pure metals. We will start by calculating what fraction of the alloy is pure gold,

Fgold =Vgold

Valloy

We just calculated Valloy. Recall from above that Vgold is

Vgold = 0.03031 malloy

[cm3

g

]

Thus

Fgold =Vgold

Valloy

=0.03031 malloy

0.078 malloy

= 0.388

The total mass of the alloy cancels. (The total mass is irrelevant. The fraction ofpure gold in the alloy is the same whether we have 2 grams of alloy or 200 grams.)

Similarly for silver,

Fsilver =Vsilver

Valloy

=0.00381

0.078= 0.049

for copper

Fcopper =Vcopper

Valloy

=0.03506

0.078= 0.449

and for zinc

Fzinc =Vzinc

Valloy

=0.00882

0.078= 0.113

38.8% by volume of 14-karat gold is pure gold, 4.9% is silver, 44.9% iscopper, and 11.3% is zinc.

6

2 Fluid Pressure in a U-tube

A u-tube is filled with water and its two arms are capped. The tube iscylindrical and the right arm has twice the radius of the left arm. The capshave negligible mass, are watertight, and can freely slide up and down thetube.

2.1 Sand

Figure 1: Fluid in a u-tube.

A one-inch depth of sand is poured onto the cap on each arm of the u-tube.After the caps have moved if necessary to establish equilibrium, is the rightcap higher, lower, or the same height as the left cap? Refer to figure 1.

The right arm has twice the radius of the left arm,

Rright = 2 Rleft

The surface areas of the two caps are

Aleft = π R2left

Aright = π R2right

The right cap has four times the surface area of the left arm,

Aright = π (2 Rleft)2 = 4 π R2

left = 4 Aleft

Sand is poured to a depth hsand = 1 in. The volume of sand on the left cap is

Vleft = Aleft hsand

7

The volume of sand on the right cap is

Vright = Aright hsand

The volume of sand on the right cap is four times greater than the volume of sand onthe left cap,

Vright = Aright hsand = 4 Aleft hsand = 4 Vleft

The density of the sand is

ρsand =msand

Vsand

The mass of the sand on the left cap is

mleft = ρsand Vleft

The mass of the sand on the right cap is

mright = ρsand Vright = 4 ρsand Vleft = 4 mleft

The mass of the sand on the right cap is four times as great as the mass of the sandon the left cap. The weight of the sand is equal to the force due to gravity,

Fg = wsand = msand g

The force due to the sand on the left cap is

Fleft = mleft g

The force due to the sand on the right cap is

Fright = mright g = 4 mleft g = 4 Fleft

The force due to the sand on the right cap is four times greater than the force due tothe sand on the left cap. Pressure is force per unit area,

P =F

A

The excess pressure due to the sand on the left cap is

Pleft =Fleft

Aleft

The excess pressure due to the sand on the right cap is

Pright =Fright

Aright

Pright =Fright

Aright

=4 Fleft

4 Aleft

=Fleft

Aleft

= Pleft

The sand added to the left cap creates the same amount of excess pressure in the leftarm as the sand added to the right cap creates in the right arm. The pressure remainsconsistent throughout the u-tube. Since the system is already in equilibrium, the heightsof the caps will not change.

8

2.2 Equal Masses

The sand is removed and a block of mass Mblock = 1.0 kg is placed on each cap(figure 2). After the caps have moved (if necessary) to reestablish equilib-rium, is the right cap higher, lower, or the same height as the left cap?

Figure 2: Fluid in a u-tube.

The forces on the left and right caps due to the respective blocks are equal,

Fleft = Fright = wblock = Mblock g

The excess pressure on the left cap due to the addition of the block is

Pleft =Fleft

Aleft

The excess pressure on the right cap is

Pright =Fright

Aright

=Fleft

(4 Aleft)=

1

4

(Fleft

Aleft

)=

1

4Pleft

The block placed on the left cap is the same weight as the block placed on the rightcap. However, the area of the right cap is four times as large as the area of the left cap.On the right side the force is thus distributed over a much larger area. The excess pressurecreated in the right arm due to the presence of the block is therefore less than the excesspressure created in the left arm. The difference in pressures is ∆P = (Pleft − Pright),

∆P = Pleft − Pright = Pleft −1

4Pleft =

3

4Pleft

Before the system reaches equilibrium the pressure in the left arm is higher than thepressure in the right arm. The force per unit area pushing downwards on the left cap isgreater than the force per unit area pushing downwards on the right cap. This meansthat the left cap is pushed downwards, while the right cap rises slightly.

9

The difference in the two pressures causes a difference ∆h in the heights of the twocaps,

∆P = ρwater g ∆h

If we assume that the radius of the left arm is Rleft = 1.0 m, what is the difference∆h in the heights of the two caps?

∆h =∆P

ρwater g

∆h =Pleft − Pright

ρwater g=

34Pleft

ρwater g

∆h =1

ρwater g

(3

4Pleft

)=

1

ρwater g

(3

4

Fleft

Aleft

)

∆h =

(1

ρwater g

) (3 Mblock g

4 π R2left

)=

3 Mblock

4 π R2left ρwater

∆h =3 Mblock

4 π R2left ρwater

Inserting values for the variables,

∆h =3 · (1.00 kg)

4 π (1.00 m)2

(1000 kg

m3

) = 0.239 · 10−3 m

If the radius of the left arm is Rleft = 1.0 meters, the right cap will rise to a height of

∆h = 0.239 mm

above the left cap. (In other words, the difference in the heights of the two caps ismuch smaller than the radius of the left arm.)

2.3 Different Masses

If a 1.0 kilogram block is placed on the left cap, how much total mass must beplaced on the right cap so that the caps are at equal height when the systemis in equilibrium?

The mass placed on the left cap is

Mleft = 1.0 kg

10

The mass placed on the right block is unknown. The forces on the left and right capsdue to the respective blocks are no longer equal, because the masses are different.

Fleft = wleft = Mleft g

Fright = wright = Mright g

The excess pressure on the left cap due to the block is

Pleft =Fleft

Aleft

=Mleft g

Aleft

The excess pressure on the right cap is

Pright =Fright

Aright

=Mright g

Aright

=Mright g

4 Aleft

If the caps are to remain at equal height, the two pressures must be equal,

Pright = Pleft

Pright =Mright g

4 Aleft

= Pleft =Mleft g

Aleft

Simplifying this expression by removing Aleft and g gives

Mright

4= Mleft

Mright = 4 Mleft

If a mass Mleft = 1.0 kg is placed on the left cap, then a mass

Mright = 4 Mleft = 4.0 kg

must be placed on the right cap in order for the two caps to remain at equal heights.

2.4 Different Heights

The locations of the two caps at equilibrium are now as shown in figure 2.The dashed line represents the level of the water in the left arm. What is themass of the water in the right arm between the dashed line and the right cap?

The density of water is

ρwater =mwater

Vwater

11

so the mass of the water in the small volume shown is

mwater = ρwater V

The volume V is the surface area of the right cap multiplied by the difference ∆h inthe heights of the left and right caps,

V = Aright ∆h = (4 Aleft) ∆h

So the mass of the water is

mwater = ρwater V = ρwater · 4 Aleft ∆h

In order to evaluate this expression we need to find the difference in heights ∆h.

The excess pressure on the left arm is equal to the pressure exerted by the block,

Pleft =Fleft

Aleft

=Mblock g

Aleft

The excess pressure on the right arm is equal to the pressure exerted by the blockplus the pressure difference ∆P due to the height ∆h of the column of water.

Pright =Fright

Aright

+ ∆P

We know the equation for a pressure difference due to the height of a column of fluid,

∆P = ρwater g ∆h

So the excess pressure on the right arm is

Pright =Fright

Aright

+ ∆P =Mblock g

4 Aleft

+ ρwater g ∆h

Since the system is in equilibrium, the pressures on the right and left arms must beequal,

Pleft = Pright

Pleft =Mblock g

Aleft

= Pright =Mblock g

4 Aleft

+ ρwater g ∆h

Simplifying this expression,

ρwater g ∆h =Mblock g

Aleft

− Mblock g

4 Aleft

=3 Mblock g

4 Aleft

∆h =3 Mblock

4 Aleft ρwater

12

So the mass of the water is

mwater = ρwater · 4 Aleft ∆h = (ρwater · 4 Aleft)

(3 Mblock

4 Aleft ρwater

)= 3 Mblock

If the mass of the block placed on the cap is Mblock = 1.0 kg, the mass of the waterin the volume depicted is

Mwater = 3 Mblock = 3.0 kg

This makes sense. The total mass which must be supported on the left side isMblock = 1 kg. The total mass which must be supported on the right side is

(Mblock + mwater) = 4 Mblock

The area of the cap on the right side is four times the area of the cap on the left side,Aright = 4 Aleft. One can see that the pressures at the dotted line in Figure 2 must beequal,

Pleft =Fleft

Aleft

=Mblock g

Aleft

=4 Mblockg

4 Aleft

=Fright

Aright

= Pright

Equal pressures on the left and right sides of the u-tube is the condition for equilib-rium.

3 Problem 15.27

A vertical tube is open at the top. The tube contains oil of density ρoil = 0.82 g/cm3

floating on top of water. The height of the oil is hoil = 6.4 cm; the height ofthe water is hwater = 6.4 cm. Find the gauge pressure at the bottom of the tube.

The total pressure at the top of the oil is P1. This is equal to atmospheric pressurebecause the tube is open to the atmosphere.

The total pressure at the interface between the oil and the water is P2. The differ-ence ∆P = P2 − P1 between the pressure at the top of the oil and the pressure at thebottom of the oil is

∆P = P2 − P1 = ρoil g hoil

So the total pressure P2 at the interface between the oil and the water is

P2 = P1 + ρoil g hoil

The total pressure at the interface just above the column of water is P2. The totalpressure at the bottom of the column of water is P3. The difference ∆P ′ between thepressure at the top of the water and the pressure at the bottom of the water is

∆P ′ = P3 − P2 = ρwater g hwater

13

So the total pressure at the bottom of the column of water is

P3 = P2 + ρwater g hwater

We can insert the expression for the pressure P2 at the interface between water andoil to obtain an expression for the total pressure in terms of known quantities,

P3 = P1 + ρoil g hoil + ρwater g hwater

The gauge pressure at the bottom of the column of water is the difference betweenthe total pressure P3 and atmospheric pressure,

Pgauge = Ptotal − Patm

Since the tube is open at the top and P1 is equal to atmospheric pressure, the gaugepressure at the bottom of the column of water is

Pgauge = P3 − Patm = P3 − P1

so our expression for the gauge pressure is

Pgauge = ρoil g hoil + ρwater g hwater

Notice that hoil = hwater = 6.4 cm, so

Pgauge = (ρoil + ρwater) g h


Pgauge =((0.82 + 1.00)

g

cm3

) (9.8

m

s2

)(6.4 cm)

Converting from cubic centimeters to cubic meters,

(1 cm)3 = (10−2 m)3 = 10−6 m3

Pgauge =

(1.82

10−3 kg

10−6 m3

) (9.8

m

s2

)(6.4 · 10−2 m)

Pgauge = 114

(10−5

10−6

) (kg m

s2

) (1

m2

)

Finally, the gauge pressure at the bottom of the column of water is

Pgauge = 1140N

m2= 1140 Pa

14

4 A Water Tank on Mars

You are assigned the design of a cylindrical pressurized water tank for a futurecolony on Mars. The acceleration due to gravity on Mars is gmars = 3.71 m/s2.The pressure at the surface of the water will be Psurface = 130 kPa, and thedepth of the water will be hwater = 14.1 m. The pressure of the air in thebuilding outside the tank will be Pair = 91.0 kPa. The area of the bottom ofthe tank is A = 2.05 m2. Find the net downward force on the bottom of thetank due to the water inside the tank, the air inside the tank, and the airoutside the tank.

The pressure at the surface of the water in the tank is Psurface. What is the difference inpressure ∆P = Pbottom − Psurface between the surface of the water and the bottom of thetank?

∆P = ρwater gMars hwater

The difference in the two pressures is thus

∆P = Pbottom − Psurface = ρwater gmars hwater

The total pressure Pbottom at the bottom of the tank due to the water and vapor inthe tank is

Pbottom = Psurface + ∆P = Psurface + ρwater gmars hwater

Pressure is equal to force per unit area,

P =F

A

This means that the force on the bottom of the tank due to the water and vapor inthe tank is

Fbottom = Pbottom A

Fbottom = Pbottom A = (Psurface + ρwater gmars hwater) A

This force is directed from the inside of the tank toward the outside. However, theair outside the tank exerts a pressure Pair on the bottom of the tank. The force whichresults from the external air pressure is directed from the outside of the tank towards theinside.

Fair = Pair A

The net force on the bottom of the tank is

Fnet = Fbottom − Fair

15

Fnet = (Psurface + ρwater gmars hwater) A− Pair A

Fnet = (Psurface − Pair + ρwater gmars hwater) A


Fnet =

(130 kPa− 91.0 kPa +

(1000

kg

m3

) (3.71

m

s2

)(14.1 m)

) (2.05 m2

)

Fnet =

(39 kPa + 52311

kg m

s2

1

m2

) (2.05 m2

)

Fnet =(39000 Pa + 52311

N

m2

) (2.05 m2

)

Fnet =(39000

N

m2+ 52311

N

m2

) (2.05 m2

)

Fnet =(91311

N

m2

) (2.05 m2

)So the net force on the bottom of the water tank is

Fnet = 187187 N ≈ 187 kN

5 Problem 15.47

A garage lift has a piston with diameter D = 48 cm supporting the load. Com-pressed air with a maximum pressure of Pmax = 500 kPa is applied to a smallpiston at the other end of the hydraulic system. What is the maximum massthe lift can support?

Pressure is force per unit area,

P =F

A

The maximum force which the hydraulic system can support is then

Fmax = Pmax A

The area A of the piston of diameter D and circular cross section is

16

A = π r2 = π(

D

2

)2

=π D2

4

So the maximum force is

Fmax = Pmax A = Pmax ·π D2

4

This maximal force can support a maximum weight wmax = mmax g,

Fmax = wmax = mmax g

So the maximum mass mmax which the lift can support is

mmax =Fmax

g=

(Pmax

π D2

4

)g

=π D2 Pmax

4 g

Inserting numeric values,

mmax =π (48 cm)2 (500 kPa)

4 ·(9.8 m

s2

)and rearranging units

mmax =π (0.48 m)2 (500 · 103 Pa)

4 ·(9.8 m

s2

) = 9.23 · 103

(m2 Pa

s2

m

)

mmax = 9.23 · 103

(m2 N

m2

s2

m

)= 9.23 · 103

(m2

m2

kg m

s2

s2

m

)

The maximum mass which the hydraulic lift can support is

mmax = 9230 kg

6 A Submerged Ball

A ball of mass mball and volume Vball is lowered on a string into a fluid ofdensity ρfluid. Assume that the object would sink to the bottom if it werenot supported by the string. What is the tension T in the string when theball is fully submerged but not touching the bottom, as shown in the figure 3?

Three forces act on the ball. The tension in the string acts in the upwards direction.The buoyant force exerted on the ball by the fluid also acts in the upwards direction.The weight of the ball acts downwards. Since the ball is in static equilibrium, the sumof these forces is zero.

17

Figure 3: Submerged ball

T + FB − wball = 0

The tension in the string is thus

T = wball − FB

The weight wball of the ball is

wball = mball g

The density of the fluid is

ρfluid =mfluid

Vfluid

A volume of fluid Vfluid = Vball is displaced by the ball. The mass of the fluid displacedby the ball is then

mfluid = ρfluid Vfluid = ρfluid Vball

The weight of the fluid displaced by the ball is

wfluid = mfluid g = ρfluid Vball g

Archimedes’ Principle states that the buoyant force FB exerted on the ball by thefluid is equal to the weight of the fluid displaced by the ball,

FB = wfluid = ρfluid Vball g

So the tension T in the string is

T = wball − FB = mball g − ρfluid Vball g

T = (mball − ρfluid Vball) g

18

7 Problem 15.50

A glass beaker measures 10 cm high by 4.0 cm in diameter. Empty, it floatsin water with one-third of its height submerged. (a) How many 15 gram rockscan be placed in the beaker before it sinks?

7.0.1 Volume of beaker

What is the volume Vbeaker of the beaker?

Vbeaker = Ah

where h is the height of th beaker and A is the area of its base. We know the heightis h = 10 cm. The diameter of the beaker is D = 4.0 cm, so the radius r is

r = D/2 = 2.0 cm

The cross sectional area is then

A = π r2 = 12.6 cm2

The volume is

Vbeaker = Ah = (12.6 cm2) (10 cm) = 126 cm3

Vbeaker = Ah = 126 cm3

7.0.2 Volume of water

What volume of water is displaced by the beaker when it floats in water with one thirdof its height submerged?

Vwater =1

3Vbeaker

What volume of water can be displaced by the floating beaker and the added rocksbefore the beaker sinks?

V ′water =

2

3Vbeaker = 83.8 cm3

7.0.3 Mass of water

The density ρ of water is

ρ = 1000kg

m3= 1

g

cm3

What is the mass of the water which can be displaced by the beaker and the addedrocks before the beaker sinks?

19

Density is mass per unit volume,

ρ =m

V

So the mass m of the displaced water is

m = ρV ′water =

(1

g

cm3

)(83.8 cm3) = 83.8 g

mwater = 83.8 g

7.0.4 Mass II

Archimedes’ Principle states that the buoyant force on an object is equal to the weightof the fluid displaced by the object. The buoyant force supporting the rocks is thus

FB = wwater = mwater g

Each rock has mass m1 = 15 g. Each rock has weight

w1 = m1 g

How many rocks of weight w1 can be supported by the buoyant force FB? In otherwords,

FB = N w1

where N is the appropriate number of rocks, and we need to find N .

FB = wwater = mwaterg = N w1 = N m1g

mwaterg = N m1g

The acceleration g due to gravity cancels out.

mwater = N m1

N =mwater

m1

=83.8 g

15 g= 5.5

N = 5.5

The beaker can support 5 rocks before it sinks.

20

Figure 4: Submerged sphere in a beaker

8 Submerged Sphere in a Beaker

A cylindrical beaker of height h = 0.100 m and negligible weight is filled to thebrim with a fluid of density ρfluid = 890 kg/m3. When the beaker is placed ona scale, its weight is measured to be W1 = 1.00 N. (See figure 4). A ball ofdensity ρball = 5000 kg/m3 and volume Vball = 60.0 cm3 is then submerged in thefluid, so that some of the fluid spills over the side of the beaker. The ball isheld in place by a stiff rod of negligible volume and weight. Throughout theproblem, assume the acceleration due to gravity is g = 9.81 m/s2.

8.1 Weight

What is the mass mball of the ball?

Density is mass per unit volume,

ρ =m

V

So the mass mball is

mball = ρball Vball

mball =

(5000

kg

m3

)(60.0 cm3)

Changing cubic centimeters into cubic meters,

1 cm3 = 10−6 m3

mball = 5

(103 kg

m3

)60.0 (10−6 m3)

21

mball = (5 · 60.0)(103 · 10−6

)kg = 300 · 10−3 kg = 0.3 kg

mball = 0.3 kg

What is the weight wball of the ball?

wball = mball g

wball = (0.3 kg)(9.8

m

s2

)

wball = 2.94kg m

s2= 2.94 N

8.2 Buoyant Force

What is the buoyant force which acts on the ball?

What volume of fluid is displaced by the ball?

Vdisplaced = Vball = 60.0 cm3 = 60 · 10−6 m3

The density of the fluid is the mass per unit volume,

ρfluid =mdisplaced

Vdisplaced

So the mass of the volume Vdisplaced of fluid is

mdisplaced = ρfluid Vdisplaced

mdisplaced =

(890

kg

m3

) (60.0 · 10−6 m3

)

mdisplaced = 0.0534 kg

What is the weight of this volume of fluid?

wdisplaced = mdisplaced g = 0.524 N

What is the buoyant force which the fluid exerts on the ball?

Archimedes’ Principle states that the buoyant force on the ball is equal to the weightof the fluid displaced by the ball,

FB = wdisplaced = 0.524 N

22

8.3 Apparent Weight

What is the reading W2 of the scale when the ball is held in this submergedposition? Assume that none of the water which spills over stays on the scale.

The weight of the system is equal to the weight of the fluid remaining in the beakerplus the buoyant force which the fluid exerts on the ball.

W2 = wremaining + FB

What is the weight of the fluid which remains in the beaker?

wremaining = wfluid − wdisplaced

We are told that the initial weight of the fluid is

W1 = 1 N

so

wremaining = 1.00 N− 0.524 N = 0.476 N

So

W2 = wremaining + FB = 0.476 N + 0.524 N

W2 = 1.00 N

The initial weight of the beaker full of fluid was W1 = 1.00 N. A volume Vball of fluidwas removed, so the weight of the fluid remaining in the beaker decreased. However, thefluid remaining in the beaker exerted a buoyant force on the submerged sphere equal tothe weight of the displaced fluid. This means that the apparent weight W2 of the fluidplus the submerged sphere is equal to the original weight of the fluid.

8.4

What is the force Frod applied to the ball by the rod? Take upward forces tobe positive (e.g., if the force on the ball is downward, your answer should benegative).

The weight wball of the ball is directed downwards. The buoyant force FB is directedupwards. We will assume that the force applied by the rod is also directed upwards.Since the ball is stationary, the sum of these forces must be zero.

Frod + FB − wball = 0

Frod = wball − FB = 2.94 N− 0.524 N = 2.42 N

Frod = 2.42 N

The rod applies a force in the upward direction.

23

8.5 Apparent Weight

The rod is now shortened and attached to the bottom of the beaker. Thebeaker is again filled with fluid, the ball is submerged and attached to therod, and the beaker with fluid and submerged ball is placed on the scale.What weight W3 does the scale now show?

Figure 5: Sphere attached to base of beaker.

Just as before, the weight of the fluid which remains in the beaker is

wremaining = 0.476 N

Just as before, the buoyant force on the ball is equal to the weight of the fluid displacedby the ball,

FB = wdisplaced = ρfluid Vdisplaced g = 0.524 N

Just as before, sum of the force exerted by the rod, the buoyant force, and the weightof the ball is zero.

Frod + FB − wball = 0

Neither the buoyant force nor the weight of the submerged ball have changed, so

Frod = wball − FB = 2.94 N− 0.524 N = 2.42 N

The rod still exerts an upward force on the ball.

Frod = 2.42 N

The weight of the fluid in the beaker acts downwards on the scale. The fluid in thebeaker exerts an upwards buoyant force on the ball, and the ball exerts a downwards forceof equal magnitude on the fluid in the beaker. This downwards force is transferred tothe scale. Finally, the rod exerts an upwards force on the ball, which exerts a downwards

24

force of equal magnitude on the rod. This force is also transferred to the scale. So theapparent weight W3 is

W3 = wremaining + FB + Frod

W3 = 0.476 N + 0.524 N + 2.42 N = 3.42 N

The net downwards force on the scale is the apparent weight W3,

W3 = 3.42 N

9 Flow Velocity of Blood

Arteriosclerotic plaques forming on the inner walls of arteries can decreasethe effective cross-sectional area of an artery. Even small changes in the ef-fective area of an artery can lead to very large changes in the blood pressurein the artery and possibly to the collapse of the blood vessel.

A healthy artery has cross sectional area A0. In such an artery the blood flows withvelocity v0 = 0.14 m/s. The density of the blood is ρ = 1050 kg/m3.

9.1 Continuity Equation

Plaque has narrowed an artery to one-fifth of its normal cross-sectional area.(This represents an 80% blockage). Compared to normal blood flow velocityv0, what is the velocity of blood as it passes through this blockage?

The healthy artery has cross sectional area A0. The cross sectional area A1 of the blockedartery is one fifth of the area of the healthy artery,

A1 =1

5A0

So the cross sectional area of the healthy artery is five times that of the blocked artery,

A0 = 5 A1

We will assume that the density of the blood remains constant. The continuity equa-tion states that the product of the cross sectional area of the artery and the velocity ofthe blood flow is constant,

A0 v0 = A1 v1

The velocity of the blood flow in the blocked artery is then

v1 =A0

A1

v0 =5 A1

A1

v0 = 5 v0

The cross sectional area of the blocked artery is much smaller than the cross sectionalarea of the healthy artery; the blood flows much more quickly through the blocked arterythan it does through the healthy artery.

25

9.2 Energy per Unit Volume

The familiar formula for kinetic energy of an object of mass m is

E =1

2m v2

The kinetic energy per unit volume of a quantity of blood of mass m is

E =E

V

We can write this expression in terms of mass, volume, and velocity,

E =E

V=

12m v2

0

V

The density of the blood is the mass per unit volume,

ρ =m

V

We can thus rewrite the expression for the kinetic energy per unit volume in terms ofdensity and velocity,

E =E

V=

1

2

(m

V

)v2

0 =1

2ρ v2

0

If we insert the numeric values for a healthy artery, we obtain

E =1

2ρ v2

0 =1

2

(1050

kg

m3

) (0.14

m

s

)2

E = 10.29

(kg m

s2

) (1

m2

)= 10.29

N

m2= 10.29 Pa

By what factor does the kinetic energy per unit volume of blood changeas the blood passes through the blocked artery?

The blood in the healthy artery is the same as the blood in the blocked artery. Wewill therefore assume that the blood in the healthy artery and the blood in the blockedartery have the same density ρ.

The speed of the blood in the healthy artery is v0. The speed of the blood in theblocked artery is v1.

The kinetic energy per unit volume of the blood in the healthy artery is E0,

E0 =1

2ρ v2

0

The kinetic energy per unit volume of the blood in the blocked artery is E1,

26

E1 =1

2ρ v2

1

Since v1 = 5 v0, we can rewrite the expression for E1,

E1 =1

2ρ v2

1 =1

2ρ (5 v0)

2 = 25(

1

2ρ v2

0

)= 25 E0

The ratio R of the kinetic energy per unit volume of the blood in the blocked arteryto the kinetic energy per unit volume of the blood in the healthy artery is

R =E1

E0

This ratio is

R =E1

E0

=25 E0

E0

= 25

9.3 Bernoulli’s Equation

As the blood passes through this blockage, what happens to the blood pres-sure?

We are looking for the change ∆P in the blood pressure when it flows from the healthyartery into the blocked artery,

∆P = P1 − P0

Bernoulli’s equation will give us the relationship between the pressure, the density,and the velocity. Bernoulli’s equation can be used to relate the flow of the blood in thehealthy artery to the flow of the blood in the blocked artery,

P0 +1

2ρ v2

0 + ρ g h0 = P1 +1

2ρ v2

1 + ρ g h1

We can assume that the blocked and healthy arteries are at the same vertical position,

h0 = h1

so Bernoulli’s equation simplifies,

P0 +1

2ρ v2

0 = P1 +1

2ρ v2

1

We know that v1 = 5 v0, so

P0 +1

2ρ v2

0 = P1 +1

2ρ (5 v0)

2 = P1 + 25(

1

2ρ v2

0

)Rearranging this expression,

27

P1 − P0 =1

2ρ v2

0 − 25(

1

2ρ v2

0

)(P1 − P0) is just the change ∆P in the pressure, which is the quantity we need to

find.

∆P = (P1 − P0) =1

2ρ v2

0 − 25(

1

2ρ v2

0

)= −24

(1

2ρ v2

0

)so our final equation for the change in the blood pressure when the blood flows into

the blocked artery is

∆P = −12 ρ v20

Inserting numeric quantities,

∆P = −12

(1050

kg

m3

) (0.14

m

s

)2

= −247

(kg m

s2

) (1

m2

)= −247

N

m2= −247 Pa

∆P = −247 Pa

When the blood flows from the healthy artery into the blocked artery, the blood pres-sure decreases by 247 Pascals.

As the blood velocity increases through a blockage, the blood pressure in that section ofthe artery can drop to a dangerously low level. In extreme cases, the blood vessel cancollapse due to lack of sufficient internal pressure.

9.4

9.4.1 Velocity

The arterial plaque now blocks 90% of the artery. By what factor does thevelocity of blood increase as the blood passes through this blockage? Com-pare the blocked artery to the initial, healthy artery.

The healthy artery has cross sectional area A0. The blocked artery has cross sectionalarea A2. We are told that 90% of the cross section is blocked, so

A2 =1

10A0

and A0 = 10 A2. We found an expression previously for the velocity as a function ofthe cross sectional area,

v2 =A0

A2

v0 =10 A2

A2

v0 = 10 v0

The velocity of the blood increases by a factor of ten as it passes through the blockedportion of the artery,

v2 = 10 v0

28

9.4.2 Kinetic Energy

By what factor does the kinetic energy per unit volume of blood change asthe blood passes through the blocked artery?

The kinetic energy per unit volume of the blood in the healthy artery is E0,

E0 =1

2ρ v2

0

The kinetic energy per unit volume of the blood in the blocked artery is E2,

E2 =1

2ρ v2

2

Since v2 = 10 v0, we can rewrite the expression for E2,

E2 =1

2ρ v2

2 =1

2ρ (10 v0)

2 = 100(

1

2ρ v2

0

)= 100 E0

The ratio R of the kinetic energy per unit volume of the blood in the blocked arteryto the kinetic energy per unit volume of the blood in the healthy artery is

R =E2

E0

= 100

9.4.3 Pressure

What is the magnitude of the drop in blood pressure ∆P as the blood passesthrough the blocked artery?

We are looking for the change ∆P in the blood pressure when it flows from the healthyartery into the blocked artery,

∆P = P2 − P0

Bernoulli’s equation will give us the relationship between the pressure, the density,and the velocity,

P0 +1

2ρ v2

0 + ρ g h0 = P2 +1

2ρ v2

2 + ρ g h2

Because h0 = h2 Bernoulli’s equation simplifies,

P0 +1

2ρ v2

0 = P2 +1

2ρ v2

2

We know that v2 = 10 v0, so

P0 +1

2ρ v2

0 = P2 +1

2ρ (10 v0)

2 = P2 + 100(

1

2ρ v2

0

)Rearranging this expression,

29

∆P = P2 − P0 =1

2ρ v2

0 − 100(

1

2ρ v2

0

)

∆P = −99(

1

2ρ v2

0

)so our final equation for the change in the blood pressure when the blood flows into

the blocked artery is

∆P = −99

2ρ v2

0

Inserting numeric quantities,

∆P = −99

2

(1050

kg

m3

) (0.14

m

s

)2

= −1019kg m

s2

1

m2= −1019

N

m2= −1019 Pa

∆P = −1019 Pa

It is easy to imagine that such a drastic drop in pressure could catastrophically affectthe functionality of an artery.

10 Water Flowing from a Tank

Water flows steadily from an open tank as shown in figure 6. The elevation of point 1is 10.0 meters, and the elevation of points 2 and 3 is 2.00 meters. The cross-sectionalarea at point 2 is 0.0480 square meters; at point 3, where the water is discharged, it is0.0160 square meters. The cross-sectional area of the tank is very large compared withthe cross-sectional area of the pipe.

10.1 Discharge Rate

Find the discharge rate of water flowing from an open tank through a pipe ofvariable cross section.

The first question is, what quantity do we need to find?

The discharge rate R is the volume of water per unit time which flows out of thetank,

R =dV

dt

Consider a pipe of constant cross sectional area A and length x. The volume of waterin the pipe is the area times the length,

30

Figure 6: Water Flowing from a Tank

V = A x

If the cross section A is constant, then the volume of water which flows per unit timeis

dV

dt= A

dx

dt

But dx/dt is just the speed v with which the water flows,

dx

dt= v

So the discharge rate R is

R =dV

dt= A

dx

dt= A v

To calculate the discharge rate at point 3 where the water flows out of the pipe, weneed to know the cross sectional area A3 and we need to calculate the speed of the waterv3 when it exits the pipe.

10.1.1 Continuity Equation

For a liquid with a constant density ρ the continuity equation is

v1 A1 = v3 A3

This equation relates the speed with which the water level is falling at point 1, thecross sectional area A1 of the tank, the speed with which the water flows out of the tankat point 3, and the cross sectional area A3 of the pipe at point 3.

31

We know that the cross sectional area at point 3 is

A3 = 0.0160 m2

However, we need this information yet. All we need to know is that “the cross-sectionalarea of the tank is very large compared with the cross-sectional area of the pipe,” or

A1 >> A3

We will rewrite the continuity equation,

v1 = v3A3

A1

Since A1 >> A3 we see that

v1 ≈ 0

In other words, if you were watching the water flow out of the tank you would notnotice very much change in the water level at point 1. Although water is gushing out ofthe pipe at point 3, the water level at point 1 is not dropping very fast. This makes sensebecause it is a big tank, and only a very small fraction of the water in the tank is flowingout through the pipe.

10.1.2 Bernoulli’s Equation

Bernoulli’s equation relates the pressure, the speed, and the height of the water atpoint 1 to the pressure, the speed and the height at point 3.

P1 +1

2ρ v2

1 + ρgy1 = P3 +1

2ρ v2

3 + ρgy3

We can simplify this expression. First, recall that v1 ≈ 0.

P1 + ρgy1 = P3 +1

2ρ v2

3 + ρgy3

Next, recall that the tank is open. What does this imply about the pressure P1? Ifthe tank is open to the atmosphere, then the pressure P1 must be equal to atmosphericpressure.

What about the pressure at point 3? The pipe is open and the water is flowing out.This means that the pressure at point 3 is also equal to atmospheric pressure. (Thiswould not be true if we were discussing point 2! Point 2 is not open to the atmosphere.)So

P1 = P3

and the two pressures cancel,

32

ρgy1 =1

2ρ v2

3 + ρgy3

Recall that in order to solve for the volume flow rate dV/dt we need to find thespeed v3 with which the water flows out of the pipe. We can now solve for v3,

1

2ρ v2

3 = ρgy1 − ρgy3 = ρg (y1 − y3)

ρ v23 = 2 ρg (y1 − y3)

The density ρ of the water cancels,

v23 = 2g (y1 − y3)

So the speed v3 with which the water flows out of the pipe is

v3 =√

2g (y1 − y3)

We were told that y1 = 10.0 m and y3 = 2.0 m. If we use g = 9.8 m/s2 we get

v3 =

√2(9.8

m

s2

)(10.0 m− 2.0 m) =

√√√√156.8

(m2

s2

)

v3 = 12.5 m/s

10.1.3 Discharge Rate

We are not quite done. We need to calculate the discharge rate R, or the volume of waterwhich flows out of the pipe at point 3 per unit time. We noted previously that

R =dV

dt= A3 v3

So

R =dV

dt= (0.0160 m2)

(12.5

m

s

)and our final answer is

R =dV

dt= 0.20

m3

s

10.2 Gauge Pressure

What is the gauge pressure at point 2?

The gauge pressure at point 2 is just the difference between the total pressure P2

and atmospheric pressure,

Pgauge = P2 − Patm

33

10.2.1 Bernoulli’s Equation

Find the gauge pressure at point 2 in terms of v2, ρ, and the acceleration gdue to gravity.

Let us use Bernoulli’s equation to relate quantities at point 1 to quantities at point 2.Bernoulli’s equation is

P1 +1

2ρ v2

1 + ρgy1 = P2 +1

2ρ v2

2 + ρgy2

We know that v1 ≈ 0, so

P1 + ρgy1 = P2 +1

2ρ v2

2 + ρgy2

We can rearange this expression to solve for P2,

P2 = P1 + ρgy1 −1

2ρ v2

2 − ρgy2

P2 = P1 + ρg(y1 − y2)−1

2ρ v2

2

Since y1 = 10.0 m and y2 = 2.00 m,

y1 − y2 = 10.0 m− 2.00 m = 8.0 m

So the total pressure P2 at point 2 is

P2 = P1 + 8 ρg − 1

2ρ v2

2

To get the gauge pressure we simply subtract the atmospheric pressure

Pgauge = P2 − Patm

Since the tank is open, the pressure at point 1 is equal to atmospheric pressure,

P1 = Patm

So we can simply subtract P1. The gauge pressure in terms of v2, ρ, and the acceler-ation due to gravity g is thus

Pgauge = 8 ρg − 1

2ρ v2

2

(Remember that the constant 8 in this equation has units of meters.)

34


Find the fluid speed at point 2.

We will use the continuity equation to relate the cross sectional area A and the speed vat points 2 and 3.

v2 A2 = v3 A3

We know the cross sectional areas A2 = 0.0480 m2 and A3 = 0.0160 m2. We havealready calculated the speed v3 with which the water flows out at point 3. So we cansolve for the speed v2 of the water when it is flowing through the pipe at point 2,

v2 = v3A3

A2

v2 =(12.5

m

s

) (0.0160 m2

0.0480 m2

)

The speed of the water at point 2 is

v2 = 4.17m

s

10.2.3 Gauge Pressure

We calculated that the gauge pressure in terms of v2, ρ, and the acceleration due togravity g is

Pgauge = 8 ρg − 1

2ρ v2

2

The density ρ of water is

ρ = 1000kg

m3

We now have all of the quantities which we need to solve for the gauge pressure atpoint 2.

Pgauge = (8 m)

(1000

kg

m3

) (9.8

m

s2

)− 1

2

(1000

kg

m3

) (4.17

m

s

)2

Pgauge = 69705kg

m s2= 69705 Pa = 69.7 kPa

11 Problem 15.56

In figure 7 a horizontal pipe of cross-sectional area A is joined to a lower pipe of cross-sectional area A/2. The entire pipe is full of liquid with density ρ, and the left end is atatmospheric pressure Patm. A small open tube extends upward from the lower pipe.

35

Figure 7: Water flowing in a pipe

11.1 Hydrostatic Equilibrium

Find the height h2 of liquid in the small tube when the right end of the lowerpipe is closed, so the liquid is in hydrostatic equilibrium. Express your answerin terms of h1, v, g.

Let us call the left end of the pipe point 1 and the top of the vertical tube point 2.

Bernoulli’s equation states that

P1 +1

2ρ v2

1 + ρgh1 = P2 +1

2ρ v2

2 + ρgh2

How can we simplify this? First note that the left end of the pipe is at atmosphericpressure, P1 = Patm. The small vertical tube is also open, so P2 = Patm. Since

P1 = P2

the pressure cancels, and Bernoulli’s equation becomes

1

2ρ v2

1 + ρgh1 =1

2ρ v2

2 + ρgh2

This system is in hydrostatic equilibrium - the fluid is not flowing. In other words,the speed v of the fluid is zero,

v1 = v2 = 0

Bernoulli’s equation is now

ρgy1 = ρgy2

The density ρ of the fluid and the acceleration g due to gravity cancel, and we have

36

h1 = h2

When the system is in hydrostatic equilibrium the height of the fluid in the smallvertical pipe is equal to the height of the fluid in the large pipe. (You might have beenable to guess this without referring to Bernoulli’s equation at all.)

11.2

Find the height h2 of liquid in the small tube when the liquid flows with speedv1 in the upper pipe. Express your answer in terms of h1, v, g.

Let us call the point at the base of the slim vertical pipe point 3. We will also calla point close to the right-hand exit of the horizontal pipe point 4. The right hand sideof the horizontal pipe is no longer closed, so fluid will flow out of the pipe with speed v4.


First we will use the continuity equation to calculate an expression for the speed v4 ofthe fluid at point 4.

v1 A1 = v4 A4

v4 = v1A1

A4

We have been told that A4 = A1/2, so

v4 = v1A1(A1

2

) = 2 v1

v4 = 2 v1

The cross sectional area of the small pipe is half of the cross sectional area of thelarge pipe; the speed of the fluid in the small pipe is twice the speed of the fluid in thelarge pipe.

11.2.2 Bernoulli’s Equation I

Let us use Bernoulli’s equation to relate the various quantities at the top and the bottomof the slim vertical pipe (point 2 and point 3).

P2 +1

2ρ v2

2 + ρgh2 = P3 +1

2ρ v2

3 + ρgh3

We can simplify this. First we should note that no fluid is flowing at point 2 orpoint 3, so v2 = v3 = 0.

37

P2 + ρgh2 = P3 + ρgh3

We will define the height h4 of the narrow horizontal pipe to be zero. Then we willmake the approximation that h3 ≈ h4. In other words, the diameter of the horizontalpipe is relatively narrow, and point 3 is just inside the vertical pipe.

P2 + ρgh2 = P3

The approximation h3 ≈ h4 is a statement that there is not enough height differencebetween point 3 and point 4 to cause a change in pressure. This means that the pressuresat point 3 and point 4 are also very similar,

P3 ≈ P4

I will thus rewrite the equation in terms of the presure at point 4 rather than thepressure at point 3.

P2 + ρgh2 = P4

We now solve this equation for the height h2,

ρgh2 = (P4 − P2)

h2 =(P4 − P2)

ρg

Since the large pipe and the small vertical pipe are both open to the atmosphere, weknow that P1 = P2 = Patm. So finally

h2 =(P4 − P1)

ρg

11.2.3 Bernoulli’s Equation II

The problem with this expression for the height h2 of the fluid in the tube is that it isnot in terms of h1, v, g. We need to use Bernoulli’s equation one last time in order tofind the pressure P4 in terms of these desired variables.

We will use Bernoulli’s equation to relate the various quantities at point 1 and point 4,

P1 +1

2ρ v2

1 + ρgh1 = P4 +1

2ρ v2

4 + ρgh4

Recall that h4 = 0 and v4 = 2v1.

P1 +1

2ρ v2

1 + ρgh1 = P4 +1

2ρ (2v4)

2

38

P1 +1

2ρ v2

1 + ρgh1 = P4 +1

2ρ (4v2

1)

P1 +1

2ρ v2

1 + ρgh1 = P4 +4

2ρ v2

1

Now we solve for the pressure P4,

P4 = P1 +1

2ρ v2

1 + ρgh1 −4

2ρ v2

1

P4 = P1 + ρgh1 −3

2ρ v2

1

Now we take this expression for P4 and plug it into the equation for h2,

h2 =(P4 − P1)

ρg

h2 =

(P1 + ρgh1 − 3

2ρ v2

1 − P1

)ρg

The pressure P1 cancels,

h2 =

(ρgh1 − 3

2ρ v2

1

)ρg

The density ρ of the fluid cancels out and we can absorb the denominator g into thenumerator,

h2 = h1 −3

2gv2

1

We have thus found the height h2 of the fluid in the tube in terms of h1, v, and g.

h2 = h1 −3 v2

1

2g

39

6 b assignment 1

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