Fluid MechanicsFluid Mechanics SOLUTIONS TO PROBLEMS Section 15.1 Pressure P15.2 Let 4 F g be its weight. Then each tire supports F g, so P= F A = F g 4A yielding F g =4AP=4(0.0240

Chapter 15 1 Fluid Mechanics

SOLUTIONS TO PROBLEMS Section 15.1 Pressure

P15.2 Let Fg be its weight. Then each tire supports

Fg

4,

so P =

FA=

Fg

4A

yielding Fg = 4AP = 4 0.024 0 m2( ) 200 ! 103 N m2( ) = 1.92 ! 104 N

P15.3

P =FA=

50.0 9.80( )! 0.500 " 10#2( )2

= 6.24 " 106 N m2

P15.4 The Earth’s surface area is 4!R2 . The force pushing inward over this area amounts to

F = P0A = P0 4!R2( ) . This force is the weight of the air:

Fg = mg = P0 4!R2( ) so the mass of the air is

m =

P0 4!R2( )g

=1.013 " 105 N m2( ) 4! 6.37 " 106 m( )2#

$%&'(

9.80 m s2 = 5.27 " 1018 kg .

Section 15.2 Variation of Pressure with Depth P15.10 The pressure on the bottom due to the water is Pb = ! gz = 1.96 " 104 Pa

So, Fb = PbA = 5.88 ! 106 N

On each end, F = PA = 9.80 ! 103 Pa 20.0 m2( ) = 196 kN

On the side, F = PA = 9.80 ! 103 Pa 60.0 m2( ) = 588 kN

2 Fluid Mechanics

*P15.11 The fluid in the hydraulic jack is originally exerting the same pressure as the air outside. This pressure P0 results in zero net force on either piston. For the equilibrium of piston 2 we require

500 lb = P ! P0( )A = P ! P0( )" 1.5 in.

2#$%

&'(

2

Let F1 represent the force the lever bar exerts on piston 1. Then similarly

F1 = P ! P0( )" 0.25 in.

2#$%

&'(

2

We ignore the weights of the pistons, sliding friction, and the slight difference in fluid pressure P

due to the height difference between points 1 and 2. By division

F1500 lb

=0.25 in.1.5 in.

!"#

$%&

2

F1 =500 lb

36

We say the hydraulic lift has an ideal mechanical advantage of 36. Next for the lever bar we ignore

weight and friction, assume equilibrium, and take torques about the fixed hinge.

!" = 0 F1 2 in.( ) ! F 12 in.( ) = 0 F =

F16

The lever has an ideal mechanical advantage of 6. By substitution,

F =

500 lb36 !6

= 2.31 lb

Section 15.3 Pressure Measurements P15.12 (a) We imagine the superhero to produce a perfect vacuum in the straw. Take point 1 at the

water surface in the basin and point 2 at the water surface in the straw:

P1 + ! gy1 = P2 + ! gy2

1.013 ! 105 N m2 + 0 = 0 + 1 000 kg m3( ) 9.80 m s2( ) y2 y2 = 10.3 m (b) No atmosphere can lift the water in the straw through

zero height difference.

P15.15 !P0 = "g!h = #2.66 $ 103 Pa : P = P0 + !P0 = 1.013 " 0.026 6( ) # 105 Pa = 0.986 # 105 Pa

Chapter 15 3

Section 15.4 Buoyant Forces and Archimedes’s Principle P15.18 Fg = m + !sV( ) g must be equal to Fb = !wVg

Since V = Ah , m + !sAh = !wAh

and A =

m!w " !s( )h

FIG. P15.18

P15.19 At equilibrium F! = 0 or Fapp + mg = B where B is the buoyant force.

The applied force, Fapp = B ! mg

where B = Vol !water( ) g

and m = Vol( )!ball .

So, Fapp = Vol( ) g !water " !ball( ) = 4

3#r3 g !water " !ball( )

FIG. P15.19

Fapp =

43! 1.90 " 10#2 m( )3 9.80 m s2( ) 103 kg m3 # 84.0 kg m3( ) = 0.258 N

P15.21 (a) P = P0 + ! gh

Taking P0 = 1.013 ! 105 N m2 and h = 5.00 cm

we find Ptop = 1.017 9 ! 105 N m2

For h = 17.0 cm , we get Pbot = 1.029 7 ! 105 N m2

Since the areas of the top and bottom are A = 0.100 m( )2 = 10!2 m2 we find

Ftop = PtopA = 1.017 9 ! 103 N

and Fbot = 1.029 7 ! 103 N

FIG. P15.21

(b) T + B ! Mg = 0

where B = !wVg = 103 kg m3( ) 1.20 " 10#3 m3( ) 9.80 m s2( ) = 11.8 N

and Mg = 10.0 9.80( ) = 98.0 N

Therefore, T = Mg ! B = 98.0 ! 11.8 = 86.2 N

(c) Fbot ! Ftop = 1.029 7 ! 1.017 9( ) " 103 N = 11.8 N

which is equal to B found in part (b).

4 Fluid Mechanics

P15.23 (a) According to Archimedes, B = !waterVwater g = 1.00 g cm3( ) 20.0 " 20.0 " 20.0 # h( )[ ] g

But B = Weight of block = mg = !woodVwood g = 0.650 g cm3( ) 20.0 cm( )3 g

0.650 20.0( )3 g = 1.00 20.0( ) 20.0( ) 20.0 ! h( ) g

20.0 ! h = 20.0 0.650( ) so h = 20.0 1 ! 0.650( ) = 7.00 cm (b) B = Fg + Mg where M = mass of lead

1.00 20.0( )3 g = 0.650 20.0( )3 g + MgM = 1.00 ! 0.650( ) 20.0( )3 = 0.350 20.0( )3 = 2 800 g = 2.80 kg

P15.25 The balloon stops rising when !air " !He( ) gV = Mg and !air " !He( )V = M ,

Therefore, V =

M!air " !He

=400

1.25e"1 " 0.180 V = 1 430 m3

Section 15.7 Bernoulli’s Equation P15.30 Volume flow rate = A1v1 = A2v2

20.0 L60.0 s

1 000 cm3

1 L!

"#$

%&= ' 1.00 cm( )2 vhose = ' 0.500 cm( )2 vnozzle

(a) vhose =

333 cm3 s3.14 cm2 = 106 cm s

(b) vnozzle =

333 cm3 s0.785 cm2 = 424 cm s

P15.33 Apply Bernoulli’s equation between the top surface and the

exiting stream.

P0 + 0 + ! gh = P0 +

12! vx

2 + ! gh

vx2 = 2g h0 ! h( ) ! vx = 2g h0 " h( )

x = vxt : y = h = 1

2g t2

! t = 2h

g

and x = vx

2hg= 2g h0 ! h( ) 2h

g: x = 2 h h0 ! h( )

FIG. P15.33

Chapter 15 5 *P15.34 Assuming the top is open to the atmosphere, then

P1 = P0 .

Note P2 = P0 .

Flow rate = 2.50 ! 10"3 m3 min = 4.17 ! 10"5 m3 s . (a) A1 >> A2 so v1 << v2

Assuming v1 = 0 ,

P1 +! v1

2

2+ ! gy1 = P2 +

! v22

2+ ! gy2

v2 = 2gy1( )1 2 = 2 9.80( ) 16.0( )[ ]1 2 = 17.7 m s

(b) Flow rate = A2v2 =

! d2

4"

#$%

&'17.7( ) = 4.17 ( 10)5 m3 s

d = 1.73 ! 10"3 m = 1.73 mm