2.41 The system in Fig. P2.41 is at 20C. Determine the pressure at point A in pounds per
square foot.
Solution: Take the specific weights of water and mercury from Table 2.1. Write the
hydrostatic formula from point A to the water surface:
lbf
 6 
 10 
 5
p A  (0.85)(62.4 lbf/ft 3 )  ft   (846)    (62.4)    patm  (14.7)(144) 2
 12 
 12 
 12 
ft
2
Solve for p A  2770 lbf/ft
Ans.
Fig. P2.41
2.64 Gate ABC in Fig. P2.64 has a fixed hinge at B and is 2 m wide into the paper. If the water
level is high enough, the gate will open. Compute the depth h for which this happens.
Solution: Let H  (h  1 meter) be the depth down to the level AB. The forces on AB and BC
are shown in the freebody at right. The moments of these forces about B are equal when the gate
opens:
 MB  0   H(0.2)b(0.1)
or: H  0.346 m,
 H
 H
    (Hb)  
h  H  1  1.346 m Ans.
 2
 3
This solution is independent of both the water density and the gate width b into the paper.
Fig. P2.64
2.87 The bottle of champagne (SG  0.96) in Fig. P2.87 is under pressure as shown by the
mercury manometer reading. Compute the net force on the 2-in-radius hemispherical end cap at
the bottom of the bottle.
Solution: First, from the manometer, com-pute the gage pressure at section AA in the champagne 6 inches
above the bottom:
 2 
 4 
p AA  (0.96  62.4) 
ft   (13.56  62.4) 
ft  patmosphere  0 (gage),
 12 
 12 
or: PAA  272 lbf/ft 2 (gage)
Then the force on the bottom end cap is vertical only (due to symmetry) and equals the force at
section AA plus the weight of the champagne below AA:
F  FV  pAA(Area)AA  W6-in cylinder  W2-in hemisphere

 (272) (4/12)2  (0.96  62.4) (2/12)2 (6/12)  (0.96  62.4)(2 /3)(2/12)3
4
 23.74  2.61  0.58  25.8 lbf Ans.
Fig. P2.87
2.114 The uniform rod in the figure is hinged at B and in static equilibrium when 2 kg of lead
(SG  11.4) are attached at its end. What is the specific gravity of the rod material? What is
peculiar about the rest angle   30?
Solution: First compute buoyancies: Brod  9790(/4)(0.04)2(8)  98.42 N, and Wlead 
2(9.81)  19.62 N, Blead  19.62/11.4  1.72 N. Sum moments about B:
 MB  0  (SG  1)(98.42)(4cos30)  (19.62  1.72)(8cos30)  0
Solve for SGrod  0.636
Ans. (a)
The angle  drops out! The rod is neutrally stable for any tilt angle! Ans. (b)
2.129 The iceberg of Prob. 2.128 may become unstable if its width decreases. Suppose that the
height is L and the depth into the paper is L but the width decreases to H  L. Again with S 
0.88 for the iceberg, determine the ratio H/L for which the iceberg becomes unstable.
Solution: As in Prob. 2.128, the submerged distance h  SL  0.88L, with G at L/2 above the
bottom and B at h/2 above the bottom. From Eq. (2.52), the distance MB is
MB 
Io
sub
LH3 /12
H2
 L SL 


 MG  GB  MG   
 2 2 
HL(SL) 12SL
Then neutral stability occurs when MG  0, or
H2
L
 (1  S), or
12SL 2
H
 [6S(1  S)]1/2  [6(0.88)(1  0.88)]1/2  0.796
L
Ans.
2.143 The tank of water in Fig. P2.143 is full and open to the atmosphere (patm 
15 psi  2160 psf) at point A, as shown. For what acceleration ax, in ft/s2, will the pressure at
point
B
in
the
figure
be
(a) atmospheric; and (b) zero absolute (neglecting cavitation)?
Solution:
(a) For pA  pB, the imaginary ‘free surface isobar’ should join points A and B:
tan  AB  tan 45  1.0  ax /g, hence ax  g  32.2 ft/s2
Ans. (a)
(b) For pB  0, the free-surface isobar must tilt even more than 45, so that
pB  0  pA  gz  ax x  2160  1.94(32.2)(2)  1.94ax (2),
solve ax  589 ft/s2
Ans. (b)
This is a very high acceleration (18 g’s) and a very steep angle,   tan1(589/32.2)  87.
Fig. P2.143
P2.153 A cylindrical container, 14 inches in diameter, is used to make a mold for forming
salad bowls. The bowls are to be 8 inches deep. The cylinder is half-filled with molten plastic, 
= 1.6 kg/(m-s), rotated steadily about the central axis, then cooled while rotating. What is the
appropriate rotation rate, in r/min?
Solution: The molten plastic viscosity is a red herring, ignore. The appropriate final rotating
surface shape is a paraboloid of radius 7 inches and depth 8 inches. Thus, from Fig. 2.23,
2 R 2
2 (7 /12 ft ) 2
8
ft 

12
2g
2 (32.2 ft / s 2 )
rad 60
r
Solve for   11.2

 107
Ans.
s
2
min
1.83* A two-dimensional unsteady velocity field is given by u  x(1 2t), v  y. Find the
time-varying streamlines which pass through some reference point (x o,yo). Sketch some.
h  8 in 
Solution: Equation (1.44) applies with time as a parameter:
dx
dx
dy dy
1


 , or: ln(y) 
ln(x)  constant
u x(1  2t) v
y
1  2t
or: y  Cx1/(12t) , where C is a constant
In order for all streamlines to pass through y  yo at x  x o, the constant must be such that:
y  yo ( x/ xo )1/(1 2t)
Ans.
Some streamlines are plotted on the next page and are seen to be strongly time-varying.
P3.117
Water at 20C, in the pressurized
(3)
tank of Fig. P3.117, flows out and creates a
vertical jet as shown. Assuming steady
Air
H?
75 kPa-gage
(1)
water
frictionless flow, determine the height H
to which the jet rises.
85 cm
(2)
Fig. P3.117
Solution: This is a straightforward Bernoulli problem. Let the water surface
be (1), the exit plane be (2), and the top of the vertical jet be (3). Let z2 = 0 for convenience.
If we are clever, we can bypass (2) and write Bernoulli directly from (1) to (3):
p3
V2
p1
V2
 1  z1 
 3  z 3 , or :
g
2g
g
2g
75000
 0 _ 0.85m  0  0  H
(9.81)(998)
Solve
H  7.66 m  0.85 m  8.51 m
Ans.
If we took an intermediate step from (1) to (2), we would find V22/2g = 8.51 m, and then going from
(2) to (3) would convert the velocity head into pure elevation, because V3 = 0.
P4.74
Consider the two-dimensional incompressible polar-coordinate velocity potential
  B r cos   B L 
where B is a constant and L is a constant length scale. (a) What are the dimensions of B?
(b) Locate the only stagnation point in this flow field. (c) Prove that a stream function exists and then
find the function (r, ).
Solution: (a) To give  its correct dimensions of {L2/T}, the constant B must have the dimensions of
velocity, or {L/T}. Ans.(a)
(b) Calculate velocities in polar coordinates:
vr 

 B cos 
r
;
v 
1 
BL
  B sin  
r 
r
At first it doesn’t look as if we can find a stagnation point, but indeed there is one:
r  L ,   90 : v  0 , vr  0
Ans.(b)
As discussed later in Chap. 8, this is the velocity potential of a Rankine half-body.
(c) With the velocities known, check the continuity equation:
B cos  B cos 
1 
1 v
(r v r ) 
 0 

 0
r r
r 
r
r
Yes, satisfied
Continuity is satisfied. Find the stream function from the definition of (r, ):
vr 
1 
 B cos 
r 
Integrate :

BL
  B sin  
r
r
  B r sin   B L ln r  const
;
v  
Ans.(c)
C2.3 Professor F. Dynamics, riding the merry-go-round with his son, has brought along his Utube manometer. (You never know when a manometer might come in handy.) As shown in Fig.
C2.3, the merry-go-round spins at constant angular velocity and the manometer legs are 7 cm apart.
The manometer center is 5.8 m from the axis of rotation. Determine the height difference h in
two ways: (a) approximately, by assuming rigid body translation with a equal to the average
manometer acceleration; and (b) exactly, using rigid-body rotation theory. How good is the
approximation?
Solution: (a) Approximate: The average acceleration of the manometer is R avg2 
5.8[6(2 /60)]2  2.29 rad/s toward the center of rotation, as shown. Then
tan( )  a/g  2.29/9.81  h/(7 cm)  0.233
Solve for h  1.63 cm
Ans. (a)
(b) Exact: The isobar in the figure at right would be on the parabola z  C  r22/(2g), where C
is a constant. Apply this to the left leg (z1) and right leg (z2). As above, the rotation rate is  
6.0*(2 /60)  0.6283 rad/s. Then
h  z2  z1 
2 2 2
(0.6283)2
(r2  r1 ) 
[(5.8  0.035)2  (5.8  0.035)2 ]
2g
2(9.81)
 0.0163 m Ans. (b)
This is nearly identical to the approximate answer (a), because R >> r.