15–114.
30
60
45
The fire boat discharges two streams of seawater, each at a
flow of and with a nozzle velocity of 50 ms.
needed to secure the boat. The density of seawater is
.rsw =1020 kg>m3
>0.25 m3>s
SOLUTION
lected from the larger reservoir (the sea), the velocity of the sea water entering the
control volume can be considered zero. By referring to the free-body diagram of the
control volume (the boat),
Ans. T =40 114.87 N =40.1 kN
T cos 60° =225(50 cos 30°) +225(50 cos 45°)
©Fx=dmA
dt
A
vA
B
x+dmB
dt
A
vB
B
x;;
+
Determine the tension developed in the anchor chain
15–115.
The chute is used to divert the flow of water,.
If the water has a cross-sectional area of , determine
the force components at the pin Dand roller Cnecessary
for equilibrium. Neglect the weight of the chute and weight
of the water on the chute..rw=1 Mg>m3
0.05 m2
Q
=0.6 m3
>
s
0.12 m
C
A
*15–116.
SOLUTION
dna,nehT.
. Applying Eq. 15–26, we have
Equation of Motion :
Ans.:
+©Fx=max;105.64 =200aa=0.528 m>s2
©Fx=dm
dt (vBx–vAx);
–F=7.546(–14 –0)
F=105.64 N
=1.22(6.185) =7.546 kg>s
dm
dt =raQQ =vBA=14cp
4
A
0.752
B
d=6.185 m3>s=14 m>s
The 200-kg boat is powered by the fan which develops a
slipstream having a diameter of 0.75 m. If the fan ejects air
with a speed of 14 m/s, measured relative to the boat,
determine the initial acceleration of the boat if it is initially at
rest.Assume that air has a constant density of
and that the entering air is essentially at rest. Neglect the drag
resistance of the water.
ra=1.22 kg/m3
0.75 m
597
15–117.
The nozzle discharges water at a constant rate of 2ft3
>
s. The
cross-sectional area of the nozzle at A is 4 in2, and at B the
cross-sectional area is 12 in2. If the static gauge pressure due
to the water at B is 2 lb
>
in2, determine the magnitude of
force which must be applied by the coupling at B to hold the
nozzle in place. Neglect the weight of the nozzle and the
water within it.
g
w
62.4 lb
>
ft3.
B
SOLUTION
dm
dt
=pQ =
a62.4
32.2
b
(2) =3.876 slug
>
s
(vBx)=
Q
A
B
=
2
12
>
144 =24 ft
>
s
(v
By
)=0
(vAy)=
Q
A
A
=
2
4
>
144 =72 ft
>
s
(vAx)=0
FB=pB AB=2(12) =24 lb
Equations of steady ow:
S
+
ΣFx=
dm
dt
(vAx –vBx);
24 –Fx=3.876(0 –24)
Fx=117.01 lb
+
c
ΣFy=
dm
dt
(vAy –vBy);
F
y
=3.876(72 –0) =279.06 lb
F=
2
Fx
2+Fy
2=
2
117.01
2+279.062=303 lb Ans.
Ans:
F=303 lb
SOLUTION
p
15–118.
The blade divides the jet of water having a diameter of 4 in.
If one-half of the water flows to the right while the other
half flows to the left, and the total flow is Q
=
1.5 ft3
>
s,
determine the vertical force exerted on the blade by the jet,
gv
=
62.4 lb
>
ft3.4 in.
15–119.
SOLUTION
. Also,.
Applying Eq. 15–25 we have
Ans.
Ans. Fy=4.93 lb
©Fy=©
dm
dt
A
youty–yiny
B
; Fy=3
4 (0.9689)(10.19) +1
4 (0.9689)(–10.19)
©Fx=©
dm
dt
A
youts–yins
B
;–Fx=0–0.9689 (10.19)
Fx=9.87 lb
dm
dt =rw Q=62.4
32.2 (0.5) =0.9689 slug>sy=Q
A=0.5
p
4
A
3
12
B
2=10.19 ft>s
The blade divides the jet of water having a diameter of 3 in.
If one-fourth of the water flows downward while the other
three-fourths flows upwards, and the total flow is
, determine the horizontal and vertical
components of force exerted on the blade by the jet,
.gw=62.4 lb>ft3
Q=0.5 ft3>s3 in.
SOLUTION
*15–120.
The gauge pressure of water at A is 150.5 kPa. Water flows
through the pipe at A with a velocity of 18 m
>
s, and out the
pipe at B and C with the same velocity v. Determine the
horizontal and vertical components of forceexerted on the
elbow necessary to hold the pipe assembly inequilibrium.
Neglect the weight of water withinthe pipeand the weight
of the pipe. The pipe has a diameter of50mm at A, and at B
and C the diameter is 30mm. rw
=
1000 kg
>
m
3
.
B
A
C
4
53
18 m/
s
v
v
15–121.
B
A
45
3
vB 25 ft/s
vA 12 ft/s
SOLUTION
Ans.
Ans.Fy=1.9559 =1.96 lb
Fy=0.06606(25) +0.03171a4
5b(12) – 0
+c©Fy=dmB
dt vBy+dmA
dt vAy–dm C
dt vCy
Fx=19.5 lb
40(p)(0.375)2–Fx=0–0.03171(12)a3
5b–0.09777(16.44)
:
+©Fx=dmB
dt vB s +dmA
dt vAs–dm C
dt vCs
vC=16.44 ft>s
vC(p)a0.375
12 b2
=12(p)a0.25
12 b2
+25(p)a0.25
12 b2
vC AC=vA AA+vB AB
dm C
dt =0.03171 +0.06606 =0.09777 slug>s
dmB
dt =62.4
32.2 (25)(p)a0.25
12 b2
=0.06606 slug>s
dmA
dt =62.4
32.2 (12)(p)a0.25
12 b2
=0.03171 slug>s
>2
flows out of the pipe at A and B with velocities vA = 12 ft>s
and vB = 25 ft>s, determine the horizontal and vertical
components of force exerted on the elbow necessary
weight of water within the pipe and the weight of the pipe.
The pipe has a diameter of 0.75 in. at C, and at A and B the
diameter is 0.5 in. gw = 62.4 lb>ft3.
The gauge pressure of water at C is 40 lb in . If water
to hold the pipe assembly in equilibrium. Neglect the
602
SOLUTION
(S
+)
s=s0+v0t
20 =0+vA cos 30°t
(+
c
)
v
=
v
0+act
–(vA sin 30°)=(vA sin 30°)–32.2t
Solving,
t=0.8469 s
vA=vB=27.27 ft>s
At B:
dm
dt
=rvA=
a62.4
32.2 b
(27.27)
a2
144 b
=0.7340 slug
>
s
+ RΣF=
dm
dt
(vA–vB)
–F=0.7340(0 –27.27)
F=20.0 lb
Ans.
15–122.
The fountain shoots water in the direction shown. If the
water is discharged at 30° from the horizontal, and the
cross-sectional area of the water stream is approximately
2 in2, determine the force it exerts on the concrete wall at B.
g
w
=
62.4 lb
>
ft3.
v
A
A
B
3 ft
20 ft
30
Ans:
F=20.0 lb
603
15–123.
SOLUTION
Ans.F=22.4 lb
F=0+(12 –0)a10(6)
32.2 b
©Fx=m
dv
dt +vD>t
dm t
dt
A plow located on the front of a locomotive scoops up snow
at the rate of and stores it in the train. If the
locomotive is traveling at a constant speed of 12 ,
determine the resistance to motion caused by the shoveling.
The specific weight of snow is .gs=6 lb>ft3
ft>s
10 ft3>s
Ans:
F=22.4 lb
604
*15–124.
The boat has a mass of 180 kg and is traveling forward on a
river with a constant velocity of 70 , measured relative
to the river.The river is flowing in the opposite direction at
5 . If a tube is placed in the water, as shown, and it
collects 40 kg of water in the boat in 80 s, determine the
horizontal thrust Ton the tube that is required to overcome
the resistance due to the water collection and yet maintain
the constant speed of the boat. .rw=1 Mg>m3
km>h
km>h
SOLUTION
Ans.T=0+19.444(0.5) =9.72 N
©Fs=m
dv
dt +vD>i
dm i
dt
vD>t=(70)a1000
3600 b=19.444 m>s
dm
dt =40
80 =0.5 kg>s
TvR 5 km/h
Ans:
T=9.72 N
605
15–125.
45
AB
Water is discharged from a nozzle with a velocity of
and strikes the blade mounted on the 20-kg cart. Determine
the tension developed in the cord, needed to hold the cart
stationary, and the normal reaction of the wheels on the
cart. The nozzle has a diameter of 50 mm and the density of
water is rw=1000 kg>m.
3
12 m>s
SOLUTION
Referring to the free-body diagram of the control volume shown in Fig. a,
;
;
Equilibrium: Using the results of and and referring to the free-body diagram
of the cart shown in Fig. b,
;Ans.
;Ans.N=396 NN–20(9.81) –199.93 =0+c©Fy=0
T=82.8 N82.81 –T=0:
+©Fx=0
Fy
Fx
Fy=199.93 N
Fy=7.5p(12 sin 45° –0)+c©Fy=dm
dt [(vB)y–(vA)y]
Fx=82.81 N
–Fx=7.5p(12 cos 45° –12)
:
+©Fx=dm
dt [(vB)x–(vA)x]
Ans:
T=82.8 N
N=396 N
606
SOLUTION
vs=
dm
dt
a1
rA
rb
=
a6.24
104(0.03)
b
=2.0 m
>
s
ΣFx=
dm
dt
(vT2–vS2)
–F=6.24(–2 cos 60°–0)
F=6.24 N
Ans.
ΣFy=
dm
dt
(vT2–vS2)
–P=6.24(0 –0.5)
P=3.12 N
Ans.
15–126.
A snowblower having a scoop S with a cross-sectional area of
As
=
0.12 m3 is pushed into snow with a speed of vs
=
0.5 m
>
s.
The machine discharges the snow through a tube T that has a
cross-sectional area of AT
=
0.03 m2 and is directed 60° from
the horizontal. If the density of snow is rs
=
104 kg
>
m
3
,
determine the horizontal force P required to push the blower
forward, and the resultant frictional force F of the wheels on
the ground, necessary to prevent the blower from moving
sideways. The wheels rollfreely.
P
T
60
Ans:
F=6.24 N
P=3.12 N
607
15–127.
SOLUTION
. Also,.
Applying Eq. 15–26 we have
a
Ans. d=2.56 ft
+©MO=dm
dt (dOB yB–dOA yA
B
;
30a0.5 +d
2b=0.2360 [4(56.59) –0]
dm
dt =ra Q=0.076
32.2 (100) =0.2360 slug>sy=Q
A=100
p
4 (1.52)=56.59 ft>s
The fan blows air at . If the fan has a weight of
30 lb and a center of gravity at G, determine the smallest
diameter dof its base so that it will not tip over.The specific
weight of air is .g=0.076 lb>ft3
6000 ft
3>
min
0.5 ft
1.5 ft G
Ans:
d=2.56 ft
608
*15–128.
The nozzle has a diameter of 40 mm. If it discharges water
uniformly with a downward velocity of 20m
>
s against the
fixed blade, determine the vertical force exerted by the
water on the blade. rw
=
1 Mg
>
m
3
.
SOLUTION
dt
=rvA=(1000)(20)(p)(0.02)2=25.13 kg
>
s
+
c
ΣFy=
dm
dt
(
vB[placeholder] –vAy
)
F=(25.13)(20 sin 45°–(–20))
F=858 N
Ans.
40 mm
Ans:
F=858 N
15–129.
A
B
650 mm
600 mm
250 mm
30
The water flow enters below the hydrant at Cat the rate of
.It is then divided equally between the two out-
lets at Aand B.If the gauge pressure at Cis 300 kPa, deter-
mine the horizontal and vertical force reactions and the
moment reaction on the fixed support at C.The diameter
of the two outlets at Aand Bis 75 mm, and the diameter of
the inlet pipe at Cis 150 mm. The density of water is
.Neglect the mass of the contained water
and the hydrant.
rw=1000 kg>m3
0.75 m3>s
SOLUTION
610
15–130.
Sand drops onto the 2-Mg empty rail car at 50 kg
>
s from a
conveyor belt. If the car is initially coasting at 4 m
>
s,
determine the speed of the car as a function of time.
2000 +50
t
4 m/s
2000 +50tf
15–131.
Sand is discharged from the silo at Aat a rate of with
a vertical velocity of onto the conveyor belt, which is
moving with a constant velocity of .If the conveyor
system and the sand on it have a total mass of and
center of mass at point G,determine the horizontal and ver-
tical components of reaction at the pin support Broller sup-
port A.Neglect the thickness of the conveyor.
750 kg
1.5 m>s
10 m>s
50 kg>s
volume shown in Fig.a,
;
Ans.
Writing the force steady flow equation along the xand yaxes,
;
Ans.
;
Ans.By=3716.25 N =3.72 kN c
=50[1.5 sin 30° –(–10)]
By+4178.5 –750(9.81)+c©Fy=dm
dt [(vB)y–(vA)y]
Bx=|–64.95 N| =65.0 N :
–Bx=50(1.5 cos 30° –0):
+©Fx=dm
dt [(vB)x–(vA)x]
Ay=4178.5 N =4.18 kN
750(9.81)(4) –Ay(8) =50[0 –8(5)]+©MB=dm
dt (dvB–dvA)
1.5 m/s
30
A
G
B