*15–132.
Sand is deposited from a chute onto a conveyor belt which
is moving at 0.5 m
>
s. If the sand is assumed to fall vertically
onto the belt at A at the rate of 4 kg
>
s, determine the belt
tension FB to the right of A. The belt is free to move over
the conveyor rollers and its tension to the left of A is
FC=400 N
.
SOLUTION
dm
0.5 m/s
F
B
FC = 400 N A
15–133.
The tractor together with the empty tank has a total mass
of 4 Mg.The tank is filled with 2 Mg of water.The water is
discharged at a constant rate of with a constant
velocity of ,measured relative to the tractor.If the
tractor starts from rest, and the rear wheels provide a
resultant traction force of 250 N, determine the velocity
and acceleration of the tractor at the instant the tank
becomes empty.
5 m>s
50 kg>s
SOLUTION
(1)
The time taken to empty the tank is .Substituting the result of t
into Eq. (1),
Ans.
Integrating Eq. (1),
Ans.
=
4.05 m
>
s
v=-10 ln
A
120 –t
B
20
40 s
Lv
0
dv=L40 s
0
10
120 – tdt
a=10
120 – 40 =0.125m>s2
t=2000
50 =40 s
a=dv
dt =10
120 – t
;
+©Fs=m
dv
dt –vD>e
dm e
dt ;
250 =
A
6000 –50t
B
dv
dt –5(50)
F
15–134.
A rocket has an empty weight of 500 lb and carries 300 lb of
fuel. If the fuel is burned at the rate of 15 lb
>
s and ejected
with a relative velocity of 4400 ft
>
s, determine the maximum
speed attained by the rocket starting from rest. Neglect the
effect of gravitation on the rocket.
SOLUTION
d
v
dm
e
v
15–135.
SOLUTION
Ans.pressure =452 Pa
pressure =(0.35) –15(9.81) =1.83(0 –(–6))
+c©F
y=dm
dt ((vB)y–(vA)y)
dm
dt =rAAvA=1.22(0.25)(6) =1.83 kg>s
Apower lawn mower hovers very close over the ground.
This is done by drawing air in at aspeed of through
an intake unit A,which has across-sectional area of
and then discharging it at the ground, B,
where the cross-sectional area is If air at A
is subjected only to atmospheric pressure,determine the
air pressure which the lawn mower exerts on the ground
when the weight of the mower is freely supported and no
load is placed on the handle.The mower has amass of
15 kg with center of mass at G.Assume that air has a
constant density of ra=1.22 kg>m3.
AB=0.35 m2.
AA=0.25 m2,
6m>s
A
v
A
B
G
*15–136.
The rocket car has a mass of 2 Mg (empty) and carries 120 kg
of fuel. If the fuel is consumed at a constant rate of 6 kg
>
s
and ejected from the car with a relative velocity of 800 m
>
s,
determine the maximum speed attained by the car starting
from rest. The drag resistance due to the atmosphere is
FD
=
(
6.8v
2
)
N, where v is the speed in m
>
s.
SOLUTION
dv
v
15–137.
If the chain is lowered at a constant speed
v=
4 ft
>
s,
determine the normal reaction exerted on the floor as a
function of time. The chain has a weight of 5 lb
>
ft and a
total length of 20 ft.
dm
i
dt
=mv
ΣFs=m
d
v
dt
+vD
>
i
dm
i
dt
R–mg(vt)=0+v(mv)
R
=
m(gvt
+
v
2
)
R=
5
32.2
(
32.2(4)(t)+(4)2
)
R=(20t+2.48) lb
Ans.
v 4 ft/s
15–138.
The second stage of a two-stage rocket weighs 2000 lb
(empty) and is launched from the first stage with a velocity
of The fuel in the second stage weighs 1000 lb.If
it is consumed at the rate of and ejected with a
relative velocity of determine the acceleration of
the second stage just after the engine is fired. What is the
rocket’s acceleration just before all the fuel is consumed?
Neglect the effect of gravitation.
8000 ft>s,
50 lb>s
3000 mi>h.
SOLUTION
619
15–139.
The missile weighs 40 000 lb.The constant thrust provided
by the turbojet engine is Additional thrust is
provided by two rocket boosters B.The propellant in each
booster is burned at a constant rate of 150 lb/s, with a
relative exhaust velocity of If the mass of the
propellant lost by the turbojet engine can be neglected,
determine the velocity of the missile after the 4-s burn time
of the boosters.The initial velocity of the missile is 300 mi/h.
3000 ft>s.
T=15 000 lb.
SOLUTION
At a time t,,where .
(1)
,,,,ereH
,.
Substitute the numerical values into Eq. (1):
Ans.vmax =580 ft s
vmax =a15 000 +9.3168(3000)
9.3168 b1na1242.24
1242.24 –9.3168(4) b+440
v0=300(5280)
3600 =440 ft>st=4s
vD>e=3000 ft>sc=2a150
32.2 b=9.3168 slug>sm0=40 000
32.2 =1242.24 slug
v=aT+cvD>e
cb1nam0
m0–ct b+y0
Lv
v0
dv =Lt
0aT+cvD>e
m0–ct bdt
T=(m0–ct)dv
dt –vD>ec
c=dme
dt
m=m0–ct
:
+©Fs=mdv
dt –vD>e
dme
dt
B
T
Ans:
vmax =580
ft>s
620
*15–140.
The jet is traveling at a speed of 720 km
>
h. If the fuel is
being spent at 0.8 kg
>
s, and the engine takes in air at
200 kg
>
s, whereas the exhaust gas (air and fuel) has a
relative speed of 12 000 m
>
s, determine the acceleration of
the plane at this instant. The drag resistance of the air is
FD = (55
v
2), where the speed is measured in m
>
s. The jet
has a mass of 7 Mg.
SOLUTION
720 km
/
h
Ans:
a
=
24.2 m
>
s
2
15–141.
The rope has a mass per unit length. If the end length
is draped off the edge of the table, and released,
determine the velocity of its end Afor any position y, as the
rope uncoils and begins to fall.
y=h
m¿
SOLUTION
Ans.v=
C
2
3gay3–h3
y
2b
dmi
+T©F
s=mdv
dt +vD>i
dmi
dt
yh
A
15–142.
The 12-Mg jet airplane has a constant speed of 950
when it is flying along a horizontal straight line.Air enters
the intake scoops Sat the rate of .If the engine burns
fuel at the rate of 0.4 and the gas (air and fuel) is
exhausted relative to the plane with a speed of 450 ,
determine the resultant drag force exerted on the plane by
air resistance.Assume that air has a constant density of
.Hint:Since mass both enters and exits the plane,
Eqs.15–28 and 15–29 must be combined to yield
©Fs=m
dv
dt –vD>e
dme
dt +vD>i
dmi
dt .
1.22 kg>m3
m>s
kg>s
50 m3>s
km
>
hv 950 km/h
S
623
15–143.
The jet is traveling at a speed of 30° with the
horizontal. If the fuel is being spent at and the engine
takes in air at whereas the exhaust gas (air and
fuel) has a relative speed of determine the
acceleration of the plane at this instant. The drag resistance
32 800 ft>s,
400 lb>s,
3 lb>s,
500 mi>h,
SOLUTION
Ans.a=dv
dt =37.5 ft>s2
–(15 000) sin 30° –0.7(733.3)2=15 000
32.2 dv
dt –32 800(12.52) +733.3(12.42)
a+©F
s=mdv
dt –vD>e
dme
dt +vD>i
dmi
dt
v=vD>i=500 mi>h=733.3 ft>s
dme
dt =403
32.2 =12.52 slug>s
dmi
dt =400
32.2 =12.42 slug>s
500 mi/h
30
ft>s.The jet has a weight of 15 000 lb. Hint: See Prob. 15–142.
of the air is FD = 10.7v lb, where the speed is measured in
22
Ans:
a
=
37.5
ft
>
s
2
624
*15–144.
SOLUTION
When the four engines are in operation, the airplane has a constant speed of
.Thus,
Referring to the free-body diagram of the airplane shown in Fig. a,
When only two engines are in operation, the exit speed of the air is
Using the result for C,
Solving for the positive root,
Ans.vp=165.06 m s=594 km h
577440.0 vp 2+2vp–1550 =0
:
+©Fx=dm
dt
CA
vB
B
x–
A
vA
B
x
D
;
¢
0.044775 dm
dt
≤
A
vp 2
B
=2dm
dt
C
–vp+775
B
–0
D
a:
+b
ve=-vp+775
C=0.044775 dm
dt
:
+©Fx=dm
dt
CA
vB
B
x–
A
vA
B
x
D
;
C(222.222)=4 dm
dt (552.78 –0)
a:
+b
ve=-222.22 +775 =552.78 m>s:
vp=c800(103) m
hda 1 h
3600 s b=222.22 m>s
A four-engine commercial jumbo jet is cruising at a
constant speed of in level flight when all four
engines are in operation. Each of the engines is capable of
discharging combustion gases with a velocity of
relative to the plane. If during a test two of the engines, one
on each side of the plane, are shut off, determine the new
cruising speed of the jet.Assume that air resistance (drag) is
proportional to the square of the speed, that is,,
where cis a constant to be determined. Neglect the loss of
mass due to fuel consumption.
FD=cv2
775 m>s
800 km>h
Ans:
vP=594
km>h
625
15–145.
SOLUTION
dv
dme
a
The 10-Mg helicopter carries a bucket containing 500 kg of
water, which is used to fight fires. If it hovers over the land in
a fixed position and then releases 50 kg>s of water at 10 m>s,
measured relative to thehelicopter, determine the initial
upward accelerationthe helicopter experiences as the water
is being released.
Ans:
a
=
0.0476
m
>
s
2
626
15–146.
SOLUTION
At a time t,,where . In space the weight of the rocket is zero.
(1)
The maximum speed occurs when all the fuel is consumed, that is, when
.
.,,,ereH
Substitute the numerical into Eq. (1):
Ans.vmax =2068 ft>s
vmax =4400 ln a24.8447
24.8447 –(0.04658(200))b
vD>e=4400 ft>sc=1.5
32.2 =0.04658 slug>sm0=500 +300
32.2 =24.8447 slug
t=300
1.5 =200 s
v=vD>e ln
¢
m0
m0–ct
≤
Lv
0
dv =Lt
0
¢
cvD>e
m0–ct
≤
dt
0=(m0–ct) dv
dt –vD>e c
c=dme
dt
m=m0–ct
+c©Fs=dv
dt –v
D>e
dme
dt
A rocket has an empty weight of 500 lb and carries 300 lb
of
fuel.
If
the fuel is burned at the rate of 1.5 lb
/
s and ejected
with a velocity of 4400 ft
/
s relative to the rocket, determine
the maximum speed attained by the rocket starting from
rest.
Neglect the effect of gravitation on the rocket.
15–147.
SOLUTION
Ans.F=(7.85t+0.320) N
=2[9.81(0.4)t+(0.4)2]
F=m(gvt +v2)
F–mgvt =0+v(mv)
+c©Fs=m dv
dt +vD>i (dm i
dt )
dmi
dt =mv
mi=my =mvt
dv
dt =0,
y=vt
Determine the magnitude of force Fas a function of time,
which must be applied to the end of the cord at Ato raise
the hook Hwith a constant speed Initially the
chain is at rest on the ground. Neglect the mass of the cord
and the hook. The chain has a mass of .2 kg>m
v=0.4 m>s.
v 0.4 m/s
H
A
*15–148.
The truck has a mass of 50 Mg when empty.When it is unload-
ing of sand at a constant rate of ,the sand flows
out the back at a speed of 7 ms,measured relative to the
truck, in the direction shown. If the truck is free to roll, deter-
mine its initial acceleration just as the load begins to empty.
Neglect the mass of the wheels and any frictional resistance to
motion. The density of sand is .rs=1520 kg>m3
>0.8 m3>s5 m3
Applying Eq. 15–29, we have
;
Ans. a
=
0.104 m
>
s2
0 =57.6(103)a–(0.8 cos 45°)(1216):
+©Fs=mdv
dt –vD>e
dme
dt
45
7 m/s
a
15–149.
SOLUTION
At a time t,,where .
Here,.
Ans.F=(m0–m¿v)(0) +v(m¿v)=m¿v2
vD>i=v,dv
dt =0
c=dmi
dt =m¿dx
dt =m¿vm =m0+ct
:
+©F
s=mdv
dt +vD>i
dmi
dt
The car has a mass and is used to tow the smooth chain
having a total length land a mass per unit of length If
the chain is originally piled up, determine the tractive force
Fthat must be supplied by the rear wheels of the car,
necessary to maintain a constant speed while the chain is
being drawn out.
v
m¿.
m0
v
F