Solution 4.42
π
2
Solution 4.43
π
== ==
2
22
40.0873ft , 4 0.349ft
cBC
AAA
=6530 lbT
T/2
B
C
Solution 4.44
′
=
=−
=⋅
=⋅
or 9.69KN
()
1122 N m
or 1.122 KN m
T
Mmvdvd
M
Solution 4.45
°
=Δ
′
:
xx
Fmv
()
()
π
π
==
==
22
22
0.1 0.00785
0.25 0.0491
4
o
Am
Am
°°
== =
0.125 0.00785 15.92 m/s
vQA
υ
m
o
g
A
V
M
Solution 4.46
1
;
0( )
yy
Fmv mpAv
vvv
′
Σ=Δ ′=
Δ=−−=
Solution 4.47
Solution 4.48
π
′==
2
4
A
d
mpAv p u
3′
υ
y
250 mm
+
M
O
υ
BB
u
υυ
Solution 4.49
32.2
xx
Solution 4.50
°
=
60 3.6
or 5.56 KN
60000
P
υo
W = 24000 lb
Solution 4.51
==′ −
22
(0)
o
MM mvd
0.32 /1000 0.32 0.971 1000
0.32 0.351 0.671 kw
o
PM
P
ω
=+ =+
=+ =
Solution 4.52
π
ρ
θθ
×
=′Δ ′= = =
=× =
2
0.040
: 1.206 240 0.364 kg/s
4
0.364 480sin 174.6sin N
22
Fmvm Av
F
For vane:
()( )
()( )
θθ θ
θ
θ
θθ
=−=
×=
==
0
0.100
0 :174.6sin cos 6 9.81 0.240sin 0
22sin
87.3 0.100 6 9.81 0.240sin
sin 0.618, 38.2
M
Assumption: Entire air stream is diverted downward along the vane, with no flow toward 0.
240 mm
υ
2
θ
–
2
Δυ = 2υ sin
θ
–
= 480 sin
Solution 4.53
mg = weight of helicopter
πρ
Power = rate of increase of kinetic energy
()
22 2
21
11
2222
2
vv
Pmvv mvmvmg
mg mg
Pr
πρ
=′−=′=′=
=
υ = 0
Solution 4.54
φ
cos
r
p
vwb
φ
=
M
M
()
()
22
o22
cos
=2sin
4
cos
If =0… = 42sin
Qr
M
PQ w r b rb
A
Qr
Mw
Ar b r
b
φφ
φ
ωφ
−++
=++
O
r
M
ω
υ
p
r
Solution 4.55
() ()
=
′=
oo
1.206 1.10
95 1020 96900 N
aa
mu
y
Solution 4.56
()
==
′
18 2000 10.3106 slugs/sec
32.2 3600
air
m
4
(2) tension in pipe at B
(4) weight of bend
Solution 4.57
Ry
R
x
C
y
Solution 4.58
()
()
()
==
66
66
2.04 10 9.81 20.0 10 N
mg
Solution 4.60
Σ= +
=
60 ft/sec
xv
Fmmu
u
() () ( )
()
a ater on; 621 2 – 2.48 60 cos30
1242 –129 1113 lb
b ater off; 0, 1242 lb
W
W
P
mP
°
=
==
==
y
20,000 lb
2
Solution 4.61
Eq. 4.21: ′+ − − =
mu PA mg R mv
0.1/ 32.2 0.00345slugs/sec
()
5.90 g
From 5P 4.12:
o
mg
Solution 4.62
Solution 4.63
π
°
Σ= +
Σ=− =−
=−
=
=
=−
×
=−
=−
=−=
−= − =
2
2
20.6(9.81)
202 N
20.6 kg
0.5 /
0.030
2(1000) (2.5)
4
3.53kg/s
2.5sin 20 0 0.855 m/s
So 202 20.6(0.5) 3.53(0.855), 209 N
y
y
Fmvmu
FPmgP
P
m
vms
mpAv
u
PP
y
P
υ
20° 20°
2.5 m/s
mg = 20.6(9.81) N
t = 0
Solution 4.64
== =
−=− =
=+
−
2
2
const., weight of descending links
so
For x v P
pgx R p
Rpgx
vx pv
pv
Solution 4.65
o
m
==
0.511
u
b
R
Solution 4.66
2
2
32.2
400 lb sec /ft at 4 sec.
220 32.2 6.83 lb sec/ft 1.5 mi/hr 2.20 ft/sec
2.20 10 cos 60 2.80 ft/sec
So180 400 6.83( 2.80), 0.498ft/sec
m
t
m
u
vav
°
=
=− =
==− =
=− =−
=+− ==
Solution 4.67
x
υ
W
60°
10 ft
–––
sec
Solution 4.68
2
3
1.603m/s
S
o 2880 16.4 (10 ) 375(77.8), (deceleration)
a
vv
=−
=+ =
mg