Solution 4.1
() ()
()
+++
==
++
==− + =
==−
232 2
23
22 2
46
2
i
i
i
ii
oi
o
i
o
i
mdj m di dj m di
mr
rmmmm
dij
H xm mvdk mvdk mvdk
HM
rv
Fdk
Solution 4.2
=−
6
Goi
dFi
Solution 4.3
()
()()
+−+ + + ++
=
23 2.51.5 5253
i
i
mbi mbi bj bk mbibjbk
mr
() () ()
==+ →=
+
4 3 2 16.5
22
5
ii
Tmv
mv mv mv T mv
() () ()
()
() () ()
()
°°
−−+
=− + −
==+−+=−+
62.5 53 52
15 2.07 1.536
3 1.5 2 5 2 2 3 14.50 4 3
o
oo
mv b k mv b i mv b k
Hijkmvb
HM FbFbiFbjFbk ij Fkb
Solution 4.4
M = m + 3m + 5m = 9m
()
()
2
3.61 2.17 9 39
0.778 2 6
G
FF
bi b j bk m j k
mm
HijkF
−+ + × − +
=
+
−b
Solution 4.5
Solution 4.6
Δ
G
Solution 4.7
()
−
Δ
==
Δ
=++
=
1
,
13.67 3.65 4.30 4.27 5.30 5.36
0.1
23610N.m
0.7 N.m
av
o
ooo
oav
H
MHM t
ijk
M
Solution 4.8
()
=
−= −
=− =
1
60 55
32.2
60 1.708 58.3 lb
ii
Fma
T
T
a
A
= 5 ft/sec
2
a
C
= 3 ft/sec
2
T
W
C
= 15 lb
Solution 4.9
()
==−
=
−
−
+− − − = +
=−− =− +
+−
=− = =
:cos15
:sin15
:14 5.5 4 9.5 2 4 9
9.5 sin15 7 cos15
SOLVING…
1.288lb, 35.5 l
…
b, 25.6 lb
left
xxxBBCC
yyyEABCAACC
DdEBA C BBAA
CC CC
xyE
FmaDmama
F maD N mgmgmg ma ma
MmaN mgmgmgmam
ma ma
DDN
mAg
2‘2.5‘5.5‘
3‘
Solution 4.10
()()
=++×+
=− ⋅
2
0.225 0.8 0.6 8.66 5
0.971 kg m /s
kij ij
k
Solution 4.11
Σ=
yy
Fma
Solution 4.12
=+×
2
oG
HH pmv
250 N250 N
y
10 g10 g
500 N
Solution 4.13
Let Pp = power to move 10 people
fr
550
2.2 hp
T
hus, 1.364 0.327 2.2 3.24 hp
P
P
=
=−+=
()
==
2
4
4,
mr w
Mt m rw r t M
Solution 4.15
Σ== Σ =Δ
2
,
o
oo o o
dH
MH Mdt H
dt
Solution 4.16
For the system as a whole
0
Solve and get 1.015 m/s, 1.556 m/s
yB A
AB
vv
=
==
Solution 4.17
Solution 4.18
Σ=0
x
F for system so Δ=0
x
G
()
()
+−
×+ ×− ×
−+
+
+
==
+
=
3
3
44 10
130 2 100 1 150 1.5 30 32.2
44 10
130 100 150 0
3
260 100 225 0.355 mi/hr
130 100 150
032.2
v
v
−
=−=
% loss of energy = 100 100 1
if f
ii
TT T n
TT
()()
()
2
22 2
1130 100 150 0.355 47.96
29
100 1 100 1
1957.5
130 2 100 1 150 1.5
29
95.0%
n
n
++
=− =−
×+
=
×+ ×
Ax
y
2 mi/hr 1 mi/hr 1.5 mi/hr
Solution 4.19
l
s
Solution 4.21
x
45°
−
′
Δ
=Δ +
12
g
UTV
x
Solution 4.22
Σ=Δ=
0so 0
xx
Fdt G
1.5 kg
F
x
0.3 m
υ
x
ω
Solution 4.23
00
for system so
25
FG
Σ= Δ =
Solution 4.24
30 km/h
s = r (2θ) = 18 m
G
rθ
Solution 4.25
System is conservative so Δ=+Δ 0
g
TV
=
Δ=
2
2
1 50,000
232.2
0
g
v
V
22
−=
=
=cos5 3.3
3&sin
32.2 32.
54.33ta
2
n5 0.379ssvv v
v
Substitute into (1) and get
()
==
22
2
2
2
15.39 ft/sec , 3.92 ft/sec
vv
x
υ
.
5°
Solution 4.26
===
1
125 m/s 7s 60 kg
o
vtm
=
53125 28.8m/s
5
o
o
Motion of C after explosion:
()
=→−=+→=−
=+ → = + →=
+
565 485 6 174.9 m/s
ccc
CPxc x x
xvt v v
x
Conservation of momentum: =i
i
mv m v
() ()()
=→=++→
=−
60 28.8 10 30 41.1 20 296 541m/s
iA A
iA A
iA A
y
z
i
zi z
yy y
z
mv m v v V
=+ = + →=
22
A
APzA A A A A
Solution 4.27
()
θ
+=−
121
1
12
1cos
22
c
mmvmg
m
mm
m2
Solution 4.28
()
20.5
225 J
=
=
0.5 2
m
32.2
0.5 m
BA
AT = 0
V
e
= kx
2
x = 0.5 m
xʹ
1
–
2
T = Σ m
i
υ
i
2
1
–
2
υ
B
υ
A
Solution 4.30
()
′′
−+
=
af
muv mu
T
Solution 4.31
0.0753
Qm g
µ
′=
Solution 4.32
Eq. 4.18: xx
Fmv
′
=Δ
3
=
=
0.01598 slugs/sec
130 190.7 ft/sec
mg = 4.6 (9.81) kN
x
y
x
F
=3.05 lb
Solution 4.33
Resistance R equals net thrust T where ()Tmuv=′−
0.082
()
== − =
3600
84.5 41.8 19.44 1885 N
RT
Solution 4.34
=
40 m/s,
v
mg
υ
Solution 4.36
xx
Fmv
′
Σ=Δ
=or 32.6 kNT
Solution 4.37
Final velocity of diverted water is ′=+
rel
vuv
()( )
π
′=
=+
rel
2
1000 0.025 20 10
Av
mP
u
y
x
υ + u
υ′
υ
2
R
1
R
2
T
Solution 4.39
()
=−
12
2
1cos
2
Q
υ
t
n
1
Solution 4.40
Ball and stream just under it:
For water stream
Δ+Δ=
×
22
0:
10,
2 32.2
g
VT
Solution 4.41
′
Σ=ΣFmu
0.5 lb
650 m/s