Solution 3.92
60 50 110 lb
R
F
=+=
() ()
3600
b : 110 3600 sin3.43 0.05g
g
xx
Fma F
Σ= − +− °=
Solution 3.93
For 75 mm, &xUT==Δ
2
11
y
mg
Solution 3.94
()
2
22 2
5
a+ ,
24
ABCD
R
LR RLRLL
=====
v
()
BD
Solution 3.95
At impact
:
Solution 3.96
Power output = rate of doing work
()() ()()
300 9.81 2 100 9.81 4
1962 ( )
1.962
JSW
KW
=−
=
=
po
Solution 3.97
()
()
2
2
2
1
5280
57.33ftsec
3600
7.33 ft
2, 0.538
250 sec
tan 0.1 5.71
B
B
v
vasa
θ
−
==
===
==°
==− °=
=
90
ma: 90 sin5.71 (0.538)
32.2
10.46 lb
F
F
F
ft-lb
10.46(7.33) 76.7 sec
or 76.7 / 550 0.1394 hp
PFv
P
== =
==
F
N
10
1
90 lb
θ
Solution 3.98
:launch
Work expression used:
Solution 3.99
()
2
22
6
5280
10.85 10 lb
=
6
10.85(10 )
NFR
Solution 3.100
Solution 3.101
22
1
(
a) : 0 2 mg , 4 g
AAB B BB
Tv T R mvv R
−
+= + = =
:mg
nnc
c
FmaN m
R
Σ= − =
(c) Call stopping point E:
AAE E
Tv T
−
+=
mg
t
Solution 3.102
–1
= 1.146°
m
3
–––
Solution 3.105
The power output of the drive train is
90
560 14000 W
PF
== =
in
0.70
so the motor output 20 KWP=
Solution 3.106
in
Solution 3.107
() ()
()
4
22
0
2
2
0
148
;3 60 0 12
232.2
32.2
32.2 (64 480) 60.82 ft sec , 7.80 ft sec
288
UT x xdx v
vv
=Δ − + = −
=+= =
=
Solution 3.108
−
=Δ =
12
0:
UT
6 (9.81) N
Solution 3.109
0
00
0.5
0.289 m
B
AB
S
SS
==
Generally :
222
With 0.14N mm , 500 mm , 1.2 ,
m
pAK===
2
0.1140 m
A
S
=
00.433 0.289 0.1445 m
BB
S
S
A
120°
A
B
30°
Solution 3.110
1
6
−
Solution 3.111
12 g
0
e
UTVV
−
′==Δ+Δ +Δ
Δ
22
2
1200 0.8 0.6 0.4 0.8 0.4
2
20 J
3
so 0 23.5 20,
2
1.537 m/s
V
v
v
Δ
=+−−−
=
=−+
=
Solution 3.112
Establish datum @ A
(
a)
+
=+
BB
AA
TV
TV
or
54.2 mm
δ
=
Solution 3.113
For the system,
where the datum is the initial position and h is the drop distance. Note that the spring
Solution 3.114
datum at
,
AA AB BB
TVU TV A
−
′
++ =+
Solution 3.115
datum at
(4) (4) 3
49 0 in.
.
AA AB BB
TVU TV A
L
−
′
++ =+
=−−
=
Solution 3.116
12 g
2
2
0
1mg 0
22
UTV
r
mv h
r
−
′=Δ +Δ =
−−=
2
mg
:2
nn
v
FmaN m
r
Σ
=+=
A
Br
N
mg
h
Solution 3.117
22(0.3)
== =
b
θ
Solution 3.118.
(a) g0TV
Δ
+Δ =
2
2
2
22
15 110 12 18 12
5sin6010sin600
2 32.2 2 32.2 18 12 12
3.
ft/s4ec8
Vv
v
++°−°=
=
(b) For entire interval g
0, 0
e
TVV
Δ
=Δ+Δ=
22 2
1
0.510 in.
x
=
Solution 3.119
1
(90 20 )
2 sin 0.287m
2
Ld
°− °
=
=−=
=
We may ignore the equal and opposite potential energy changes associated with two of the
masses.
Solution 3.120
T
L
1
L
2
20°
O
Solution 3.121
The additional stretch in the spring runs at twice the drop of the cylinder.
2.54 ft/se
datum at initial
n
c dow
TVU TV
v
′
++ =+
=
Now, left State be the stop position:
11 133 3
22
12 122
TVU TV
xd xd
−
′
++ =+
+
Solution 3.122
12 ge
UTVV
−
′=Δ + Δ+Δ for system
s
Solution 3.123
22
13 9 12 4.5
12 12
15
6mg up
B
N=
2 lb
NB
Solution 3.125
g
2
0
0
1900 0.4 1 sin
e
VVT
T
V
θ
++
=
ΔΔΔ=
Δ=
Δ−
Solution 3.126
()
()
3
0
2 9.825 6371.10
2g
21.5
3
3
R
R
v=
=
3 kg 3 kg
4 kg
0.2 m
Solution 3.127
TV TV
+
+=
()()
1
0.2 cos60 0.280 0.38
0mg
2
0.280m
mg mg
=
=− − =−
V
2.30 m/s=
A
v
Solution 3.128
()
()
g 9.825 3600 / 1000 127.3 10 km/h==
Thus
22
23
111
26300 km/h
B
v
=
datum
O
0.2 m
Solution 3.129
12 1 g1 2 g
2
0 so UTVTV
′
−++==
12
232.2
42
30.0 ft/sec or 20.4 mi
v
=/hr
Solution 3.130
()
For motion from 60 to 180
0.981 m/s , 0.990 m/s
=° = °
==
vv
θθ
Solution 3.131
13.50 m
ππ
=
22 2 2
21 2 1
2
2
2
2
22
2
90(1000) 4(9.81)(13.50)
3600
625 529.9 95.07(m/s)
9.75 m/s or 35.1km/h
−+ = =−
=−
=− =
==
mV V r V V r
V
VV
r = 15 m
r = 15 m
r
–
υ
Solution 3.132
Constant total energy is AA
gAP
g
ET V T V=+ =+
Solution 3.133
22 2
22
g
0.9 , 0,
10.9
0; ( ) mg 0
0.9 19.62 2
AB
xx
xy xxyy V y x V
yy
TV mxy y
xyy
+= += =−= =
Δ+Δ = + + − =
=−
0.81 2
1.8 0.6
dy
a
nd 0.81 2 30
=−=
x
Solution 3.134
max
Solution 3.135
0
VTV
′
=Δ +Δ =