Solution 3.136
2
0.01294
v
=
10 lb
Solution 3.137
2
With 0.4 m, 3kg, 40 ,and g 9.81m/s ,
we find 111.9 N/m
bm
k
θ
===°=
=
22
(
b) With 100 N/m and 25 in :
11
mg mg cos
22
S
olveto obtain 0.522 m/s.
==°∗
=+ +
=
k
bmv b k
v
θ
θδ
Solution 3.138
For the unit 0
UTV
′=Δ +Δ =
( 2mg sin mg cos ) ( mg 0)
mg ( 2sin cos 1)
g
Vrr r
r
θθ
θθ
Δ
=− − −− +
=− −+
45
0.748,
0.865 g
Vr
°
=
=
2
0.908
g
max
Vr
=
cos 0.2 0.8 1 or 0.6, 0 or 126.9
max
θθθ
=±= − = = °
Solution 3.139
22
Solution 3.140
11
tan 5.71
θ
−
==°
1500 (9.81) N
x
r
Vg = O
2 m
θ
θ
Solution 3.141
2
12 9.6 5 m/s
At 2s : 18 12 m/s
vi jk
tvi j
=− +
==−
T
hen 1.2(12 9.6 5 )
14.40 11.52 6 kg m/s
== − +
=− +⋅
Gmv i j k
ijk
222
14.40 11.52 6 19.39kg m/s=++=⋅
G
:1.2(1812)
21.6 14.4 N
FGRmv i j
ij
Σ= = = −
=−
Solution 3.142
Conservation of system linear momentum:
0.075(600) 50.075 , 0.899 m/s
+
→= =
ff
vv
1
Solution 3.143
I. Collision
GG
→=+
mg
Solution 3.146
/
31
322
mc m c
vvv
ij
=−
=+
cx
Solution 3.147
0.6
81 8
(a) (30) (60)(0.4)
32.2 2 32.2
v
−+ =
ʃN dt ʃF
x
dt
ʃmg dt
x
y
Solution 3.148
()
1.218m/s down
=
v
Solution 3.150
The horizontal location of the system mass center is
mx m x
+
()
1
11
2
w
efin
d 2.90 m/s, 0.483m/s
l
f
e
t
==−
xx
x
1
Solution 3.151
()()
1200 9.81 N
2
3.6 3.6
t
Solution 3.152
()()
Δ
3200 3400
3200
0; 30
gg
xx
v
G
+
Δ
=
=
Solution 3.153
No difference between cases (a) + (b).
11
v
T
500 N
(1200) (9.81) N
N (y)
υ
Solution 3.154
32.2 60
Solution 3.155
()
/2
12
0
cos mg sin
k
k
mv F kt F kt d v
t
m
π
µ
+
+−−=
g
k
F
µπ
()
µ
(independent of k)
θ = 0.5°
+=
mυ
1
mυ
2
ʃ
8N dt
ʃ
L dt
ʃ
8μ
k
N dt
ʃ
mg dt
Friction is of no
consequence here
+=
mυ1mυ2
ʃ
F dt
ʃ
mg dt
θ = kt
Solution 3.156
()()
[
]
()()()()
2
2
0; 0.140 600 0.140 3 0.100 0
11
0.140 600 0.140 0.300 190.9
Gv
E
Δ= − +× =
Δ= − +
Solution 3.157
3
3
17970 m h
k/
v
=
Solution 3.158
+=
ʃ
Pdt
ʃ
mg dt
20 kg
Solution 3.159
0: mg sin30
0
FN P
+
↑=−+ °=
2.02s
=
t
2
12
:
t
mv Fdt mv
+
→+=
s
Solution 3.160
() ()
4
0
90 2.64
so 90000 0.00264 , 5
9.5 N
4
f
av av
Fdt F
===
mg P
Solution 3.162
1
tan 0.1 5.71 ,sin 0.0995
Fdt m v
θθ
−
==°=
=Δ
32.2 30
704 lb (tension)
P
=
Solution 3.163
()()()
+
12
2
2
100
0 60 15 10 0.35 100 15 10 32.2
40.2 ft/se
c
mv Fdt mv
v
v
→=+
+−− −=
=
(There is no net horizontal impulse before t = 10 sec!)
()()()
0 60 15 10 0.35 100 10 0
18.57 sec
t
t
++ − − − =
=
+=
ʃ
F dt
ʃ
N dt
ʃ
mg dt
15,000 lb
(Trailer)
F
Solution 3.164
()
(
a
)
=′
AA A B
mv mm v
2
5.56 0 55.6 m/s
0.1
Δ−
== =
Δ
B
v
at
Solution 3.165
0
0.075 ft or 0.900 in.
d
=
Solution 3.166
0.20
4.819
x
Fdt Ft m v
v
==Δ
25°
150 ft/sec
mg ≈ 0
Solution 3.167
yυ2 = 130 mi/hr
Solution 3.169
R
y
Δt
y
υ
2
= 22 m/s
20°
Solution 3.170
System:
(m
B
+ m
S
)g
Solution 3.171
Set up x–y axes at launch point of A.
2.94 m
=
()
F
or B: 2g 2(9.81) 6tan 40 2.94
6.40 m/s
=− =− °−
=
vh
C
‘ 3.06 3.83 m/s
=−
vij
Subsequent projectile motion:
0
0
: 6 6 3.06( 0.653)
x
xx vt d t
=+ +=+ −
Solution 3.172
vvv
=+
2
υc = absolute velocity
of car.
15°
x
Solution 3.173
+= = = ⋅
2
0
H 2(7)(9.20) 128.7 kg m /s
mvd
Solution 3.174
Hrmv
=×
Solution 3.175
(
a) 3.5( cos30 sin30 )
12.99 7.5 kg m/s
== − °− °
=− − ⋅
Gmv i j
ij
0
2
(
b)
2(cos15 sin15 ) ( 12.99 7.5 )
21.2 kg m /s
=× =×
=°−°×−−
=− ⋅
HrmvrG
ij ij
k
22
11
(c) T (3)(5) 37.5J
22
mv== =
2 kg
Solution 3.176
(
a) 2g
2g , mg
B
vr
Hmrvmr rH r
=
== =
Solution 3.177
()()
00
AP
HH
=
Solution 3.178
2
12
000
about -axis :
+=
t
HMdtH z
Solution 3.179
()()
2
12
1
000
5..
22
0
about -axis :
02 21.20.4 ( )
+=
+= =
t
t
s
HMdtH z
tdt H mr
θθ
Solution 3.180
Angular momentum about central axis is conserved so
2
d
Solution 3.181
2
22 22
323, 4kg
4(6 6) (9 18 ) (12 6 )
=−− =
=−+− +−
rtitjtkm
tti t tj t tk
22
72 3 2 260 N m
=+=⋅M
mg