Solution 3.253
()()
00
:
22
xx
Fmakxx mxa
mm
=−−=−


=+ − =

22
0
0
0 for max. and hence max
kx
dx kx x
ax
Thus
()
22 2
2222
0000
00000 0
2
l0
2
2
o
rel max
max
re max
kx ma ma m a ma
k
vxa x xx x x
mkmkk k


=++−−++−=




=
Solution 3.254
For accel. a vertically down,
aa
=°
o
a
y
n
Solution 3.255
32 80 112 J for both c+ases
bc
TTT
==+=
Solution 3.256
()
0
0
+.In–dir., 0,
so cos 0
rel t
t
Fmaa t F
al a
θθ
==
=− =
In –dir.,
nn
nFma
−
0
00
2
cos
2
dl
a
dd
l
θθ
θ
θθ θ θ
=

0.8 m/s
m
b
= 10 kg
m
= 250 kg
θ = 90°
θ
.
= 3 rad/s
0.8 m/s
θ
ℓ = 0.8 m
υb
υcυc
υb
υb/c
.
(b) (a)
n
t
O
ℓθ
.
2
Solution 3.257
0
:mgsin sin
()
tt
Fma ml a
θθ θ
=− = +
a
o
θ
..
ℓθ
.
2
Solution 3.258
=
66.3
or 5
rel r
rel
P=0.1206 hp
50
Solution 3.259
0, elevator is Newtonian frame
(a)
a
=
0
1
// /
22
12 1
2
5g 5
g
24
0 2 4
25g85g
BE BE BE
h
vv ate t
ee ge
eh
heh
′′
=+ =− +
=− +
=− =
SBOa =
g
Solution 3.260
rel rel
UT
=Δ
0
Where d is the horizontal distance traveled by the block.
y
υ
o
mg
Solution 3.261
Solution 3.263
90.2 N
E
quil. of forces at B:
T
=
T
T
30°
υ
Solution 3.264
11
xx
Fdt m v
=Δ
()( ) ()
7451m/s
1 : 6371 600 6771 1
0.0295
max
rae e
e
=
=+ + = +
=
Solution 3.266
out
in
550 550
587
:0.90 , 652 hp
P
eP
PP
+
==
=
Fx
Solution 3.267
2
1g;
c
Acy
yyvt t
=+ −
2
113
R
mi
n
:
mg , g
R
R
n
c
n
c
F
v
ma
m
v
Σ
===
Solution 3.268
AB
C
y
υc
G
x
62″
180 lb
n
mg
Solution 3.269
3000(2 ) 314.2 rad/s
60
π
ω
==
()
2
3
3
39.5 10 N
=
Solution 3.270
2
(
130(0.866) 1 .1
2)62
stt
−= −
Solve (1) & (2)toobtain 2.15 , 37.3 fttss==
ft
sin60 g 30(0.866) 32.2(2.15) 43.3 sec
y
vu t
=−= − =−
°
y
s
30°
u = 30 ft/sec υnυnυn
‘
│υy│
N
G
0.4 N
Solution 3.271
222
Solution 3.272
:
xx
Fma
=
k
(M + m)g
y
Solution 3.273
60 0.457m/s
180
B
a
=


=
Solution 3.274
112 2
2
1
TU T
−+=
+
mg t
15°
B
T
e
t
θ
.
= 5
–––
rad
–––
π
Solution 3.275
()()
2g 2 32.2 6 19.66ft/sec
vh
== =
g
2
23
0
120
2 32.2
20 1.667 ft lb where isininches
12
e
g
TV V
V
δδδ
Δ
+Δ +Δ =
Δ− =−
−
=
2
υ