Solution 6.62
, but 0 so 0
G
MI I M
α
== =
Hence no friction force and 0
s
µ
=
θα
==
;mgsin
xx A
Fma mr
G
r
mg
α
A
Solution 6.63
y
Solution 6.64
450 mm 0.45
30 kg 275 mm
os
r
mk
µ
==
=
=
6
If Roll w/oslip,
If Roll w/slip, N
o
fk
ar
F
α
µ
=−
=
Solution 6.65
450 mm 0.45
30 kg 275 mm
os
r
mk
µ
==
=
=
6
If Roll w/oslip,
If Roll w/slip, N
o
fk
ar
F
α
µ
=−
=
:mgsin
Fma FTma
θ
=+−=
ma
x
α
ω
Solution 6.66
()
22
;
OO
MI madImkr
α
=+ = −
Solution 6.68
212
Kinematics:
GA
GA
aaa
=+
=−
1.143g
A
a
L/2
Solution 6.69
Rf is lateral force at both front wheels, Rr is similar force at both rear wheels Car angular
acceleration is CCW
a
R
α
=
(a) 0:
υ
=
nnfr
R
(1)
22
fr
Rd Rd
R
f
R
r
d
d
G
n
a
Solution 6.70
2
2
14
23
G
r
Imrm
π
=−
If no slip,
α
α
ππ
=−=−
1
2
12
4116
:329
yGy Gy
GG f
r
MIFr N mr
Cart:
:
xGxf
Fma FMa=−=
Solving,
()
()
()
()
()
()
()
()
()
()
π
π
ππ
α
πππ
−
=−
=
=
+
+
+
+−− −+
=− ==
+++
122
8mg
8g
8g
33
33
39
32 mg 32 g 8 g
mg CW
93 93 33
Gx
f
Gy
M
mM ax
a
FmM
mM
mM
mM mM mM
Na
mM mM rmM
x
mg
y
+
Ff
Ff
4r/3π
G1
N1
A
A
a
1
a
2
α
Solution 6.71
2
11
15 2 42.5ftsec
ar
α
== =
()
()
3
: 58.7 0.417 , 16.30
2
Solve and get 145.4 lb, 161.7 lb
GG BA BA
AB
M I TT TT
TT
α
=−= −=
==
Solution 6.72
(A)
θ
=− −+=
:gsin
xGxA AGx
Fma m FTma
(1)
Gx
(B)
+
↓= −=
:g
xxB BB
Fmam Tma
(5)
==− =
29.7 N 64.5
17.73 N
0.275
B
S
TF
A
G
C
F
TT
x
+
x
y
r = 0.15
m
Solution 6.73
aa a a
==+
32.2
Solution 6.74
α
=
2
7.50 rad s
A
yyy y
A
y
G
5(9.81) N
0.4 m
G
= 2.0 α
mrα
mrω
2
= 5(0.4) 2
2
= 8.0 N
ma
A
= 5(4)
= 20 N
1
12
Iα = 5(0.8)
2
α
= 0.267 α
20 lb
Solution 6.75
03
Gy
b
a
α
=−
Solve Eqs. 1–4 to obtain
()
22
22
12 g CW
3mg (up)
73
A
b
bh
Tbh
+
=
+
T
A
T
B
= 0
Solution 6.76
CG
aaa
rr
α
−
==
;
xx
FmaFma
==
3
2to the left
3
GC
aaaa=−=
1123
d
222
a
mg
a
FC
r
Gx
a
α
(a
B/A
)
n
(a
B/A
)
t
3
Solution 6.77
AB A
()()
=+ +
2
22
B A BA BA
nt
aa a a
From diag.
()
== 2
4912ms
3
BA t
a
2
0.6 21.8, 36.4 N
BB
==
A
x
A
y
CO
0.2 m
0.5 m
ma
A
Solution 6.78
ω
ω
α
α
==
== 2
2rads
4rads
BC AB
B
C
AB
k
k
()()
αω
=× − = × + °+ − °
2
4 0.7 0.175 cos 45 0.7 0.175 sin 45
a
rrk i j
()( ) ()
12
4 3.96 0.175 sin 45 , 3.55 N m CCW
B
M
−°=⋅
Solution 6.79
; 13.79
xx
FmaB
α
==
………………………………… (2)
()
α
=°= °=
2
2.4 cos 40 2.4 3.23 cos 40 5.93 m s
A
a
M
B
B
C
0.175 m
0.175 m
4(9.81) N
G
2
45°
B
ma
x
ma
y
Iα
40°
a
A
a
y
= 1.2α sin 40°
a
x
= 1.2α cos 40°
a
B
= a
A
+ (a
B/A
)
t
)
t
= 2.4α
CB
y
Solution 6.80
43
7
52 5 4
24
mL m L
+
Equation 4.13:
MH mr
ρ
=+×
m
5
O
L/2
L
/2
mg
r
G
+
x
y
Solution 6.81
2
OGO
ma ma ma ma mr mr
ωα
=+ =+ +
()
()
()() ()
()
θαα
θ
αθθ
=+−
=+
2
1
mg 6 sin 12 6 6 3 6 cos
12
1gsin 3cos
8
O
mm m
mrω
2
Iα
mg
ma
O
x
O
y
6‘
O
x
y
a
O
= a
θ
Solution 6.82
Kinetics:
=°+°=
()()
sin30 cos30 1.640 0.543
Gx Gy A
aia j a i j k ri rj
α
+= °− °+× −
α
===
53.6 N 53.1 N 4.78 rad s CCW
AB
NN
2
2
α
Solution 6.83
µµ
===
;,g
g
W
FmaW aa
2
2
so time to acquire
g
k
tr
rr
υυ υ
ω
αµ
== = =
Equate t’s and get
2
2
22
00
0
222
,and2,
r
kas
υ
υυ
υυυυ
−== =−