Solution 3.48
22
:mg 1
Σ
==
n
vv
Fm m
2
2
gg
,
ys
ss
rr
ωω
µµ
==
Solution 3.50
2
5280
mg
1
n
2
N
O
µ
S
N
n
Solution 3.51
2
Solution 3.53
()
B 163.8 ft sec 111.7 mi hr
B
v
Solution 3.54
Solution 3.55
2
:
600 3.6
nn
Fma
Σ=
2
:
600 3.6
nn
Fma
Σ=
n
90 (9.81) N
B:
NB
Solution 3.56
()
()
2
max
27.8
200 9.81 cos 10 200 sin 10
300
2020 N
0.70 2020 1415 N
yy
s
N
N
FN F
µ
−°=°
=
== = >
Crate does not slip.
P
Rr
δ
θ
N
Solution 3.58
Top view:
Solution 3.60
30
=°
θ
0.820 m
0.644 0.910 m s
22 22
=
===
r
v
v
θ
()()()
2
2
or 0.820 0.349 2 0.644 0.785
1.835 m s
=+
=
a
a
()
2
: cos 45 mg sin 30 0.2 1.835 2
rr
Fma N
Σ= − °− °=
()
1.754 N vertical slot
N
=−
Solution 3.61
2
:e
Gm m V
FmaF m
Σ= = = +
t
O
θ
υ
L = 0.6 m
Solution 3.62
If 60 , 30
β
θ
=°=°
32.2
θ
Solve Eqs. (1) and (2) to obtain
0.0979 lb
F
Solution 3.63
r
F and F
θ
are the r– and
θ
-components of the total friction force F.
2
22
8.29 N
25.7 N
0.5
r
s
F
F
θ
µ
=
=
F
O
R
θ0.1 lb
θ
r
Solution 3.64
Point A:
2
:
22
nn
Fma
Σ=
195.8 N
B
B
N
=
(Note static normal of magnitude
()
mg 75 9.81 736
N
N== = .)
Solution 3.65
Σ
2
6
2
3000
; 2000 6 sin 30 2000
1
6910,
6.00 m s
Σ
=°=
=
=
nn
Fma P
P
v
(assumes velocity of rocket is parallel to its longitudinal axis)
75 (9.81) N
t
n
n
N
B
R = 9.6 kN
Solution 3.66
θ
40 0.698 rad s
==
θ
r
θ
Solution 3.68
6 rad s constant
θ
==
s
Solution 3.69
2
32.2 12
2.52 lbT=
()
2:
Fmamr r
θθ
θθ
Σ= = +
mg
F = µ
k
N
mg
P
N
r
θ
N
θ
Solution 3.70
The distance traveled from A to C is
π
cA tcA
()
0 60 2 231 , 16.77 ft / sec
3600 tt
aa
=+ =−
Speed at B:
22
22
32.2
32.2
3000 66.3
:32.2 250
xx
ttt
nn
v
Fm F
ρ
Σ= =
Solution 3.71
;
ttt
FmaFmr
α
Σ= =
44
πα π α
Solution 3.72
x
F
t
n
()
0
00
22
0
2
00
:0
2
rr
tt
Fma mrr
rr r
dr
rr
r
vr r e e
ωω
θ
θω
ω
ω
θω
−
Σ= = −
==
=
== = +
00
00
tt
r
ωω
ω
−
00
00 0
cos
cos
rr h t
vr ht
θ
ω
ωω
=
=
With numbers,
0.1 cos
0.1 cos
rht
vht
θ
=
=
Solution 3.73
()
2
0
2
0
;mgcos
mg cos 2mg 1 cos
mg 3 cos 2 g
nn
v
Fma Nm
R
m
Nv
R
v
R
θ
θθ
θ
Σ= −=
=−−−
=−−
10
2
cos 33g
v
R
β
−
=+
For 1
0
2
0, cos 48.2
3
v
β
−
== =°
Solution 3.74
min
1
tan
10,450
tan 43.4
15,209 4159
ar
α
−
=−
==°
−
15,208mi;= so
()
()
1.133 10
15,208 5280 sec
θ
==
()
15,208 5280 sec
r
b = 10,450 mi
Solution 3.75
22
2
Switch to polar form:
22
2
x
y
O
θ
+
Solution 3.76
2
0: mg
:
yy
nnn
FN
v
FmaNm
r
Σ
==
Σ
==
0
1ln g
µ
2k
rg
µ
Solution 3.77
Solution 3.78
22
11
F
y
mg
n
Σ
−
Solution 3.79
22
11
;2 0
UTkxmv
=Δ = −
Solution 3.80
5.62 m s
B
v
=
Solution 3.81
22
164.4
f
Solution 3.82
θ
=
550
Solution 3.83
[]
1
2
6
15280
65 32.2 0.6 cos 3.43 sin 3.43 0
s
−
+°+°=
−
[]
65 32.2 0.6 cos 3.43 sin 3.43 0
2 3600
s
+− °+ °=
Solution 3.84
2
1
2(9.81) N
N
y
Solution 3.85
For collar, 12 0UT
Solution 3.86
Solution 3.87
2
AABB
TU T
−
+=
where
dR R
=+ −=
So
()()()()
20 0.618 1.6 0.8 9.81 1.6 0.8
v⋅− =
6 ft
(a) (b)
Active-force diagrams
for entire system
Solution 3.88
Wh
Pt
=Δ
()
120 9 550 0.393 hp
Solution 3.89
For system
12
−=Δ
UT
π
Solution 3.90
Power PFr=⋅
()()
()()
2
4
40 20 36 8 2.4 1.5
40 20 36 8 9.6 24
320 192 864 992 W
ts
Pijkitjtk
Pijkijk
=
=−−⋅+−
=−−⋅+−
=−+=
Solution 3.91
()
Net power required 30 140 24 33,000
3.05 hp
=
=
Power required 3.05
Mechanical efficiency 0.764
Power supplied 4.00
===
–
r
–
F
–
r
∙
θ
4 (9.81) N