y
Solution 3.1
0 : mg 0, mg
Σ= − = =
Σ= − =
y
FN N
0
4.59 mx=
Solution 3.2
()
max
127.0 N
0.2 474 94.8N :motion
=
== = <
s
x
F
F
NF
µ
x
max
23.0 N,FFF=− < so no motion: 0a=
(c) 300 N;P= Equilibrium check yields 173.0 NF=−
max ,FF> so motion ,.
k
FF=
() ( )
2
: 300 0.15 474 50 9.81 sin15 50
2.04 m s
Σ
=− − °=
=
xx x
x
Fma a
a
20 000g 10 000g
Solution 3.3
()
2
3.58 ft sec up
a
=
Solution 3.4
Solution 3.5
Woman and package:
50 lb each
60 (9.81) N
N
tr
Solution 3.6
()
22
21 2 1
2
−= −
vv axx
Solution 3.7
:mg sin 40 mg cos 40
6.31
F
or constant accel.
7.51
xx k
k
Fma ma
µ
µ
Σ
=°−°=
=−
0
2
1
2
0.0395
k
k
svt at
µ
=+
=
Solution 3.8
Bob
:sin mgsin (1)
Σ
xx x
FmaT ma
αθ
Σ
=+=
1500 (9.81) N
µ
k
mg cos 40°
x
mg
Mg
y
P
Solution 3.9
2
:
1
2.31 m s
3.6
xx
x
xu
Fma
a
Σ=
=
:
3.6
Σ=
xx
xd
Fma
Solution 3.10
Entire system:
300 000 (9.81) N
x
0.5°
300 000 (9.81) N
(M + m)g
Solution 3.11
()
750,000
: 4 40,000 32.2
Fma a
+←Σ = =
Solution 3.12
Let m be the mass of each car and 2m that of the locomotive.
100
100
32.2
Solution 3.13
() ()
100
;1cos30 0.25 100sin30 5
32.2
xx
FmaP N
Σ= + °− − °=
1.866 0.25 65.53
PN
−=
102 mg
ya = 5 ft/sec
2
x
P
100 lb 30°
Solution 3.14
()
−
=
362
9
2.84 10 km
Solution 3.15
()
()
2
: 53.7 26 2 cos 30 15
1.3
90 m s
Σ= − + + =
=→
xx
Fma a
a
50 (9.81) N
Solution 3.17
A
Solution 3.18
F = retarding force (drag, tire rolling resistance, etc.)
Solution 3.19
2400 lb
8‘2‘6‘12”
F
B
F
A
TT
2 T
a/2 = 10 ft
––––
sec
2
mg
Solution 3.20
1
2
BA
LS S
=+
=−+
() ()
2
11
1: 0.5 0.25m/s
22
AB
aa
=− =− =−
32.2
Solve and get
2
2.93 ft sec
a
=
T
3
Solution 3.22
Check for motion by assuming static equilibrium.
Solution 3.23
0.532 m s
=
a
T90 mm 160 N
yx
675
––––
2mm
R
Solution 3.24
Solution 3.25
()
()
2
23
300
:120 10 3.6
==
Dkv k
17.28 vxx−==
1
–
4lb
Solution 3.26
y
D
–1
250
Solution 3.28
532.2
xx B
Solution of Eqs. (1)–(3):
2
2
8.32 ft sec
36.1 ft sec
11.21 lb
A
B
a
a
T
=
=−
=
Solution 3.29
0
00 0
2
:
10 200 2
5 100
0.490 m/s
550 0
=−=
−=
=−
=
−=
x
FmxPkxmx
xx
xx
v
xx
x = 0 (initial condition) or x = 0.10 m or 100 mm
T
Eq. pos.
Solution 3.30
:cos sin
Fma F N m
a
θθ
Σ= − + =
12
xx
So
() ()
12 12
0.0577 g 0.745 gmm P mm+≤≤ +
y
Solution 3.31
00
2
0
2
0
g
yh v
yv
dv
m
vdv
dy kv
m
==
=
=−
+
Solution 3.32
sin120 sin 30 sin 30
AB
lss
°°°
==
With 0.289 m
AB
ss== and 0.4 m s : 0.4 m s==−
AB
vv
Diff. again:
()
22
22 22 0
A A A B B B AB A B AB AB
v sa v sa vv sa as vv+ ++ + +++ =
With 22
2 m sec , 2.37 m s==−
AB
a
a
60
: cos30 2(2) (Kinetic 1)
: cos30 3( 2.37) (2
:
)
sAc
B
FmaFT
Fma T
+
+
°
⎯
⎯→Σ = − °=
Σ= − °= −
Solution of Eqs. (1) and (2): 11.11 N
8.21 N
c
F
T
=
=
S
A
B
T
R
A
R
B
mg
Solution 3.33
2
02
gg
R
= (all pertaining to moon)
mgT
A
Solution 3.35
2
425mm
100 mm s
=
=
A
A
s
v
22 2 2
2
11
−
() ()()
113
22 22 2222
222
1
250 250 250
AA AAA AAA
ss sss sss
−−−
−
++ + − +
BB
Slider B:
1
1
250
tan 425
610 250
tan 40.3
425
B
s
α
−
−
−
=
−
==°
S
C
250
mm
x
y
T0.5 (9.81) N
Solution 3.36
2
=
Gm
Fx
B
Solution 3.38
22
1.4
: 0.120 0.200
nn
v
Fm R
r
Σ= =
2
5280
g32.2
v
ρ
==
0.120 (9.81) N
N = O
t
n
Solution 3.42
g = surface gravitational acceleration on earth
2
;mg , g
Σ
===
nn
Fma mr r
ωω
Solution 3.43
0: sin30 mg 0
y
FN
Σ
=°−=
Solution 3.44
()
=−0.0504 lbN
Ans. Magnitude N = 0.0504 lb
0.2 lb
Slider:
ω
y
Solution 3.45
mg (1)
y
T
=°
Solution 3.47
2
15 11.84 ft sec
v
35°
5 m
y
T