BB
()
()
2
:0 9
:g 10
x
yB
xGxB
yGyf B
FmaBN
FmaBF m ma
ωω
=−=
=+−=−
L
=
,max
35.3 lb
A
f
F
=
,max
10.88 lb
B
f
F
=
16.06 rad sec CCW
B
α
()
()
2
4.01 ft sec Down
4.01 rad sec CW
y
G
a
α
aiaj aj k Lij
GG B
xy
+=−+×− −
()
⎡
⎣
⎢⎤
⎦
⎥
2cos sin
N
B
α
B
r
k
B
x
m
g
B
Solution 6.181
υψ θ ω φ ω
=− =
: cos 0.6 cos 0.5 cos 0.653 m s CCW
BAB
AB
j
22
2
nn
BBC BCBC
BBC B
2
B A AB B A AB B A
aa r r
αω
=+ × −
()
ω
== = =
2
22
0 0.2 3 1.8 m s
tn
AA OA
a
aOA
()()()
2
sin cos 0.531 cos sin 1.8 cos sin
t
B
aij ij ij
ψψ ψψ θθ
−+ +− − =−−++
t
A
B
500 mm
200 mm
x
y
+
θ
ψ
υ
A
υ
B
a
At
a
Bt
a
Bn
a
An
ϕ
ω
22
t
BBC BCBC
()
2
0: 0.200 sin 0
OAB
MM F
θφ
=− +=
C
B
m
= 8 kg
k
c
= 450 mm
O
y
C
y
C
x
ψ
ϕ
Solution 6.182
For
θ
=
θ
max/2 = 19.95° [State (3)]
θθ
22
max max
11
b
C
m
1
= 1.2 kg b/4
CB
G
υ
h
Solution 6.183
υ
ωω
=Δ = =
222
111
:4
C
W
TT I
()
44.91 1 cos cos ftsec
19.66 cos 1 cos ft sec
θθ
θθ
=−
=−
υAmax = 7.57 ft/sec
θ = 48.2°
0
2
4
6
8
Velocity of end A, ft/sec
Solution 6.184
αθ
=
2
23
3gsin
2
OO
L
()
ωω θθω θ
==−
00
sin , 1 cos
2
3g 3g
4
dd
LL
LL
()
θθ
=− + −
2
22 2
3g 12cos 3cos
4
3g 3 cos 1 cos 1
O
y
G
n
t
θ
L
––
2
L
––
–
L
–
L
2
Increasing in Oy from
θ
= 70.5° to
θ
= 90° reflects the fact that, in
a
θ
is increasing, showing the
0
–1.5
–0.5
0.5
20 40
∠θ°
O
y
/mg
O
x
/mg
60 80 100
Solution 6.185
12 96
θ
ωω θ
θ
== +
=
0
d
t
θπ
=
2
y
Solution 6.186
22
2
+− =
(1)
533 2 2
2
ωθ θ
or
()
ωθ
=−+−−
(2)
Solution 6.187
Solution 6.188
y
+
aBt
aBn
aAn
A
OA = 80
O
B
240
θ
ωOA, αOA
200
ϕ
ψ
70
()()()
ψψα φφω φφ
−− +× − − −
2
2
cos sin 0.240 cos sin 0.240 cos sin
tn t
n
BAB AB
a
ijk ij ij
2
tnt n
A A B B AB AB
12
AB AB
(12)
22
RAA=+
(13)
max 3.93 N m at 145.6M
80 mmOA =
Oy
Ox
O
M
θ
a
max 91.7 N at 184.2
BC
A
0
1
2
3
4
Plot of crank torque M for one complete crank revolution at 60 rev/min
100
90
Plot of pin forces for one complete crank revolution at 60 rev/min
Crank rotation angle,
θ
, deg
0 40 80 120 160 200 240 280
RB RA
320 360
Solution 6.189
aBt
aAt
B
ψ
υB
Ax
Ay
A
M
θ
()
θθ ψψαφωφ
−− =− − + −
2
: sin cos sin cos 0.240 sin 0.240 cos 5
n
tn t n
BAB AB
A A B B AB AB
ia a a a
GAABGAABGA GA
αφ φ φψ φψ
=−− −
2
: 0.120 cos 0.120 sin 0.120 cos sin 0.120 sin cos
111
12
y
yGBC AByABGy
G G y x BC BC
AB AB
MI A A F F
(12)
22
RAA=+
(13)
max 3.61 N m at 360M
A
y
F
BC
m
AB
g
A
x
G
B
ϕψ
120 mm 120 mm
x
y
+
max
73.7 N at 185.3
BC
F
θ
0
–4
–2
–1
0
1
2
3
4Plot of crank torque M for one complete crank revolution from 0 to 60 rev/min
Input torque, M, N
40 80 120 160 200 240 280 320 360
100
90
Plot of pin forces for one complete crank revolution from 0 to 60 rev/min
Crank rotation angle, θ, deg
0 40 80 120 160 200 240 280
RB RA
320 360
Solution 6.190
a
Bt
2and 0.120 cos sin
30.100 cos sin
()
()
()
()
αϕϕ ψ α
=−=
=−−=
=−−=
3,
3,
2
:13
:g 14
1
:0.2 sin cos 0.1 gcos 15
3
xx
yy
xGxxBCG
yGyyBCBCG
C C x y BC BC BC
FmaCBma
FmaCBm ma
MI B B m mBC
mBCg
100 mm
By
Bx
G2
Cy
C
B
