Solution 2.161
From Prob. 2.160…
/
.
==
AB r
r
aae
()()
2
2
2 2
2.73 2.78 . sin19.75 cos19.75 1.691 m/s
−− − °+ °=−
ij i j
e
θ
B
30°
B
t
υ
B
θ
𝛽
Solution 2.162
υBoυAo
STOP
00
2
2 2 2 120
36.7
AAA A
×
=−= − =
2
22
22
1 36.7 1
102.9 29.4 76.3 58.7 18
AB AA A BB B
xx xvt atxvt at
tt tt
′′
=→+Δ− Δ=+Δ− Δ
+Δ− Δ= +Δ− Δ
2
1
()
58.7 18 1.232 36.5 ft/sec
mi
BBB
vvat
′
Solution 2.163
C
y
x
υA/B
υB
350 m
/
30 km/hr 8.33 m/s
==
B
ABAB
vvvv
31.3
=
AC
α
or 12.82 s
74.3
=
AC
α
If sitting still,
0
350 70 cos
CAxAC AC
xxvt t
α
=+ → =
Solution 2.164
Use
2
gg
o
R
=
2
2
3959
For B,
g 32.23 0.733 ft/sec
3959 22,300
B
==
+
()
/
0.733 29.2
BA B A
aaa i j
=−=+ −−
Solution 2.165
45
100 cos 21sin30
45
t
α
=+ °
21t
100t
45′
30°
Solution 2.166
A
r = 300 m
θ = 45°
υ
A
= 60 km/hr υ
A
= υ
B
+ υ
A/B
45°135°
β
υ
A/B
υ
A
υ
B
Solution 2.167
Solution 2.168
Find flight time t:
=+ − =+ °−
0
22
0
1g : 7 3 100sin30 16.1
2
y
y
yvt t t t
Solve to obtain 0.0822 sec (discard) and =3.02t sec
Range
()
0
00 100 cos30 3.02 262 ft
x
Rx vt=+ =+ ° =
Fielder must run 262−200 = 41.8 ft
in (3.02−0.25) sec
==
41.8 15.08 ft/sec
2.77
B
v
Velocity components of ball when caught:
()
()
0
0
/
100 cos30 86.6 ft/sec
g 100 sin30 32.2 3.02 47.4 ft/sec
86.6 47.4 15.08
71.5 47.4 ft/sec
xx
yy
AB A B
vv
vv t
vvv i j i
ij
== °=
=−= °− =−
=−= − −
=−
Solution 2.169
/
=−
BA B A
vvv
()
()( )
/
2
1.2sin30 12000 2 120.3 0.00579
=+
°= + −
BA
arr
θ
θθ
θ
()
32
0.1660 10 rad/s
−
=
θ
θ
r
30°
A
x
Solution 2.170
()
/
(a) 50 50 50 50 m/s
AB A B
vvv i j ij
=−= −− = +
()
2
/
//
2
sin 45
2 70.7
AB
BA BA
n
r
v
aa
ρ
=°=
β
A
x
t
aA/B
Solution 2.171
Length of cable =++3constant
AB
LSS
Differentiate:
6constants
0 6 so… 6
AB
AB B A
Lyy
Lvv vv
=++
== + → =−
Solution 2.173
Cable length 2 3 constants
AB
LS S=++
02 3 (1)
AB
vv
=+
A
S
A
S
A
Solution 2.174
A
B
L
Solution 2.176
Length of cable is
velocity 20
Solution 2.177
Solution 2.179
/
2
/
(6) 4 ft/sec
33
3 2 1ft/sec
AB
BA B A
BA B A
aa
vvv
== =
=−=−=
Solution 2.180
3
3 degrees of freedom
C
1
, C
2
, C
3
are constant lengths
CB
y
A
y
D
y
2
y
1
()
111 1
;02
BA B BA
Ly y yy C y yy
=++ −+ = +−
Eliminate +++=
12
&&get4 8 4 0
AB CD
yy v v vv
C
D
A
Solution 2.181
Retain , , , .
BB AA
xx y y
With (1), (2), and (3):
[][]
22 00(5)
AA
BAB AA
A
yy
xRyx Ry y
x
−− + +−− −=
At position shown, sin60
cos60
A
A
xR
yR
=− °
=− °
With 2 m/s, 1.070
BB
xxR==
(from (4)), then
from (5) we obtain 2.39 m/s
A
y=−
()()
cos 60 2.39
From (2) sin60
1.382 m/s
AA
A
A
R
yy
xxR
−°−
=− =− −°
=
22
Then 2.76 m/s
AAA
vxy=+=
Solution 2.183
Cable length is
22
2
AB
Ly y b
=+ +
AB
yb
Solution 2.184
22
222
2
ss xs
Lx x s
=+ + =+ +
BA
xs
xs
B
y
A
2b
Solution 2.185
Let A be a point on cable 1, and let B be the
2(2)
WB
Ovv
=−
Combine (1) & (2): 6 0
AW
vv+=
Solution 2.186
Cable is reeled in at a constant rate l.
1
−
M
A
2
b b
Solution 2.187
Solution 2.188
()
222
2
22cos
21cos
sbb
b
θ
θ
=+
=+
Eliminate
θ
& get
() ()
2
21 cos
cos sin tan
B
AB
y
l
sv
vl v
bb
θ
θθθ
+
==
B
b
θ
(υ
)
υ
B
n
B
υ
B
Solution 2.189
SA = 425 mm
Length of cord L = 1050 mm
2
222
22
or
()
22 2
2
250 02
AA B B
AA
Sv S v
Sv
+−
+=
B
S
B
C
250
mm
Solution 2.190
From the solution to Prob. 2.189, the velocity constraint equation is
1
12
22 2 2
2
250 250 250 0
Sv S Sv S v S S
−
−
+ + +− ×+− =
3
AB AAB B
−
B
Solution 2.191
−
=−+
0.4 2
86
t
se tt
dt
2
0.4
t
()
()
0.4 0.617
3.2 6 2 0.617
7.27 m/s
ve
−
=− − +
=−
Solution 2.192
10.60 3.48 0.85
t
=