Problem 11.56
B
a
Given
Xa YB
E.EE
x
d
74
Is
c
n
Collars A and B start from rest so at
ᵗx atePIE
mm pfrt.to
time
d Xa
Xa Xa
d 24A Xp
4
2X
0
PserposfYfErtdownanwdarads
the length of the cable which is
i
4
Xc
be the position
constant is given by
B
c
Differentiate for velocity and again for acceleration
25A TB 4T
26A Of 48C
a
We can now
solve for acceleration of block C
22A ñB 4a
4
E
28A IB
At
Tec
32
4 1.5
J
If
Bdx
FEB
E
ñ
a
TB
TB
16
TB XB
XB
constant acceleration
f
Idt
52
L
Ñ
TBo
a
IB
Is
16mm s
Using the acceleration found above for block
b
fift's
feat fax
E at
Vc
4t
f
du
t
42
for its distance
8
F
fia.at fidu
4
we can solve
4t
at
v
2
3
2
213
dx̅
dt
t
ft
x
ft
Xc
3
É
Xc
Xc 0
xc 7.875m
Problem 11.125
Given
EEEi's
8m
f
Horizontal motion of the ball
Fit
Tx Tx x̅B x̅o Txt exot
a
EE
At ymax 8m
2
5
219.81
0
8
E
556.96
ix
12.53m s
0
i
from
Ty
0
12.534
0 12.53
we
8 0
At the time of the catch y
0
Vy Zay T
4.9055
9 81
2
o
0
i
from
J T.tvy.tt fayt
12.53 4.9052
474
2.55s
53
get
we
get
Motion of
the deck
ix
txottb xp
xottxottf
p notionofbanreiativet.th
JB Jp
TBD
x̅
x
Box
apt
t.ie
IB ID
ix t
ñpt
IBD
Tx tapt
Box
TBDx E f
dean
ix
At the time of the catch d x̅Bp
a
d
age
t
thrown
qq.EE
IIg
E
Ñpt
1
21 0.3
2.55
0.9756
1
b The relative velocity of the ball with respect to the deck is given
Box
TBDx
by
apt
0.3 2.55
TBDx 0.765mn
Problem 11.142
If
400m
Ta
JB
A
450km
h
aa 8m52
300m
at
ten
a
A
af
160
v
rm
3ms
tal and normal components
450km h
197
0s
540kmh
197
0s
8m s
feet 3m15
125m15
150m s
I
a
Velocity of B relative to A
TBA
TB
ÑB TBet
a
break the velocities down into
VBA
VBCOS60 VA
Vola
15000560
VBA
129.9m s
129.90 m s
TBA 1200
TBA
1505in60 0
VBA
UBlax 200m15
and y componets
Vesin Got Vay
VBA
125
x
Tt29.9
ÑBA 238.48m
b Acceleration of B relative to A
Tpa
Ip Ga
et
IB
en
are
of TB is given need to find
are
air 75m15
Tangental component
In
Bla
guff
abt
app
abtcosbo
Galax
300560
aBax
58.45m52
GBA
EBIA
en
normal component
x
and y components
abncoszotteaxtblay abtsin60 abns.in30 day
750530
58.451 34.90 m s
452 34.92
ÑBA 68.08m
aBn
da
Break acceleration down into
ABlax
et
8
GBA
35in60 755in30 0
GBa
34.9mm
Problem 11.158
Normal Acceleration of the Moon
Earth
r
Efsatellite
r
384103km
moon
63okm
E
ñn
g
in
a.at EY
In
2.7 103m sa
j
We know that an is also given
by
at
where p
r
in this case
can
i
we
find the speed of
the Moons orbit using this equation
8
a
32
air
I
7 105384 106
w̅ 1018.23m s
8
3665.630116
The speed of the Moon relative to the Earth is
3665.63km h
Problem 11.163
fr
Transverse and Radial components
a
zoom
n
IT
min
e
9
13
so
0
8.33m's
zoom
so
0.0349rad s
0 25 2
velocity of parasailer relative to boat
Jam
rertrifeo
TaB
Oer
200 0.0349
VAB 6.9816ms
1850m
3 05
Acceleration of parasailer relative to boat
2
r
zero eo
TAB CO 200 0.0349
ert 200
i
TAB
r
er
20 0.0349
o
0.2436hm52
GAB
Convert to rectangular components
Ja
6.9815in30 i
Tay
3.4911 6.0460 m s
Since
i
Jay TA JB
JA 8.3331 3.4911
JA
and b
c
ñA
Gay 0.2436L cos30 0.24365in300
QaB10.2110 0.12180 m s2
6.98120530
1.8241 6.0460
TB
0
at
we
can
write
Ja JAB JB
6.0460
m s
GAB
0.21101 0.12180 m s2
TA T 4
Ñ 13.28n
6.0462
velocity of parasailer
Tea
IA
10512182
0.2436m1
accelerationofparasailer
0
