Problem 11.

56

Given
Xa YB
B 74
a
d
x E.EE Is
c
n

Collars A and B start from rest so at time 4 0

be the position
ᵗx atePIE mm pfrt.to PserposfYfErtdownanwdarads

i the length of the cable which is constant is given by

d Xa Xa Xa 2X Xc B
d 24A Xp 4 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 a J If constant acceleration

4 28A IB
E
FEB
Bdx
f Idt

E At TB
TB 52 L
Tec 32 16 TB XB XB Ñ TBo

ñ 4 1.5
a

IB Is 16mm s
b Using the acceleration found above for block we can solve for its distance

fift's F 8
fia.at fidu feat fax
4 E at du
f 4t t dt dx̅

at
42 t v ft x
É
Vc 4t 3
2 2
ft Xc Xc

213 3 Xc 0
xc 7.875m

Problem 11.125

Given

EEEi's
8m

f Horizontal motion of the ball

Tx Tx x̅B x̅o Txt exot

a
Fit
EE
At ymax 8m 5 0 i from Ty Vy Zay T we
get
2
0 8 219.81 8 0

E 556.96
ix 12.53m s

At the time of the catch y o i from J T.tvy.tt fayt we get

2
0 0 12.534 9 81 0 12.53 4.9052
4.9055 53
0 12.53
474
2.55s
Motion of the deck

ix
txottb xp
xottxottf
p notionofbanreiativet.th

TBDx E f IIg
dean
t.ie qq.EE
JB Jp TBD IB ID E
thrown
Box x Tx tapt x̅ ix t ix t age
Box apt IBD ñpt

a At the time of the catch d x̅Bp Ñpt

1
d 21 0.3 2.55

0.9756
1

b The relative velocity of the ball with respect to the deck is given by
Box apt
TBDx 0.3 2.55

TBDx 0.765mn

Problem 11.142
tal and normal components
400m If
A Ta 450km h 197 0s 125m15

450km
h JB 540kmh 197 0s
150m s

aa 8m52 300m
at 8m s
ten

a
rm
3ms
feet 3m15 I
A

af
160
v
a Velocity of B relative to A
TBA TB a ÑB TBet
break the velocities down into x and y componets
VBA VBCOS60 VA VBA Vesin Got Vay
Vola 15000560 125 VBA 1505in60 0

UBlax 200m15 VBA 129.9m s

TBA 1200 129.90 m s

TBA Tt29.9
ÑBA 238.48m

b Acceleration of B relative to A

Tpa Ip Ga IB et en are et aBn en

Tangental component of TB is given need to find normal component

In guff are air 75m15

Bla abt app da

Break acceleration down into x and y components
ABlax abtcosbo abncoszotteaxtblay abtsin60 abns.in30 day
Galax 300560 750530 8 GBA 35in60 755in30 0

aBax 58.45m52 GBa 34.9mm

GBA 58.451 34.90 m s

EBIA 452 34.92

ÑBA 68.08m
Problem 11.158

Normal Acceleration of the Moon

ñn g E
in a.at EY j
r
Efsatellite
In 2.7 103m sa
Earth r 384103km moon
We know that an is also given
63okm
by
at where p r in this case

i we can find the speed of

the Moons orbit using this equation

a 8 7 105384 106
3 05
32 air w̅ 1018.23m s 1850m

I 3665.630116
8
The speed of the Moon relative to the Earth is 3665.63km h
Problem 11.163

fr Transverse and Radial components

a

zoom

n
e
so

IT min 9 13 0
8.33m's

zoom

so
0 25 2 0.0349rad s

velocity of parasailer relative to boat

Jam rertrifeo VAB 6.9816ms
TaB Oer 200 0.0349
Acceleration of parasailer relative to boat
2
TAB i r er r zero eo
TAB CO 200 0.0349 ert 200 o 20 0.0349
GAB 0.2436hm52
Convert to rectangular components

Ja 6.9815in30 i 6.98120530 Gay 0.2436L cos30 0.24365in300

Tay 3.4911 6.0460 m s QaB10.2110 0.12180 m s2
Since Jay TA JB we can write Ja JAB JB
i JA 8.3331 3.4911 6.0460

JA 1.8241 6.0460 m s

and b c TB 0 at GAB
ñA 0.21101 0.12180 m s2

TA T 4 6.0462 Tea 10512182

Ñ 13.28n IA 0.2436m1

velocity of parasailer accelerationofparasailer