Solution 4.70
θ
Σ= − =
:sin
xx x
F ma T mg ma
2
0
S
o( )sin( )
o
dv
mu m mt g m mt dt
θ
′′ ′
−− =−
Solution 4.71
Sol. I: entire chain
p
d
p
k
p
xx
ρgx
Solution 4.72
Let 0
m= initial mass of car =3
25(10 ) kg
4(10 ) kg/ sm
0
=− ==
++
0
00
000
v
dt v
vmmt dtmmt
Solution 4.73
Linear impulse – momentum applied to this mass element:
Δ
t
F
m′
–––
υgL
Solution 4.74
Eq. 1 of the previous solution still holds, so
′
m
Combine Eqs. 1 and 2:
′Δ=Δ −Δsin 2 sin
mgL t mv m gh
Solution 4.75
22
F = O
Solution 4.76
=− =− =
++
===
+=
0000
00
;ln ln ,
or and for 2 , 2
2
v
v m px v m v m px
vv
vxLv
xv
m+−
=− ± + =
0
0
2
0
2
,
2
8
14 ,
2for + root
tx x
pp
vtp
m
pm
mv t
mm
xx
ppp
Solution 4.77
()
() ()
++
Σ
==
=
2
2234
22
13
ii
mvj mvk mvi
mr
rmm
mv
Solution 4.78
From Eq. 4.10 with P replaced by
=+×Σ
=−×Σ
0
0
0:
or
Gi
Gi
HHrmv
HH rmv
dv
()
()
=− +
77
223
7
Fd ij
Solution 4.79
12
o
F = 20 lb
x
Solution 4.82
=Δ Δ=
‘: cos20
xx
Fmv v v
υ
Solution 4.85
5.2 kg/s
m
m
′
=
−=−
Solution 4.86
120(1 0.866) 224 ft / sec
0.713 224 159.8lb
F
=+ =
=×=
Solution 4.87
Δ =Δ = =Δ =−Δ
′
0so
eg
TV U V Q
Energy is lost in the generation of heat and sound upon impact of rope with fixed guide.
y
G
r
–
30°
υ
Solution 4.88
R
2
R
1
T
Solution 4.89
Take entire chain as system (constant mass)
2
22 2 2
22
,, ( ) ()
,
[( ) ( )( ) ]
1
[( ) ( ) ]
2
3()
2
x
x gy aG pL x yx px yy
xgty at
p
gLRp xy xyxy Lx
p
agt agt Lg
pa g t p
== = −−−+
==
−= −+ −+ + +
=−+ − + +
=− + +
22
3
so ( )
2
L
g
Rpagt=+
Solution 4.90
System is conservative, so 0.
g
VTΔ+Δ=
−+ = ==
222
10, ,
xg g
p
gx pLx x x x x
()
== = =a acc ol se,
ggg g
xx xa
LL
ax
LL
()
==−
=−
b:()
(1 )
g
FmaT pLx x
L
x
T pgx L
Check from vertical part
()
−= = −
== ==
22
,1,OK.
c: ,,
22
vL
xoo
gx
pgx T px x T pgx
LL
gvgL
vdv a dx vdv xdx v gL
LL
x
yR
T
T
ρgx
ρg(L – x)
ρg(L – x)
Solution 4.91
()
()()
=
××
µ×
′== =
×
==
5000 231/ 60 1728
62.4 5000 231 21.6 lb sec/ft
32.2 60 1728
so 21.6 127.7 8 /12 1837 lb ft
QAv
mQ
g
M
Solution 4.92
With neglect of mass of pulley and weight of small portion of chain in contact with pulley
≈==
12
0so
o
MTTT
H
So
() () ()
=− + − = −
,4
hh
TpHh gpgHhTpgH
HH
Pulley and chain on it: steady flow gives
[
]
()
()
()
′
=Δ −= −− = −
===
2
22
00
:2 , 2 2 5
But , , 6
yy
vh
Fmv TRpvv vRT pv
ggg
vdv hdh v h v h
HHH
Substitute (4) and (6) into (3) and get
=−− =−
22
22
22,2
hg h
R pgH p h R pgH
HH H
pA = T
C
R
Solution 4.93
Solution 4.94
For entire rope of constant mass
() ()
−
=−=−
−
2
22
2or (2 )
g
d
xLx x
xgxLxLxg
Lx x
Substitute into (1) and get
()
[
]
[
]
−− −
−
22 2
2
x L xx x L xx
xLx
d
Differentiate 2
x and get
()
=+ −
2
2
1.
xg
Lx Substitute and get
−
2
2
x
xL
()
2
L
pgL