Solution 1.1
5.59 slugs
4.4482 N
81.6 kg
=
=
m
m
Solution 1.2
m
Solution 1.3
()
1
2
12 cos30 sin30
10.39 6
34
15 9 12
Vij
ij
Vijij
=+
=+
=−+ =−+
Solution 1.4
The weight of an average apple is
==
5lb 0.417 lb
12 apples
0.417 0.01294 slugs
W
W
Solution 1.5
()
8
2.68 10 N
−
=
Force is a vector quantity, so
()
8
2.68 10 cos35 sin35
FF i j
−
== − −
FCu
y
2R
Solution 1.6
0
1
2
h
mg mg
=
=
Solution 1.7
()
γ
=+
=+
=
== =
24
rel
2
2
abs abs
0.03382 cos
9.797 337 0.03382 cos 35
9.820 031 m/s
60 9.820 031 589 N
gg
Wmg
rel
Solution 1.8
()
()()
e
h2
823
3.439 10 4.095 10 29.9 ft/sec
Gm
gRh
−
=+
××
32.17 l
4sugs
Absolute weight at 150 miles:
h
=
()()
6.22 29.9 186.0 lb
hh
Wmg== =
Solution 1.9
se
se
22
12
,
Gmm Gmm
FF
dd
==
22
2
e1 s2
se
mx mx
hmm
−
=
−
With
()
ese
5.976 10 kg, 333,000mmm==
, and x1 and x2 as above:
() ()
85
1.644 10 or 1.644 10 kmhm=
m
Fs
e
Fe
S/C
(1496 – 2)108 m
(x2)
d12 = x12 + h2
Solution 1.10
()
SJ
r
1.770
θ
=
(We could have used the small-angle approach SJ
dr
θ
=!)
m
Solution 1.11
()
em
20
3.84398 10
×
Force exerted by sun on moon:
()
()()
()()
()
2
11 24
sm
SA 22
11 8
es em
6.673 10 5.976 10 333,000 0.0123
1.496 10 3.84398 10
Gm m
Frr
−
××
==
+×+ ×
Solution 1.12
2
1
t
t
E mgr dt
=
[
]
[]
2
2
2
2
m/s( )
U.S. : slugs.ft sec (not base)
1
SI :
b-sec ft sec 1b-ft-sec
f
kg b
t
.ase
=
=⋅=
=
E
E
Solution 1.13
2
1
Q
ρυ
=
A
B
Sunlight
r
es
r
em