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PHYS 2101
HW#8 WileyPlus Problem Solutions
1
HW #8 Solutions – or at least most of them…
(USING BOOK VALUES 9
th
Edition in problems; problem # indicated by [])
2) [136] The gravitational forces on
m
5
from the two 5.00
g masses
m
1
and
m
4
cancel each other.
Contributions to the net force on
m
5
come from the remaining two masses:
The force is directed along the diagonal between
m
2
and
m
3
, toward
m
2
. In unitvector notation, we have
.
3) [139] Both the Sun and the Earth exert a gravitational pull on the space probe. The net force can be
calculated by using superposition principle. At the point where the two forces balance, we have
, where
M
e
is the mass of Earth,
M
s
is the mass of the Sun,
m
is the mass of the
space probe,
r
1
is the distance from the center of Earth to the probe, and
r
2
is the distance from the center
of the Sun to the probe. We substitute
r
2
=
d
!
r
1
, where
d
is the distance from the center of Earth to the
center of the Sun, to find
Using the values for
M
e
,
M
s
, and
d
given in Appendix C, we take the positive square root of both sides to
solve for
r
1
. A little algebra yields
Note: The fact that
indicates that the probe is much closer to the Earth than the Sun.
4) [1310] Using Eq. 131, we find
F
AB
"
=
2
Gm
A
2
d
2
j
^
and
F
AC
"
=
–
4
Gm
A
2
3
d
2
i
^
.
Since the vector sum of all three forces must be zero, we find the third force (using magnitudeangle
notation) is
F
AD
"
=
Gm
A
2
d
2
(2.404
#
–56.3º) .
This tells us immediately the direction of the vector
r
"
(pointing from the origin to particle
D
), but to
find its magnitude we must solve (with
m
D
= 4
m
A
) the following equation:
2.404
$
%
'
(
)
Gm
A
2
d
2
=
Gm
A
m
D
r
2
.
This yields
r
= 1.29
d
.
In magnitudeangle notation, then,
r
"
= (1.29
#
–56.3º) , with SI units
understood. The “exact” answer without regard to significant figure considerations is
(a) In (
x, y
) notation, the
x
coordinate is
x
= 0.716
d
.
(b) Similarly, the
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 Spring '07
 GROUPTEST
 Physics, Force, Mass

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