358
CHAPTE R 13 GRAVITATION
Force 12 is directed in the positive direction of the y axis (Fig.
13-4b) and has only the y component F 12 . Similarly, 13 is directed in the negative direction of the x axis and has only the x
component ϪF 13 (Fig. 13-4c). (Note something important: We
draw the force diagrams with the tail of a force vector anchored on the particle experiencing the force. Drawing them
in other ways invites errors, especially on exams.)
To find the net force 1,net on particle 1, we must add
the two forces as vectors (Figs. 13-4d and e). We can do so on
a vector-capable calculator. However, here we note that
ϪF 13 and F 12 are actually the x and y components of 1,net .
Therefore, we can use Eq. 3-6 to find first the magnitude and
then the direction of 1,net .The magnitude is
ϭ 4.1 ϫ 10
Ϫ6
N.
(Answer)
Relative to the positive direction of the x axis, Eq. 3-6 gives
the direction of 1,net as
Is this a reasonable direction (Fig. 13-4f)? No, because the
direction of 1,net must be between the directions of 12 and
13 . Recall from Chapter 3 that a calculator displays only
one of the two possible answers to a tan
Ϫ1
function. We find
the other answer by adding 180°:
Ϫ76° ϩ 180° ϭ 104°,
(Answer)
which is a reasonable direction for 1,net (Fig. 13-4g).
F
:
F
:
F
:
F
:
ϭ tan
Ϫ1
F 12
ϪF 13
ϭ tan
Ϫ1
4.00 ϫ 10
Ϫ6
N
Ϫ1.00 ϫ 10
Ϫ6
N
ϭ Ϫ76Њ.
F
:
ϭ 2(4.00 ϫ 10
Ϫ6
N)
2 ϩ (Ϫ1.00 ϫ 10
Ϫ6
N)
2
F 1,net ϭ 2(F 12 )
2 ϩ (ϪF 13 )
2
F
:
F
:
F
:
F
:
F
:
Sample Problem 13.01 Net gravitational force, 2D, three particles
Figure 13-4a shows an arrangement of three particles, particle 1 of mass m 1 ϭ 6.0 kg and particles 2 and 3 of mass m 2 ϭ
m 3 ϭ 4.0 kg, and distance a ϭ 2.0 cm. What is the net gravitational force 1,net on particle 1 due to the other particles?
KEY IDEAS
(1) Because we have particles, the magnitude of the gravitational force on particle 1 due to either of the other particles is
given by Eq. 13-1 (F ϭ Gm 1 m 2 /r
2
). (2) The direction of either
gravitational force on particle 1 is toward the particle responsible for it. (3) Because the forces are not along a single axis, we
cannot simply add or subtract their magnitudes or their components to get the net force. Instead, we must add them as vectors.
Calculations: From Eq. 13-1, the magnitude of the force 12
on particle 1 from particle 2 is
(13-7)
ϭ 4.00 ϫ 10
Ϫ6
N.
Similarly, the magnitude of force 13 on particle 1 from
particle 3 is
(13-8)
ϭ 1.00 ϫ 10
Ϫ6
N.
ϭ
(6.67 ϫ 10
Ϫ11
m
3
/kgиs
2
)(6.0 kg)(4.0 kg)
(0.040 m)
2
F 13 ϭ
Gm 1 m 3
(2a)
2
F
:
ϭ
(6.67 ϫ 10
Ϫ11
m
3
/kgиs
2
)(6.0 kg)(4.0 kg)
(0.020 m)
2
F 12 ϭ
Gm 1 m 2
a
2
F
:
F
:
Additional examples, video, and practice available at WileyPLUS
on the particle. In this limit, the sum of Eq. 13-5 becomes an integral and we have
(13-6)
in which the integral is taken over the entire extended object and we drop the
subscript “net.” If the extended object is a uniform sphere or a spherical shell, we
can avoid the integration of Eq. 13-6 by assuming that the object’s mass is
concentrated at the object’s center and using Eq. 13-1.
F
:
1 ϭ ͵dF
:
,
Checkpoint 2
The figure shows four arrangements of three particles
of equal masses. (a) Rank the arrangements according
to the magnitude of the net gravitational force on the
particle labeled m, greatest first. (b) In arrangement 2, is
the direction of the net force closer to the line of length
d or to the line of length D?
d
D
m
(1)
m
d
D
(2)
(3)
(4)
d
D
m
m
D
d
CHAPTE R 13 GRAVITATION
Force 12 is directed in the positive direction of the y axis (Fig.
13-4b) and has only the y component F 12 . Similarly, 13 is directed in the negative direction of the x axis and has only the x
component ϪF 13 (Fig. 13-4c). (Note something important: We
draw the force diagrams with the tail of a force vector anchored on the particle experiencing the force. Drawing them
in other ways invites errors, especially on exams.)
To find the net force 1,net on particle 1, we must add
the two forces as vectors (Figs. 13-4d and e). We can do so on
a vector-capable calculator. However, here we note that
ϪF 13 and F 12 are actually the x and y components of 1,net .
Therefore, we can use Eq. 3-6 to find first the magnitude and
then the direction of 1,net .The magnitude is
ϭ 4.1 ϫ 10
Ϫ6
N.
(Answer)
Relative to the positive direction of the x axis, Eq. 3-6 gives
the direction of 1,net as
Is this a reasonable direction (Fig. 13-4f)? No, because the
direction of 1,net must be between the directions of 12 and
13 . Recall from Chapter 3 that a calculator displays only
one of the two possible answers to a tan
Ϫ1
function. We find
the other answer by adding 180°:
Ϫ76° ϩ 180° ϭ 104°,
(Answer)
which is a reasonable direction for 1,net (Fig. 13-4g).
F
:
F
:
F
:
F
:
ϭ tan
Ϫ1
F 12
ϪF 13
ϭ tan
Ϫ1
4.00 ϫ 10
Ϫ6
N
Ϫ1.00 ϫ 10
Ϫ6
N
ϭ Ϫ76Њ.
F
:
ϭ 2(4.00 ϫ 10
Ϫ6
N)
2 ϩ (Ϫ1.00 ϫ 10
Ϫ6
N)
2
F 1,net ϭ 2(F 12 )
2 ϩ (ϪF 13 )
2
F
:
F
:
F
:
F
:
F
:
Sample Problem 13.01 Net gravitational force, 2D, three particles
Figure 13-4a shows an arrangement of three particles, particle 1 of mass m 1 ϭ 6.0 kg and particles 2 and 3 of mass m 2 ϭ
m 3 ϭ 4.0 kg, and distance a ϭ 2.0 cm. What is the net gravitational force 1,net on particle 1 due to the other particles?
KEY IDEAS
(1) Because we have particles, the magnitude of the gravitational force on particle 1 due to either of the other particles is
given by Eq. 13-1 (F ϭ Gm 1 m 2 /r
2
). (2) The direction of either
gravitational force on particle 1 is toward the particle responsible for it. (3) Because the forces are not along a single axis, we
cannot simply add or subtract their magnitudes or their components to get the net force. Instead, we must add them as vectors.
Calculations: From Eq. 13-1, the magnitude of the force 12
on particle 1 from particle 2 is
(13-7)
ϭ 4.00 ϫ 10
Ϫ6
N.
Similarly, the magnitude of force 13 on particle 1 from
particle 3 is
(13-8)
ϭ 1.00 ϫ 10
Ϫ6
N.
ϭ
(6.67 ϫ 10
Ϫ11
m
3
/kgиs
2
)(6.0 kg)(4.0 kg)
(0.040 m)
2
F 13 ϭ
Gm 1 m 3
(2a)
2
F
:
ϭ
(6.67 ϫ 10
Ϫ11
m
3
/kgиs
2
)(6.0 kg)(4.0 kg)
(0.020 m)
2
F 12 ϭ
Gm 1 m 2
a
2
F
:
F
:
Additional examples, video, and practice available at WileyPLUS
on the particle. In this limit, the sum of Eq. 13-5 becomes an integral and we have
(13-6)
in which the integral is taken over the entire extended object and we drop the
subscript “net.” If the extended object is a uniform sphere or a spherical shell, we
can avoid the integration of Eq. 13-6 by assuming that the object’s mass is
concentrated at the object’s center and using Eq. 13-1.
F
:
1 ϭ ͵dF
:
,
Checkpoint 2
The figure shows four arrangements of three particles
of equal masses. (a) Rank the arrangements according
to the magnitude of the net gravitational force on the
particle labeled m, greatest first. (b) In arrangement 2, is
the direction of the net force closer to the line of length
d or to the line of length D?
d
D
m
(1)
m
d
D
(2)
(3)
(4)
d
D
m
m
D
d