WEBVTT
1
00:00:00.790 --> 00:00:04.160 A:middle L:90%
There are two gravitational forces acting on particle. A
2
00:00:04.480 --> 00:00:07.960 A:middle L:90%
one off the gravitational forces comes from particle B and
3
00:00:07.960 --> 00:00:10.960 A:middle L:90%
another one from particle. See, So I disabled
4
00:00:11.160 --> 00:00:15.320 A:middle L:90%
have be a there there additional force acting on particle
5
00:00:15.330 --> 00:00:18.210 A:middle L:90%
A as an effect off the mass off particle Be
6
00:00:18.579 --> 00:00:22.730 A:middle L:90%
similarly I labeled f c A. The magnitude off
7
00:00:22.730 --> 00:00:26.800 A:middle L:90%
the gravitational force acting on particle A because off the
8
00:00:26.800 --> 00:00:29.820 A:middle L:90%
mask off particle See, now we proceed to company
9
00:00:29.820 --> 00:00:33.060 A:middle L:90%
to the magnitudes off these forces the magnitude off the
10
00:00:33.060 --> 00:00:38.000 A:middle L:90%
force provoked by particle be on particle eight is equals
11
00:00:38.000 --> 00:00:42.359 A:middle L:90%
to g times the mass off particle be times the
12
00:00:42.369 --> 00:00:47.219 A:middle L:90%
mass off particle A divided by the distance between particle
13
00:00:47.229 --> 00:00:51.829 A:middle L:90%
be and particle a squared, then plugging in their
14
00:00:51.829 --> 00:00:54.429 A:middle L:90%
values that were given in the problem. We get
15
00:00:54.439 --> 00:00:58.539 A:middle L:90%
that F B eight is a close to 6.67 times
16
00:00:58.539 --> 00:01:03.060 A:middle L:90%
stand toe minus 11 times 363 which is the mask
17
00:01:03.069 --> 00:01:07.879 A:middle L:90%
off particle 80 times 517 which is the mask off
18
00:01:07.890 --> 00:01:12.019 A:middle L:90%
particle be divided by the distance between particles A and
19
00:01:12.019 --> 00:01:15.760 A:middle L:90%
me and we can see in the figure that this
20
00:01:15.760 --> 00:01:19.269 A:middle L:90%
distance is a close to 0.5 meters. So we
21
00:01:19.269 --> 00:01:23.659 A:middle L:90%
have 0.5 squared down here and this gives us a
22
00:01:23.659 --> 00:01:30.859 A:middle L:90%
gravitational force off approximately five 0.7 times 10 to minus
23
00:01:30.859 --> 00:01:34.540 A:middle L:90%
five new terms. Now for the magnitude of the
24
00:01:34.549 --> 00:01:40.629 A:middle L:90%
gravitational force between particle see and particle A, we
25
00:01:40.629 --> 00:01:46.599 A:middle L:90%
have the following F c A busy Costa G M
26
00:01:46.599 --> 00:01:51.370 A:middle L:90%
C times M A divided by the distance between particles
27
00:01:51.370 --> 00:01:53.450 A:middle L:90%
, see and particle a squared. Then by plugging
28
00:01:53.450 --> 00:01:57.000 A:middle L:90%
in the values that were given in the problem we
29
00:01:57.000 --> 00:02:00.230 A:middle L:90%
get the following FC eight is equals to 6.6 to
30
00:02:00.230 --> 00:02:07.180 A:middle L:90%
7 times stand to minus 11 times 154 kilograms times
31
00:02:07.340 --> 00:02:13.050 A:middle L:90%
363 kilograms divided by the distance between particles A and
32
00:02:13.050 --> 00:02:15.689 A:middle L:90%
C and the distance between particle A and C is
33
00:02:15.689 --> 00:02:23.490 A:middle L:90%
he goes to 0.5 plus 0.25 which is 0.725 So
34
00:02:23.490 --> 00:02:29.620 A:middle L:90%
we have 0.75 squared down here and this gives this
35
00:02:29.629 --> 00:02:36.620 A:middle L:90%
a gravitational force off approximately six 0.6 to 9 times
36
00:02:36.620 --> 00:02:39.969 A:middle L:90%
Stan to minus six mutuals. Now let me organize
37
00:02:39.969 --> 00:02:44.330 A:middle L:90%
the board Now I have to choose a reference to
38
00:02:44.569 --> 00:02:47.460 A:middle L:90%
calculate the net force So you choose that everything that
39
00:02:47.460 --> 00:02:51.389 A:middle L:90%
is pointing to the right it's positive and as a
40
00:02:51.389 --> 00:02:53.080 A:middle L:90%
consequence, everything that is pointing to the left is
41
00:02:53.080 --> 00:02:58.319 A:middle L:90%
negative. The reform the net reputational force that is
42
00:02:58.330 --> 00:03:01.840 A:middle L:90%
acting on particle A is given by, if be
43
00:03:01.849 --> 00:03:07.340 A:middle L:90%
a plus f c A. Because both of these
44
00:03:07.340 --> 00:03:10.900 A:middle L:90%
forces are pointing to the right and this Zico's to
45
00:03:10.900 --> 00:03:19.000 A:middle L:90%
5.7 times stand to minus five plus 6.6 to 9
46
00:03:19.009 --> 00:03:23.099 A:middle L:90%
times 10 to minus six, and this gives us
47
00:03:23.099 --> 00:03:29.129 A:middle L:90%
a net force off. Approximately 5.6 to 7 times
48
00:03:29.129 --> 00:03:32.780 A:middle L:90%
stand to minus five new tones on its pointing to
49
00:03:32.780 --> 00:03:36.400 A:middle L:90%
the right. For the next item, we have
50
00:03:36.400 --> 00:03:39.360 A:middle L:90%
to complete the magnitudes off the reputational forces that the
51
00:03:39.360 --> 00:03:44.750 A:middle L:90%
particle a exerts on particle be and the magnitude off
52
00:03:44.750 --> 00:03:47.770 A:middle L:90%
the gravitational force that part Cosi exerts on particle Be
53
00:03:49.340 --> 00:03:52.610 A:middle L:90%
to begin with note the following the magnitudes off The
54
00:03:52.610 --> 00:03:57.530 A:middle L:90%
reputational forces are symmetric. I mean that the magnitude
55
00:03:57.539 --> 00:04:01.050 A:middle L:90%
off the gravitational force that particle a exerts on particle
56
00:04:01.050 --> 00:04:04.330 A:middle L:90%
be Izzy coast to the magnitude off the reputation off
57
00:04:04.330 --> 00:04:09.500 A:middle L:90%
works that particle be exerts on particle A. So
58
00:04:09.669 --> 00:04:12.229 A:middle L:90%
we do not need to Copley these again. We
59
00:04:12.229 --> 00:04:15.100 A:middle L:90%
had already calculated the magnitude off the gravitational force that
60
00:04:15.110 --> 00:04:18.290 A:middle L:90%
particle be exerts on particle A. So we can
61
00:04:18.290 --> 00:04:21.819 A:middle L:90%
just use this value again because off the symmetry on
62
00:04:21.819 --> 00:04:25.930 A:middle L:90%
the magnitude stuff the gravitational force. So I'm not
63
00:04:25.939 --> 00:04:28.529 A:middle L:90%
calculating it again. The only thing that we have
64
00:04:28.529 --> 00:04:30.300 A:middle L:90%
to count leave this time is the magnitude of the
65
00:04:30.300 --> 00:04:35.000 A:middle L:90%
gravitational force exerted on particle be by particle seat because
66
00:04:35.000 --> 00:04:39.300 A:middle L:90%
we haven't completed it yet. Then we have the
67
00:04:39.300 --> 00:04:43.680 A:middle L:90%
following f C V is it goes to g times
68
00:04:43.680 --> 00:04:46.329 A:middle L:90%
the mass off particle C times the mass off particle
69
00:04:46.329 --> 00:04:50.449 A:middle L:90%
be divided by the distance between particles B and C
70
00:04:50.709 --> 00:04:54.439 A:middle L:90%
squared by the figure, you can see that the
71
00:04:54.439 --> 00:04:59.220 A:middle L:90%
distance between particles B and C is equals to 0.25
72
00:04:59.230 --> 00:05:01.550 A:middle L:90%
meters. So by plugging the other values that were
73
00:05:01.550 --> 00:05:05.050 A:middle L:90%
given by the problem, we get F C B
74
00:05:05.279 --> 00:05:11.149 A:middle L:90%
has bean equals to 6.67 times stand toe minus 11
75
00:05:11.569 --> 00:05:21.949 A:middle L:90%
times 154 times 517 divided by 0.25 squared. And
76
00:05:21.949 --> 00:05:26.800 A:middle L:90%
this gives us a gravitational force off a magnitude that
77
00:05:26.800 --> 00:05:33.439 A:middle L:90%
is approximately pate 0.497 times stand to minus five Newtons
78
00:05:33.449 --> 00:05:36.069 A:middle L:90%
. Now let me organize the Born before finishing Diz
79
00:05:36.069 --> 00:05:42.120 A:middle L:90%
item. Okay, so the Net Force that acts
80
00:05:42.129 --> 00:05:46.209 A:middle L:90%
on particle be easy Coz too f c b that
81
00:05:46.209 --> 00:05:50.959 A:middle L:90%
is pointing to the right minus f a B that
82
00:05:50.959 --> 00:05:55.829 A:middle L:90%
is pointing to the left and this is equals to
83
00:05:55.839 --> 00:06:01.800 A:middle L:90%
8.497 times stand to minus five miners. Remember that
84
00:06:01.810 --> 00:06:04.639 A:middle L:90%
F A b is. It goes to f B
85
00:06:04.649 --> 00:06:10.149 A:middle L:90%
eight and f B A is a question 5.7 times
86
00:06:10.149 --> 00:06:14.920 A:middle L:90%
stand to manage five. So we have 5.7 times
87
00:06:14.920 --> 00:06:16.819 A:middle L:90%
stand to minus five here and these gives us a
88
00:06:16.819 --> 00:06:25.480 A:middle L:90%
net force off approximately three 0.49 times 10 to minus
89
00:06:25.480 --> 00:06:29.680 A:middle L:90%
five new times to the right. Now let me
90
00:06:29.680 --> 00:06:31.660 A:middle L:90%
clear the board for the next item on the last
91
00:06:31.670 --> 00:06:34.699 A:middle L:90%
item we have to calculate the magnitudes off the reputational
92
00:06:34.699 --> 00:06:40.389 A:middle L:90%
forced exert that on particle see by both particles A
93
00:06:40.550 --> 00:06:44.250 A:middle L:90%
and particle. Be now remember that the magnitude of
94
00:06:44.250 --> 00:06:48.220 A:middle L:90%
the gravitational force it's metric than F A C is
95
00:06:48.220 --> 00:06:51.699 A:middle L:90%
equals two f c eight and we had already calculated
96
00:06:51.709 --> 00:06:55.949 A:middle L:90%
f c A on the first item. It's here
97
00:06:56.540 --> 00:07:00.889 A:middle L:90%
and also F B C is He goes to F
98
00:07:00.920 --> 00:07:04.519 A:middle L:90%
. C V and had calculated FCB on the second
99
00:07:04.529 --> 00:07:08.170 A:middle L:90%
item. Therefore, we do not need to copulate
100
00:07:08.180 --> 00:07:12.300 A:middle L:90%
anti gravitational force again and we can proceed already directly
101
00:07:12.300 --> 00:07:15.120 A:middle L:90%
to the net force. So the net force that
102
00:07:15.129 --> 00:07:18.560 A:middle L:90%
acts on particle see is it goes to as both
103
00:07:18.800 --> 00:07:23.509 A:middle L:90%
forces are pointing to the left. Both forces will
104
00:07:23.509 --> 00:07:26.850 A:middle L:90%
have a minor sign the reform. We have miners
105
00:07:26.860 --> 00:07:30.240 A:middle L:90%
f A, C minus F B C and this
106
00:07:30.470 --> 00:07:33.800 A:middle L:90%
is a close to minus F A C, which
107
00:07:33.800 --> 00:07:39.509 A:middle L:90%
is 6.6 to 9 times 10 to minus six minus
108
00:07:39.519 --> 00:07:45.259 A:middle L:90%
F B C. We just 8.497 time stand to
109
00:07:45.259 --> 00:07:48.670 A:middle L:90%
minus five and this gives us a net force off
110
00:07:48.670 --> 00:07:55.389 A:middle L:90%
approximately miners. Nine points 16 times stand to minus
111
00:07:55.389 --> 00:07:59.079 A:middle L:90%
five new terms. Off course we have a minus
112
00:07:59.079 --> 00:08:01.589 A:middle L:90%
sign here. Therefore, this net force points to
113
00:08:01.589 --> 00:08:03.149 A:middle L:90%
the left