1
00:00:03,980 --> 00:00:08,530
And now I'm in position
to talk about what
2
00:00:08,530 --> 00:00:10,650
is the maximum force.
3
00:00:10,650 --> 00:00:15,810
If I push F harder, as if I push
F and the blocks go together,
4
00:00:15,810 --> 00:00:18,480
the accelerations are the same.
5
00:00:18,480 --> 00:00:19,900
I push F harder.
6
00:00:19,900 --> 00:00:21,460
Accelerations are the same.
7
00:00:21,460 --> 00:00:25,650
I push F harder, and
I push F so hard,
8
00:00:25,650 --> 00:00:30,580
that the static friction
no longer reaches
9
00:00:30,580 --> 00:00:33,580
its maximum value.
10
00:00:33,580 --> 00:00:36,800
And if I push F
harder than that,
11
00:00:36,800 --> 00:00:39,210
I will not have--
the static friction
12
00:00:39,210 --> 00:00:42,880
can't get bigger, and the
blocks 1 and 2 will start
13
00:00:42,880 --> 00:00:45,290
to slip relative to each other.
14
00:00:45,290 --> 00:00:53,080
So my no slipping condition
is that I want two things.
15
00:00:53,080 --> 00:00:55,020
a1 equals a2.
16
00:00:55,020 --> 00:00:56,456
We'll call that a.
17
00:00:56,456 --> 00:01:06,730
And the maximum force
condition is that F static
18
00:01:06,730 --> 00:01:13,050
is equal to maximum value.
19
00:01:13,050 --> 00:01:19,380
Now, what is the normal force
between that we refer to
20
00:01:19,380 --> 00:01:20,600
in our fiction law?
21
00:01:20,600 --> 00:01:23,480
There's two normal
forces-- the ground
22
00:01:23,480 --> 00:01:25,740
and the normal force
between the surfaces.
23
00:01:25,740 --> 00:01:28,250
But the static friction
that we're talking about
24
00:01:28,250 --> 00:01:30,230
is between the surfaces.
25
00:01:30,230 --> 00:01:35,430
So that's why we use N
here and not N ground.
26
00:01:35,430 --> 00:01:40,520
And now I can solve
for this F max.
27
00:01:40,520 --> 00:01:45,690
By the way, we also have the
condition that fk is mu k.
28
00:01:45,690 --> 00:01:48,280
What normal force
are we talking about?
29
00:01:48,280 --> 00:01:53,830
N, which we've
called N ground 1.
30
00:01:53,830 --> 00:01:57,020
And now I look at my
equations, and my goal
31
00:01:57,020 --> 00:02:03,460
is to solve for F. I know
Ng1 from this equation.
32
00:02:03,460 --> 00:02:09,169
It's just equal to M1g plus N.
I Know N from that equation.
33
00:02:09,169 --> 00:02:12,600
So Ng1 is just the sum
of the masses times g.
34
00:02:12,600 --> 00:02:14,570
So I know this.
35
00:02:14,570 --> 00:02:17,904
I have f, which is m2a.
36
00:02:17,904 --> 00:02:24,550
And I can now apply my result.
37
00:02:24,550 --> 00:02:31,740
So what we'll do is we'll solve
for the a's, a1 equals a2,
38
00:02:31,740 --> 00:02:34,070
in terms of F max.
39
00:02:34,070 --> 00:02:39,990
So over here we have
that a is f over m2.
40
00:02:39,990 --> 00:02:43,360
That's from this equation.
41
00:02:43,360 --> 00:02:47,650
And now I'll substitute
a1 is equal to that.
42
00:02:47,650 --> 00:02:49,990
I'll substitute that there.
43
00:02:49,990 --> 00:02:55,579
And I get that F
Max is going to be
44
00:02:55,579 --> 00:03:01,133
equal to fk plus f static max.
45
00:03:04,680 --> 00:03:06,900
That's that one.
46
00:03:06,900 --> 00:03:16,460
Plus m1 times a1, which is
f static max divided by m2.
47
00:03:16,460 --> 00:03:21,970
And so I get that
F max equals-- now
48
00:03:21,970 --> 00:03:25,230
I'm going to substitute
in all of these values.
49
00:03:25,230 --> 00:03:27,590
It's going to look a
little complicated.
50
00:03:27,590 --> 00:03:32,070
And so I'd like to have a
little space here for that,
51
00:03:32,070 --> 00:03:33,520
to get everything in here.
52
00:03:38,980 --> 00:03:49,870
And we'll see that it's equal
to mu k m1 plus m2g, fk, mu k,
53
00:03:49,870 --> 00:03:56,770
plus m1 2mg plus f static max
times 1 plus and m1 over m2.
54
00:03:56,770 --> 00:04:01,860
But f static max is
mu N, and N is m2g.
55
00:04:01,860 --> 00:04:08,980
So I get m2g times
1 plus m1 over m2.
56
00:04:08,980 --> 00:04:14,110
And there is, if I push
any harder than that,
57
00:04:14,110 --> 00:04:17,350
block 2 will slip with
respect to block 1.
58
00:04:17,350 --> 00:04:22,050
Again, all of our terms have
the dimensions of acceleration.
59
00:04:22,050 --> 00:04:26,110
This is dimensionless-- 1,
dimensions of acceleration.
60
00:04:26,110 --> 00:04:29,150
And we did miss
one little thing.
61
00:04:29,150 --> 00:04:33,820
We missed the coefficient
of static friction.
62
00:04:33,820 --> 00:04:38,680
And there we have
it, a tricky problem.