WEBVTT
00:00:00.060 --> 00:00:02.780
This is College Physics
Answers with Shaun Dychko.
00:00:03.293 --> 00:00:07.886
In this problem, we're told to pretend that
the chain there doesn't exist or it's broken,
00:00:08.373 --> 00:00:13.780
in which case the horizontal force
on the foot of one of the sides
00:00:13.780 --> 00:00:18.333
of the sandwich board has
to be provided by friction.
00:00:18.920 --> 00:00:22.526
We're going to figure out what is the
coefficient of static friction that's needed,
00:00:22.746 --> 00:00:26.013
the minimum anyway that's
needed to make a force
00:00:26.013 --> 00:00:28.846
that's big enough to prevent
this foot from slipping.
00:00:30.686 --> 00:00:33.760
So we're going to consider just
one half of the sandwich board.
00:00:34.080 --> 00:00:37.273
It's always good to take
a complicated problem
00:00:37.273 --> 00:00:40.986
and try to reduce it into a smaller
problem that's easier to think about.
00:00:42.820 --> 00:00:45.106
We have a bunch of forces on here.
00:00:45.106 --> 00:00:48.066
We have the normal force straight
up provided by the ground,
00:00:48.273 --> 00:00:49.953
we have the friction straight to the left,
00:00:50.573 --> 00:00:53.733
we have gravity acting
at the center of mass
00:00:53.733 --> 00:00:58.446
which is the geometric center of this board
since we're told that it's a uniform board
00:00:59.133 --> 00:01:02.633
and the mass is distributed evenly
throughout the whole length of this board.
00:01:02.633 --> 00:01:06.813
So its center of gravity is
in the physical center.
00:01:10.166 --> 00:01:14.380
Then we have the hinge force which
I've drawn straight to the right
00:01:14.560 --> 00:01:17.746
and I need to explain why we
know that it's straight to the right.
00:01:17.946 --> 00:01:21.860
Because being a hinge it could conceivably
have a component that is vertical
00:01:22.226 --> 00:01:24.066
but we'll see that it is not in this case
00:01:24.626 --> 00:01:27.140
because we'll turn our
attention to this picture here.
00:01:27.573 --> 00:01:32.413
Consider this whole sandwich board
as being a single point mass
00:01:33.113 --> 00:01:37.166
and the total force downwards
on it would be
00:01:37.226 --> 00:01:40.033
two times the force of gravity on each side.
00:01:40.353 --> 00:01:45.293
So I have F g here representing the force
of gravity on one side of the board,
00:01:46.293 --> 00:01:50.866
and so if you reduce this whole
sandwich board to a dot,
00:01:52.166 --> 00:01:54.900
the total force down on it
would be two times F g.
00:01:56.226 --> 00:01:58.393
And order for it to not
accelerate vertically,
00:01:58.626 --> 00:02:03.840
the ground will have to provide a force
upwards that's of the same magnitude.
00:02:04.293 --> 00:02:07.393
So two times this normal
force in other words,
00:02:09.530 --> 00:02:16.740
which is the only force upwards on
the whole thing is going to be --
00:02:17.240 --> 00:02:20.346
well, it works out to saying that
the normal force upwards
00:02:20.346 --> 00:02:22.546
has to equal the gravity downwards.
00:02:25.593 --> 00:02:29.660
Since that's the case, there can't
be any upwards hinge force
00:02:29.666 --> 00:02:36.400
because this normal force is
accounting for the only force
00:02:36.400 --> 00:02:40.433
that is upwards on the
sandwich board as a whole.
00:02:47.626 --> 00:02:48.773
Do I need to say anymore about that?
00:02:51.526 --> 00:02:55.700
You know, if the sandwich
board was not symmetrical
00:02:56.026 --> 00:02:59.346
and maybe one side was long
and the other side was short,
00:02:59.740 --> 00:03:01.246
then things would get more complicated
00:03:01.246 --> 00:03:06.380
and there would be a vertical component
to the hinge force for sure.
00:03:06.873 --> 00:03:09.173
We'd have to be given more
information to solve it
00:03:10.866 --> 00:03:15.953
because you'd no longer be able to say
that two times F n is the force upwards
00:03:15.953 --> 00:03:19.506
because there'd be force
normal on the right side
00:03:19.506 --> 00:03:23.886
and there'd be force normal on the
left side and they would not be equal.
00:03:26.253 --> 00:03:36.993
That means you'd have F nl plus
F nr here instead of two F n
00:03:38.153 --> 00:03:39.613
but then you'd have two unknowns.
00:03:41.146 --> 00:03:42.366
So that would complicate things.
00:03:42.953 --> 00:03:45.626
But fortunately we have
a symmetrical system
00:03:45.626 --> 00:03:51.153
in which case the normal force
on each foot upwards is the same
00:03:51.560 --> 00:03:55.433
and so that's why we're able to say two F n.
00:03:57.410 --> 00:03:59.886
Yeah, so that's how we conclude
that the normal force upwards
00:03:59.886 --> 00:04:01.253
equals gravity downwards.
00:04:02.080 --> 00:04:08.060
Since that's the case there is no place
for having a vertical force on the hinge.
00:04:09.666 --> 00:04:10.233
Okay.
00:04:12.080 --> 00:04:18.900
So, well I guess I could expand
on that slightly more.
00:04:18.993 --> 00:04:21.393
Let's suppose you still were to insist that
00:04:21.393 --> 00:04:24.133
there is a vertical component
of the hinge force.
00:04:24.613 --> 00:04:29.333
You'd have to say that in the y
direction the hinge force vertical
00:04:29.886 --> 00:04:33.280
plus the normal force would
equal the gravity downwards
00:04:33.593 --> 00:04:35.466
and then you could solve
for the hinge force
00:04:35.466 --> 00:04:38.020
and say that it's gravity minus normal force.
00:04:38.020 --> 00:04:40.780
But since we've just said that gravity
and the normal force are equal,
00:04:41.346 --> 00:04:42.206
that makes this zero.
00:04:42.206 --> 00:04:46.413
So there is the more rigorous
reason for saying
00:04:47.346 --> 00:04:49.060
vertical component of hinge force is zero.
00:04:50.266 --> 00:04:51.293
Okay, there.
00:04:51.846 --> 00:04:56.513
Now, with that out of the way,
we can think about this picture
00:04:56.513 --> 00:05:01.040
and consider our three equations which we
get from the two conditions for equilibrium,
00:05:01.040 --> 00:05:06.506
one being that the translational
forces have to add up to zero
00:05:06.506 --> 00:05:08.200
in both directions, x and y,
00:05:08.533 --> 00:05:11.580
and then we'll say that the total torque
has to add up to zero as well.
00:05:13.933 --> 00:05:16.773
So to say that the horizontal forces
add up to zero is the same
00:05:16.773 --> 00:05:19.520
as saying the forces to the right
equal the forces to the left.
00:05:19.853 --> 00:05:22.953
So the hinge force to the right
equals the friction force to the left.
00:05:22.953 --> 00:05:25.226
I substituted coefficient
of static friction
00:05:25.226 --> 00:05:27.820
multiplied by normal force
in place of the friction force.
00:05:29.326 --> 00:05:33.146
Then vertically speaking we have
normal force equals gravity,
00:05:33.146 --> 00:05:35.386
we've kind of already said
that but just saying it again,
00:05:35.766 --> 00:05:42.306
and that is m g, mass being the mass
of one side of the sandwich board.
00:05:42.306 --> 00:05:45.253
It's eight kilograms divided by
two which is four kilograms.
00:05:47.286 --> 00:05:51.386
Then the total torque has to equal zero and
I've taken the pivot to be at the ground
00:05:51.386 --> 00:05:53.686
but you can take it at the hinge if you
like, doesn't really matter where.
00:05:54.040 --> 00:05:56.040
I chose to choose the pivot at the ground
00:05:56.200 --> 00:05:58.486
because then we have only
two torques to consider.
00:05:58.486 --> 00:06:02.700
We have this due to gravity and
the torque due to the hinge.
00:06:02.946 --> 00:06:06.786
If we had chosen the hinge to be the pivot,
then we'd have the torque due to gravity,
00:06:06.786 --> 00:06:08.986
the torque due to friction and the
torque due to the normal force
00:06:08.986 --> 00:06:10.986
and we'd have three terms to consider.
00:06:11.746 --> 00:06:14.180
So I think it's a little bit
simpler to have only two.
00:06:14.826 --> 00:06:19.900
So we're taking the pivot to be at the
point of contact with the ground.
00:06:20.400 --> 00:06:20.766
Okay.
00:06:21.200 --> 00:06:28.946
So we have the hinge force multiplied by
its perpendicular distance to the pivot
00:06:28.946 --> 00:06:31.733
which is this 1.3 meters height
of the sandwich board.
00:06:32.180 --> 00:06:34.260
That's going to be the torque
which is clockwise,
00:06:34.506 --> 00:06:37.766
and that's going to equal
in magnitude the torque
00:06:37.973 --> 00:06:40.886
going counter-clockwise
due to the force of gravity
00:06:41.240 --> 00:06:48.560
which will have a perpendicular
distance to the pivot of 1.1 meters
00:06:48.560 --> 00:06:52.566
which is the total width of the
sandwich board divided by four.
00:06:52.853 --> 00:06:55.620
So 1.1 divided by two would
take you to the hinge
00:06:55.786 --> 00:06:59.453
and then this center of gravity is halfway
between the center point and the foot
00:06:59.606 --> 00:07:03.340
and so we divided by four
to get 0.275 meters
00:07:03.340 --> 00:07:07.700
as the lever arm perpendicular for gravity.
00:07:09.753 --> 00:07:12.426
So we've equated those two
things, those two torques,
00:07:13.070 --> 00:07:17.733
force of gravity is m g and then
we're dividing both sides by r h
00:07:18.140 --> 00:07:20.960
the lever arm for the hinge to
solve for the force of the hinge.
00:07:21.320 --> 00:07:25.553
So that's four kilograms times 9.8 newtons
per kilogram, times 0.275 meters,
00:07:25.553 --> 00:07:29.953
lever arm for gravity, divided by
1.3 meters, lever arm for the hinge.
00:07:30.073 --> 00:07:34.086
That gives 8.29 newtons
and that is horizontal.
00:07:34.706 --> 00:07:36.300
That's the answer for part B.
00:07:36.846 --> 00:07:40.320
The answer for part A, the
coefficient of static friction,
00:07:40.466 --> 00:07:46.720
we get by rearranging this formula here
and we divide both sides by F n
00:07:48.013 --> 00:07:50.040
to get that coefficient of static friction
00:07:50.040 --> 00:07:51.733
is the hinge force divided
by the normal force.
00:07:52.146 --> 00:07:55.486
So that's 8.2923 newtons
divided by four kilograms
00:07:55.486 --> 00:07:57.173
times 9.8 newtons per kilogram,
00:07:57.173 --> 00:08:00.160
giving a coefficient of
static friction of 0.211.