[ Music ]
Okay, there are some left over, a few of you
gave me, these are left over, you can pick
up your quizzes here, leftover quizzes there
and so we go there, so let's go --
[ Shuffling Papers ]
David. David. Luke. Christian. Kristin. [inaudible]
[ Shuffling Papers ]
Okay. Now let's move on with the idea of friction.
Is there any question about the frame, or,
okay, about the frame or process? So let me
put your homework down here for next chapter,
chapter eight homework. Please write this
down.
[ Writing on Board ]
Just friction, number 3, 7, 18, 21, 27, 32,
41, 48, 52, 55, 105, 115. The last two are
the belt system. These two, 52 and 65 are
wedges. We are going to talk about them today.
And the rest of it are stand friction problem.
I measured it last time. We have three scenario.
The system can be in equilibrium. System will
be at the time of impending motion. Everybody
understand when the forces become equal to
the friction forces become [inaudible] times
N and then depending on what type of movement
you have, you have to decide, and if it is
the object is not moving, you use it the coefficient
of static friction, yes or no. If it is moving,
you have to use the coefficient of kinematic
friction, which is a little bit lower, so
they would give it to you sometimes but many
times you don't have to use it, depends on
the type of problem. Sometimes you use it,
sometimes you don't. Most of the time it is,
operation is [inaudible], so this is the problem,
typical problem of what you are going to see,
so please write this down. It's not in your
handout. The handout that I gave you is a
new problem there we are going to discuss
that also. So here we have a problem like
that, which is sort of [inaudible] incline
surface. And here we are having a pulley system.
And here is a box for example, a box A, which
has [inaudible] weight, and then this pulley
system, which is [inaudible] is also attached
to another box sitting, another box perhaps
sitting on this floor, this incline surface.
And the angle of this surface is given equal
to 30 degree. This is block B. This is block
A. For the time being, we are assuming that
this pulley is frictionless because later
on we are going to talk about friction and
the pulley as well. For the time being, there
is no friction in the pulley, so what it means
that whatever tension I have in the table
is going to transfer to the other side. So,
and then WA is given equal to 25 pounds, and
we want to calculate the weight of box B for
equilibrium. What's going to happen here.
How do you analyze this problem based on what
we talked yesterday? What is going to happen
here?
[ Inaudible Response ]
What?
[ Inaudible Response ]
Yeah, we have to [inaudible] force of box
B, is that correct or not? Because the tension,
first of all the W is given. The tension in
the cable if no friction is would be 25 pounds.
That's [inaudible]. So if I draw the [inaudible]
of this one, which has the horizontal force
of T, there is the weight, where I should
put my friction? You see, I ask that question.
Some of you said like this, which I said wrong.
Is that correct? [inaudible] depends on?
Motion.
Motion. Depends on the motion, exactly. What
motion? We haven't [inaudible]. You see, that's
exactly where it comes to this result that
I have to see what that box B is to heavy
or too light. Everybody understand that? If
box B is too heavy, the box B moves down the
slope. Is that correct? Or it's too light.
Then maybe this pushes that way, so we are
two scenario here, so we have to investigate
the impending motion. Which one? You have
to decide. So there is actually two problem
in one, is that correct or not based on what
we know now, correct? Everybody agrees. So
you have to decide way to go. Let's assuming
B moving up. So let's do that. So okay. So
if B moving up, what does it mean? [inaudible]
there, smallest value, is that correct or
not? So, you'll have to write it down. B moving
up. That means it is the lightest, it's going
that way. Is that correct or not? This or
that, now I can complete my [inaudible] yes,
you said we have to draw the [inaudible] but
now I can't decide. Anyhow, in that case,
your [inaudible] simple problem. This is your
box. We have done that in the past. The box
has a weight. You just put the weight of W,
which we don't know. That's the question mark.
There is a tension here. So we have a tension
T here. The tension is of course it was [inaudible]
--
[ Inaudible Response ]
-- pounds going that way, and then you have
here, you have to put, maybe we should do
that or in color that we know which one is
low and which one not low. So, the blue one
are, or maybe I should use red. So the red
one are going to be in unknown quantity of
[inaudible] is unknown quantity, so let's
it to red. So you can put it on top. You can
put it anywhere, but this is the equilibrium
of -- they are not [inaudible] of course there
is [inaudible] force, is that correct or not?
Yes? And what else? Now [inaudible] knowing
that the box going up the friction would be
going against the movement so the friction
is here at the time of impending motion that
I have there and that's exactly where we are.
We are [inaudible] at the time that exactly
[inaudible], is that correct? [inaudible]
between the static and dynamic. Is that correct
or not? Yeah. Now how many unknown do we have
there or seem do we have? First of all, this
is not equilibrium of rigid body. This is
equilibrium of a? Particle. Yes. How many
unknown can I have there?
Two.
How many that look there?
[ Inaudible Response ]
Actually it looks three arrows but actually
it is?
Two.
Two. Because if I have N, I have F, or I have
F, I have N. Everybody understand? Because
F and N, write it down, I gave you [inaudible]
minute, but we can discuss that a little bit
later on when we need that to be done. There
is a relationship between the F and N because
as we said it before, at this [inaudible]
impending motion, N equals 2. I'm sorry, F
equals 2. F equal to US times N. Of course
I have to give you U. I forgot to give you
U. So we put here U is [inaudible] equal to
point [inaudible] is given equal to point
35. So that's also it. Now I can finish my
[inaudible] yes?
[inaudible] when we did the [inaudible] diagram
of the --
Box B.
Box B. [inaudible] the little F was on an
angle, and then we put a baby X on --
Little F was an angle? No F wasn't not never
an angle. That was something else we discussed.
Oh, okay.
That was the relationship between horizontal
force and that has nothing to do with this.
Oh, okay.
You see that one. Okay, since you asked I
have to explain that, but that's what I put
last time. Last time look this is the box.
This is not on the incline surface. This is
on the horizontal surface. Yes or no? This
is a W. This is, and these two are equal and?
Opposite.
Opposite. It's not like that. This is a little
bit different from this problem. However,
where is the friction?
[ Inaudible Response ]
So you don't it, that's [inaudible]. Then
I put here a force what? I put here a force?
P.
P. Look the friction is here parallel to the,
parallel to the surface, yes or no? The surface
horizontal so F was here. When I put capital
F here.
Yes.
So that's not, this is not the impending motion
at this time, at the time of impending motion,
this you said you are reaching to that point
now. This is impending motion.
Okay.
At the time of impending, you know, I asked,
I have to answer that question. At the time
of, here, this is the value of F. This is
the value, this was the value of -- if you
are here, you see the value of F is less than
small F so there is equilibrium. And that's
exactly the conclusion we are going [inaudible].
However, in this problem, we are assuming
that we are at the time of impending motion,
so this F becomes now equal to F we are at
that point. Is that correct? Okay, good. So
therefore that is our system. Now you have
to choose an XY coordinate. Is that correct
or not? We have done in the past, everybody
knows that it is better to take XY coordinate
in this direction. Okay. We take XY coordinate
like that and you write [inaudible] sigma
if X equals to zero and sigma [inaudible]
XY, is that correct or not. Sigma if X equal
to zero, so I put it here. Sigma FX equal
to zero. Look what we have. We have here minus,
see this is our X [inaudible] minus?
[ Inaudible Response ]
Then I have plus F [inaudible] so that's F
[inaudible] component and this one has also
horizontal component plus what that angle
is how many degrees at angle is?
Thirty degrees.
Thirty degrees. Remember how many problem
we had like this so you should not make any
mistake there. So F plus WB plus WB, whatever
that number is, is a horizontal. This is cosign.
The other is sign of course, sign [inaudible]
so this is still [inaudible] unknown in this
equation. Cosign of 30 degrees [inaudible]
. Correct? Then you write sigma FY, sigma
FY equal to zero, so sigma FY what do we have
there. This doesn't [inaudible] anything.
[inaudible] plus N and then we have minus?
W --
WB times cosign of 30 degree equal to zero
[inaudible] also one equation and to unknown.
So the best way to solve here, calculate N,
N becomes equal to WB times cosign 30 degree,
which is 0.866, so you put here 0.866, so
N becomes equal to 0.866WB because I want
to eliminate all the other unknown. Then at
the time of impending motion, F equal to what?
[inaudible] times?
N.
N. So F equal to [inaudible], which is 0.35
times 0.866WB, so the total become here, you
can calculate that, become 0.303WB, notice
I'm going always to minimal of three digits.
That what's established long time ago. Is
that correct or not? So now everything being
expressed in terms of?
W.
W. So we plug this one in the first equation,
yes or no? So we have minus 25, minus 25.
Then we have plus F, which is 0.303WB plus
sign of 30 degrees times WB, which is 0.500WB
equal to 0, you solve for WB to be equal to
31.1 pound. So if I choose that this weight
equal to 31.1 is at the time of the impending
motion, everybody [inaudible] anything less
than that this is going to move and going
to accelerate. How, I'm going to accelerate
as I said that will be a discussion you will
going to have in your dynamic class [inaudible]
one of earliest problems you are going to
have [inaudible] object is going to accelerate
downward. Is that correct or not? Yes? However,
this is not a problem. The problem was still
I can have this one. Now you will think that,
at the beginning you will this that after
above 3.11 is going to go the other way, that's
not the case. There is some gapping between.
Everybody understand that? So the next, write
it down. This was case one. So write it down
in your notes. Know this was case one. Now
for me it is easy because now I can change
that, erase that and do it. You have to rewrite
it. Case two, B is moving?
[ Inaudible Response ]
Down. Okay. You don't have to do the whole
problem again, because many of the aspect
of that one will remain the same. Since this
is the same problem, I'm going only to change
the direction of the F. What is changing?
Everything is the same except?
[ Inaudible Response ]
This time the box is going down. This way
the friction must be going?
The other way. So let's change that. This
is for case two. You can erase that one. Put
an F going that way. Is that correct or not?
Yes?
Yes.
Okay. Look, this problem. This one, plus become
minus, just change it like that. You don't
have to rewrite that. Is that correct or not?
Because this has not changed. This has not
changed. This has not changed. The only thing
plus F becomes minus F. This equation is exactly
the same. That has not changes. Is that correct
or not? Yes? Everybody with me. So this one
has not changed. That one has not changed.
Is that correct? The only thing here instead
of having this plus, now I have to change
it to minus. So as a result, WB now becomes
much larger. For this problem, WB becomes
equal to 127 pounds. So for 127 pounds and
larger, the box B is going to move down. Now
without realizing, if this was a static problem,
which I have given you in the past, if I give
you a weight of 100 pounds, the system would
be in?
Equilibrium.
Equilibrium. Not moving either way. Everybody
understand? And sometimes they raise this
question. They said it, don't get me wrong,
sometimes the give the box a weight of box
equal to 100, and they would ask you to tell
me whether this system is in?
Equilibrium.
Equilibrium or not. But you can see that you
can use the same logic. Is that correct or
not? One way another find the [inaudible]
impending force or impending motion, how much
force do you need or how much weight do you
need, and make a decision based on that. Or,
do it differently. Do not call this one F,
call this one capital F, calculate that F,
everybody understand what I'm saying that?
And if that F is less than [inaudible] times
N, so this object is not going to?
Move.
Move. Is that correct or not? Yes? If it is
more than mew [phonetic] times N, then object
already have moved, so it is beyond [inaudible].
So you can see this problem in different scenarios,
but don't get me wrong. This is all, this
is different to that [inaudible]. You can
be below or you can be above or you can be
right at the time of impending motion. So
all of these questions will be asking you
set up homework that you have. Sometimes they
ask you to find out, they ask you to find
the amount of force for the impending motion.
Sometimes they give you the force or some
sort of a number. They ask you to determine
whether the system is in equilibrium or not.
We will see more of that coming. But these
are basic ideas. So I am introducing to you.
The second problem is in your handout. So
that one is an interesting problem because
it shows you something new and something also
important [inaudible] to recognize. So let's
put this one -- there are two boxes put on
top of each other, so there are two boxes
like that. And there are on the ground. And
the man is standing to the left, and he has
put his hand here. So he is pushing the box
at this point, very bottom of the top box.
Each box has a weight of 100 pounds. So please
write the weight of each one, 100 pounds her,
100 pounds here, and this is force P that
the man is pushing. This is like what you
have here. Perhaps piece of furniture that
the people who [inaudible] they put two boxes
on top of each other. They put it on your
driveway, and you want to push this one inside
the garage or something there, and something
going drastic to happen and somebody's going
to get mad. I'll tell you in a minute what
is going to happen because that's the question
you want to ask, yes or no? Is that correct
or not? So if I have this scenario, the height
of each box is three feet each. The depth
we don't care. The width is two feet and it
is sitting on the ground. The coefficient
of the static friction with the ground is
given equal to point, let's see how much static
per friction do we have. That's given equal
to 0.3. Yes. And then the coefficient of the
static friction between the two boxes of 0.8.
Let's say these boxes both crate and they
are lots of friction between wood and wood.
Is that correct or not? What's going to happen?
Question. What's the value of the P for impending
motion. Please write it down. It is in your
handout actually. You have your handout with
you? Yes? Oh it's there. Okay, I said anyway
you don't need it there. So what is the, what
is the value of the P for impending motion.
So what are the impending motion. What's going
to happen to this? That's what I said, something
drastic major happens, so what's going to
happen? So what's going to happen?
[ Multiple Speakers ]
First of all, you pushed this one. There is
a possibility that both box, that's what your
intention is, that both boxes move together.
So you have to investigate that. What's the
other possibility?
[ Multiple Speakers ]
Exactly. This one is holding in place the
other one is right on top of that. That's
what I said. The second and third possibility,
which is the drastic thing which you don't
to happen, especially if this a fragile thing
in it, you push it and it drops. Somebody
going to scream at you probably inside the
house. Is that correct or not? Yes. All right.
So what happens. So there is a tipping as
well. So there is three functions that you
have to consider. Tip over, this box is sliding
or two of them. Many of your problem is like
that. So you have to investigate different
type of motion. All of them together. Is that
correct or not? Yes? So let's do -- is this
such as simple problem which is very basic.
You don't have even to do any analysis [inaudible].
Now let's ask you only top box, only top box
moving. If only top box moving, this will
be your free body [inaudible], you are having
here a force P. Of course you have here 100
pound force there, and then you have normal
force, oh sorry. Let's put it in there. So
you have here a normal force of N, and you
have a friction force F, correct. And of course
P is also unknown because it was not given
in here. Of course you don't have to do any
analysis yet. N equal to how much? N equal
to?
One hundred.
One hundred pounds. What's the value of F?
[ Inaudible Response ]
Point eight times 100 is 80. And the value
of P also becomes 80. As I said, this is just
symbolic to show you for you to be able to
do your homework, all your homework is maybe
[inaudible] or maybe it is something is added
to it, but the principle is the same. Is that
correct or not? So how much force do I need
to move that box? Eighty pounds. Now, maybe
before if I apply 40 or 50, maybe the two
box move together. Is that a possibility or
maybe I did need more. Since there are two
boxes, you expect both of them need more force,
but actually it is in reverse. Do you know
why? Because looks at the coefficient of static
friction. Let's see that. Static friction
is much lower, but you don't know that. Therefore,
this is the coordinate for the top box move.
Now we go to the case two, moving both box
together. So this time I'm going to draw it
here next to it. This is case two. So case
two. Now don't do this because there is no
movement here everybody. There is no action
here, so the two boxes stay together. You
do not need to separate that. That, sometimes
I have seen a student do that. They separate
this from here assuming there is a friction
here, there is a friction. That's no, no.
Everybody understand that you are assuming
that the action happened only on the ground.
Is that correct or not? Impending motion here,
this is not moving, so the two are sticking
together as if you are mating it together.
Is that correct or not? There is no motion
there. However, in this case this is the P
[inaudible] unknown. So this is the P, and
what do we have here? We have N and we have
F. Is that correct or not? But what's the
value of F this time depends how much load
we have. The load is?
Two hundred.
No 200 pounds is not, 100 and? Hundred. Notice
that N is larger of course, larger than what
you had there. That was 100. N equal to?
Two hundred.
Two hundred. But what is the value of F now,
see that? The area of F is?
[ Inaudible Response ]
Because this is like [inaudible]. The other
one was wood on wood, so the friction was
much higher. So as I said, in this case the
force here required force which becomes how
much. This is point, what did I give you?
[ Inaudible Response ]
Three. So it is 0.3 times 200, so it is 60
pounds, that's correct. So I write it down
because that's what it is. So therefore P
becomes equal to also equal to 60 pounds.
So this never happens. Everybody [inaudible]
you increase, you put your hand there and
you increase that. At 60, both of them are?
Moving.
Moving. That's the end of it because that's
the impending motion. Is that correct or not?
However, if somebody comes and puts a block
in front of this one, everybody [inaudible]
then the other one may move. Is that correct?
Because you have to overcome the block [inaudible].
You see how this static [inaudible] or the
principle of static works. You now have gone
through a static. You can see all of this
right away. Is that correct. The third one
was something new. The third one is a tip
over. So it's going, the whole system, that's
the drastic one. The system is going to take
over so let's draw that one out here. Tipping
over is like this. So you draw the two boxes
together, again like that. This time you need
the height. Now this one was look like equilibrium
[inaudible]. This I use it as equilibrium
of a [inaudible]. When it comes to tipping
over, there are some forces that causing the
tipping over, which is the P. There are some
forces like W which keep the box in place.
Is that correct or not? So this is what you
do. [inaudible] So the tipping over will be
about this point. Is that correct or not?
Yes? So you assume. That's exactly what you
assume. You are assuming that this has been
separated from the ground. This is exaggeration.
This box has little bit, just [inaudible]
separated about this point. Therefore, you
put your N and F there. So this is your normal
force at point H, and this is your friction
force there. And then of course your T is
here. And then your 200 pounds is like that.
This is 200 pounds. This [inaudible] and that
is [inaudible] is 1 foot and 1 foot. That's
the distance. So all we have to do in order
to calculate the value of the P, we have to
take them over [inaudible] 0.8 because this
is, oops, so I already added it, about 0.8,
so therefore we are saying that of course
this N is equal to? Two hundred. Again, we
do that, so it becomes P times 3, P times,
okay, P times 3 equal to, this is very simple,
3, going that way, which is negative plus
200 times 1, which is positive equal to 0
or P become equal to 200 divided by 3 equal
266.7 pounds, and therefore that is going
also to cause this one, that to happen. So
between the three, which one is going to happen
first?
[ Multiple Speakers ]
Now what happened is this. This is really
something that you do automatically. The tipping
over does not depend on the value of the P,
I can prove that for you later on. It depends
where you put your hands here. Remember? If
the guy puts his hand at 5 foot there, the
value keeps changing because the movement
becomes larger. In that case, it's going to
tip over. That's why when you want to move
something you are afraid of your back also
or you usually bend down. Is that correct
or not? Yes, you bend your knee and then push
something. Is that correct or not? Because
you are afraid of tipping over. The higher
you put your force, the more likely to tip
it over, automatically that becomes [inaudible].
Is that understood? Now, one more question
is asked here, which you don't see it there,
which is very interesting also. The next question
that's been asked in this problem, can't a
man pushing the box, can push the box. I mean
you send a child there after to push it. She
cannot do it. Is that correct or not? So the
man should be able to? What's the answer between
these three first of all? What's the answer?
Is it -- of course it's not 66.7, it's not
80, it is there.
Sixty.
Sixty. Now we have to create a situation that
a person should be able to push 60 pounds.
Is that correct? Yeah, [inaudible] my problem
is stuck. How do I draw that man on the board,
you know. Okay, so some of you do much better
drawing that. I do, but I try my best anyway.
So, anyhow, here is the man.
[ Multiple Speakers ]
It is [inaudible]. What's the weight of the
man?
[ Multiple Speakers ]
One hundred?
One hundred fifty.
Fifty. Not even that. Worse, I have to draw
his hand too. So therefore he is pushing there.
Which direction is the force now? That way
or this way?
This way.
Only [inaudible].
[inaudible] This way?
This way.
Correct. So okay. So here is the 60 pounds.
That's right, because the box pushes toward
the man. The man pushes toward the box. Is
that correct? We went through that many, many
times. Correct? So 60 went there. And then
we come here. That's the same. So you can
choose both [inaudible] together at one piece.
So therefore here, if this was the floor,
you put here N, and you put here F. Is that
correct or not? Okay. What's the coefficient
of friction there? I forgot how much. Is it
the same 0.3 or is it different?
Point three.
Point three?
Yes.
Okay. So if this is N equal to 150 pounds,
correct or not? That's 0.3. You don't have
to calculate. F becomes equal to 45 pounds.
And this is 60, so the man cannot [inaudible].
So what is going to happen to it? [laughter]
You remember. Michael Jackson. You know, [inaudible]
. That's it. So okay. Because his shoes is
sliding like you are standing on the ice,
you want to push a car. It's impossible. Is
that correct or not? For that reason, the
owner of the project, whoever it is, asks
[inaudible] shoes that have lots of rubber
underneath [inaudible] actually I'm joking
because these are the one that it is for labor
people. See all of them have special shoes
because those are regulation actually. For
two reasons. One, they are safer because they
cannot slide. Actually, it's for safety reason
not for that, but at the same time, he can
probably, because that's why wearing that
shoes, this is a coefficient of friction.
Maybe it's more than three. It becomes more
than four, so you are, he is able to push
it without realizing it. Is that correct or
not? So there are two advantages, but really
it is for safety because they are sliding.
They are going on top of the beam construction,
so everybody has to have shoes with a larger
friction. Everybody understand? The same thing
with ladder. So these are the type of questions
I'm not going to discuss because we have much
more [inaudible] . So if you have a ladder
here, your ladder is there, the ladder has
a weight. Is that correct or not? The person
has a weight too. You put the weight of, put
it down at [inaudible] because some of your
homework maybe look that, but this is nothing
there. The [inaudible] I'm going to discuss
it in a minute. Is that correct or not? Yes?
So therefore there is the weight of the ladder.
Then there is person here with the weight
of the person with going up. They give you
this angle. You are putting it but how steep
you put the ladder there. But the ladder touches
here, touches the ground and touches the wall.
Is that correct or not? So you have friction
here. You have friction there. And the ladder
wants to slide. And you go up the ladder [inaudible]
slide so you have a friction going this way,
you have a friction there, and you can study
that yourself to see how far the man goes
up before it slides. So very heavy man wants
to go on the ladder, it will not work. No.
That's the fact. Is that correct or not? Therefore,
you have to have a friend or somebody push
the ladder that way or and that's reason they
put some rubber here. Every ladder you buy
this is rubber there to increase the friction
in order for safety again. Everybody understand?
So you go to the Home Depot, you buy a ladder,
they are not steel, they are rubber there.
And these are all part of a friction problem
that you can do that. The next one that we
want to discuss, all these problems are the
same sort, so they are all the boxes on the
slope or there are some sort of boxes like
that. [inaudible] for example that this problem
is boxes like that, and so on and so forth.
There a lot of important things that you can
worry about. You can resolve it. However,
mathematically some of them become a little
bit more complex, a few point here and there
probably you are going to ask me about that.
The next object is sometime we are using a
little piece of wood, which is cut in angle,
and we call them wedges. And the wedge, please
write it down, the wedges are the purpose
of -- okay if I have time I will do another
example as well, but let's leave it at that
for time being. Let's go to the wedge because
I want you guys to finish all the problems
there. So look at the problem on the wedges.
The wedges are generally used for leveling
something. So you are constructing let's say
a freeway and there are, part of freeway which
is on this elevation and the people are going
to pour that for concrete after you put the
beam, etc., for concrete. So you put lots
of support system [inaudible]. These boards
have certain length. Let's say at one part
there is a quarter of inch of sag, and you
want to correct that. Is that correct? Or
you don't go there, change all the support
system. What they do have usually this is
what happens. This is what I'm saying [inaudible].
So let me erase the board if I find my eraser
somewhere there. Okay. So what I'm saying
is that you can use the switches for different
purpose of course. One of them is the following.
That for adjustment of the height or the level
of the part of operation that you're doing
there. This is, I said, this is a part of
a freeway ramp under construction, and there
are a series of wooden posts here or steel
posts here, which is sitting on the ground,
and the inspector goes there to check that
one and says oh this one happens to be a little
bit lower, you want to raise it a little bit.
So by doing that, because the system is there,
so what they do, they put, if this is the
column, under the column is this column coming
here, under the column they put something
like that for example. Because two pieces
of wood, small wood like this, notice what
happened here. They are sitting on this in
this angle. Is that correct or not? Therefore,
if I push this one a little bit this way,
everybody understand, this will go higher,
is that height. If I want to lower that, I
have to push it that way. Is that correct
or not? If I push it a little bit, just not
too much, push it that way, that this becomes
adjusted. So I think the question comes up
if with have wedges like that and we want
to calculate this value of the P, we don't
want there to go create [inaudible] because
[inaudible] causes this to move out of way
so it's going to break the whole system. Everybody
understand? You want to find the value of
the P for impending?
Motion.
Motion. Is that correct or not? Just how much
force you should put there for a slight movement
there. That becomes wedges. Is that that correct
or not. That's an example of that. You will
see again problem number 52 and 55, but let's
look at example number two in your handout.
Example number two and three in your handout
are the wedges. And let's see how we can resolve
that issue. So therefore, this is simple problem,
but you have to be a little bit careful how
to handle that one. So --
[ Shuffling Papers ]
Okay. Here is a beam.
[ Writing on Board ]
Okay. The length of the beam is given to you.
Okay, we passed that, those, so let's go to
the wedge. Okay. So it is 8 meter long, so
3 meter, 3 meter, so we got 2 meter. So this
is 3 meter here, 2 meter there, 3 meter there.
And here we have a load of 4 kiloliter. Here
we have a load of 4 kiloliter, according to
the system. Here we have a load of 2 kiloliter,
and here we have a load of 2 kiloliter. Here
I just put it because the [inaudible] in front
of you, but if you don't have it, I put it
here, there. This is point A, and that is
point B, but look the way that point B is
[inaudible]. The point B is, a wedge is put
here like that. A wedge is there. And that
is a force P, which is unknown. You want to
answer that. And that is sitting on an I-beam
or something like that, which is immaterial.
That means it is in contact with the floor,
and the floor is here or something like that.
And this P and this A are in the same level.
Again, the question is after having all of
that and having the coefficient of a static
friction [inaudible] for wedge and the beam
is given equal to I believe static friction
is given equal, how much is [inaudible] 0.285,
0.285. We want to calculate the value of the
P for impending motion. In other words, level
the beam a little, correct? Now, of course
you have gone through a static in that you
know that what you should do. You should find
that, you should draw the free body diagram
of the beam and freeway diagram of the wedge,
because this is too parts. That one part will
not do it for you. So let's look at each one
separately here. So I'm going to draw the
freeway diagram of the beam first. So let's
see what we have here. Of course, that one,
will, I draw it here. Okay. If I get my pens
working now. All this pen is going without
ink. So, all that, hopefully one of these
work. Okay. This is fine. So, here we have
the beam at, you put here all the load there,
2 kiloliter here and 4 kiloliter there, 4
kiloliter there. These distances are given.
At point A you put AX and?
[ Multiple Speakers ]
AY? Okay. AX and AY. The distances are given.
It is here, so I don't have to repeat that.
Then on the wedge here, because it is sitting
on this, the beam is sitting on the wedge
and it is a rough surface, and it is moving.
Therefore, I have to put there an N, and I
have to put an F, which direction?
Right?
This is the direction of the [inaudible] . That
way. You see? Everybody says that way. More
or less. In every class. But in every class
you are wrong. [laughter] This is very important.
Actually this is one of the principles that
the only reason I'm doing that exactly for
that one. That, you have to consider the action
and the?
Reaction.
Reaction. Now everybody makes that at the
beginning mistake, which is fine. That's exactly
why I asked that question. [inaudible] that's
right. You got it. Okay. So, let's get [inaudible].
Now you have to draw the freeway diagonal
of the wedge. So my advice to you, always
draw the freeway diagram of wedge first before
the beam. Is that correct or not? Now if you
do that, so draw the freeway diagram of the
wedge underneath here, which would be something
like that. I draw it a little bit bigger because
that's small. Of course, the angle of the
wedge would be given to you. What was the
angle of the wedge for this problem? The angle
of the wedge is?
Fifteen.
Fifteen degrees. Because if there is not an
angle, this will be not be a wedge. Is that
usually a block? Is that correct or not? Yes?
It has to have that angle. Then, here is the
[inaudible]. Now you saw it need to be. This
is force P. Is that correct or not? This is
force N, correct? Now where should I put the
friction here? Now you see what [inaudible].
This is moving. Not the other one. The other
one was in place. That's reaction to this
action. Is that correct or not? Therefore
do you agree that I should put your F here?
Yes or no? Okay. So put an F here. However,
this side is also in contact, yes or no? Yes.
Therefore, I should put here another N but
not that one. So let's call this N floor and
F, and also here is friction. At the time
of [inaudible]. So call it N1, N2, or N top
or bottom or whatever you want to call it.
Now this is the scenario. As you know, you
see, since this is going to the right, this
must be going to the left. So that's why you
were wrong. Because you were looking at this
one. And without realizing, you wanted to
look at the block first. You see, that's what
you answer. Is that correct or not. Which
is typical of every class. It happens. Is
that correct or not? That is exactly why I'm
raising this question. So be careful about
that. [inaudible] two blocks there, and you
are in problem number 52, two or three blocks
on top of each other. First assume which one
is moving and then decide on that. That's
the key. So it's very simple. Yes? See this
is the motion of the block. This is the force.
This is the friction [inaudible] motion. So
here I have to put equal at, so in other words
this block tries to push this beam this way,
not the other way. Is that correct or not?
Yes? Anyhow, now can we solve the problem?
Yes.
[ Multiple Speakers ]
What? One more force [inaudible] . All right,
of course. [inaudible] all right. That because
this is symmetrical. This is 2 meter, and
this is 3 meter, and that's 2 meter, so I
do not have to work for these two. These two
must be equal and equal to each other because
symmetrical. Yes or no?
Yes.
So [inaudible] . So that's 4, 4, 8, 10, 12,
so AY must be equal to 6 kiloliter, and without
doing any calculation. This is 6 kiloliter.
As a result, AX must be equal to F. And F
equal to how much. F equal to, notice I use
the symmetrical idea to calculate AY equal
to N equal to total load, which is 12 kiloliter
divided by [inaudible] kiloliter, and I calculate
the value of F, which is 0.25 times, 0.25
times 6 kiloliter, so it gives me 1.5 kiloliter.
So what happened here now? This is known [inaudible].
This is known [inaudible]. This is equal to
6 kiloliter, and that is equal to 1.5 kiloliter.
Still I have two more unknown, yes or no?
Yes? Okay. Can we solve this problem? Yes,
we can. However, there is a little key here
that I want to discuss. This is the problem.
One side of it is the slope [inaudible]. You
can write it as sigma FX and significant FY
without any difficulty. But I am going to
ask you to do something that is to happen
in every of your homework, so you should do
that. Anytime is a slope like that, this is
[inaudible] to X, this is [inaudible] to X,
this is [inaudible] to Y, yes or no? However,
these two are an angle of? Fifteen degree
or with the horizontal or angle of 15 degree
with a vertical. Is that true or not? So this
has two component. And on X and Y, this has
two component. Usually mathematically, this
becomes much more complex. So you should reduce
these two to one force. Now what's that one
force? Let's go to through the operation of
that one. So this is how we are discussing
in general. Let's assume the surface is horizontal.
Go to your next page. Write this process down.
And you can do it on this page by the way.
This is the same thing I'm talking about.
This one. So you can write it there. If you
want not to use your notes, but underneath
draw a different picture. Those are on the
slope, which is a little bit more advanced.
And let's first get the idea of what we are
talking about. Let's take that there is a
box here and we found out there is a normal
force there. Is that correct or not? Yes?
And there is, there should be [inaudible]
Let's say assuming the friction, the box going
that way, the friction is this way. This is
just assumed that way. Is that correct or
not? When there is an F and N, we already
decided that F and N are related to each other
at the time of impending motion with the coefficient
of a static?
Friction.
Friction. Therefore, this is what you do from
now on. So, you can actually reduce this two
force into one force. Is that correct or not?
So you draw one force here, like that. Let's
call it R. R is the resultant of N and F.
And we call this angle generally V. Okay.
So look what happened there. [inaudible] of
V is equal to what? This length. This length
is F. This length is N. F divided N. However,
at the time of impending motion, F equal to
[inaudible] N. So you have [inaudible] static
times N divided by 8 [inaudible] so actually
that V depends on the coefficient of static
friction. That's why we call it angle of friction.
Is the correct? So if we have [inaudible]
equal to 0, that angle is 0 automatically.
If [inaudible] equal to 0 times [inaudible]
equal to 0, V equal to 0, that's our problem
we had before. So the N is like that. [inaudible]
increases to certain level. The angle of V
increases. Everybody understand that? So you
put this one there. Is that correct or not?
Let's see whether I have that picture here
or not. And this is matches this one. Yes.
It is matches that one, good. But it doesn't
matches that [inaudible]. Okay, I have to
draw another one. Okay. So now what happened
here now? I want to reduce this to, into [inaudible],
make it much simple for myself. Remember that
one. Where is the V. First of all, where is
this angle? What is this angle?
Fifteen.
That's angle is already 15 degree. Where is
the, where is the R. R is here. Is that correct
or not? Let's do the R here. This is the R.
Is that correct or not? This angle is V. Is
that correct or not? So what's the angle of
R with the vertical? The angle of R with the
vertical is? Here it was only [inaudible],
remember. Because [inaudible] if I shifted
15 degree, look I have to add there 15 degrees,
15 plus V. So I have to calculate V first.
So what I'm doing here, I calculate V equal
to [inaudible] minus 1 of 0.25, which equal
14.01 degree. I have to go to three digit.
Is that what you got? Yes?
[ Inaudible Response ]
Yes, I know that answer. Okay. Yes?
Yes.
Correct. All right. So the total angle is
how much, 14 point, actually it's [inaudible]
if this is 14.1, you should put it here. But
this is 14.41, so I'm going to forget about
that. Everybody see that? Three digit. So
what's the total angle here? S29. So you've
reduced all of that. This one and that one.
Everybody see what I'm talking about? You
here we come 29 degree. Now, I add these two
together. Is it always adding? No that depends
on the slope. That depends on the feet. If
the F was coming from the other side, I should
shift it. The other side. So if we come V
minus 15 degree, so that's why you have three
different options. I don't want you to look
at that one because that's out of the book.
Calculate your own based on which way you
are moving. Is that correct? Just put your
vertical line here. Just draw your art. Sometimes
you're adding. Make this in your notes. Sometimes
you have to add. Sometimes you have to subtract.
This is not always going in one direction.
Therefore, we go from there and then we calculate
now. It's become very simple. Okay. Because
now you can write sigma FX equal to 0 and
sigma because this one now has the one component
horizontal and one component vertical. You
can do it. There is no problem. Everybody
see that there? So unfortunately, I put that.
I was hoping to get to the last to the last
two problems. I don't have that, but please
between now and next class Tuesday do all
your homework except the bell problem because
we have two more very important chapters to
follow, and we have enough time. We are not
short of time, so we should be able [inaudible].
So please do everything. You have all the
questions. We can answer that. But please
come before 1:30 [inaudible].
So the homework's due Thursday?
Yes.
Thursday?
Not Tuesday.
No, it will be next Thursday.
Okay.
Because I'm not finished yet. So it is not
due. Your homework will be next Thursday,
and we are going to have a quiz we will do
next Thursday as well, so please, let's put
everything aside and let's go to the quiz
then. This is section three, okay, section
three.
[ Music ]
