[ Music ]
You've got 12, I've got 20, or whatever. You
did it in the format that I did not recognize;
whether you understand the concept or not.
If you are applying the concept, then you
most likely would understand part B. In other
words, there is no reason to do what you are
doing. Many of you added the forces and right
sigma if X is equal to that [inaudible] and
four is equal to that. Why? If you can explain
why you did that, then you can do part B.
Otherwise, it doesn't make any sense going
to even part A in that format. This is the
idea. This is the system. Yes or no? Of course,
you'll break this force with an angle, this
force with angle. You break it into component.
That's all you need. Is that correct or not?
What's the question number one, guys? Reduce
this to one force and one [inaudible] at point
A. OK. So this is it. The same system, everything
is the same. OK. This is the same. [Inaudible]
is the same. Now I show this in color in order
for you to see what you are going to do. You
are going to reduce all of them to one force
FA and one [inaudible]. Since this is a 2-D
problem, I don't have to put this in vector,
but let's put it in generally as a vector.
Is that correct or not? This is the idea now.
Now, how many forces do I have on that one?
Many forces. How many forces do I have in
system two? This is system two now. Remember
that? That is system one. Let's call it as
it is. This is system two, right? Now, how
many forces is here? Only one. Yes. Summation
of the force in the system one must be equal
to summation of the forces in system two.
Therefore, if that's the case, I'll put it
in between. Summation of forces in system
one must equal to summation of the forces
in system two. But how many forces do I have?
How many agreeing forces do I have? Only one.
Since this is FA, then all of that when added
together must be equal to? FA. Isn't that
what you used or what you were supposed to
use? Either way. Yes or no? Correct. So that's
why you're adding all of those forces together.
Otherwise, this part is the green part. This
part is the black part. Yes or no? However,
this part, you only one force because you
reduced it only to one force [inaudible] couple.
Yes or no? [Inaudible]. The second part here
is FA. Many people write [inaudible] and leave
it at that. And I asked you why? I don't have
-- so I gave you the benefit of doubt, I thought
you probably know it, and so I gave you full
credit. Is that correct or not? It still is
not complete if you did not write for me what
is FA and what is MA. Everybody understand
that? So FA, which is, because this has reduced
to once force, so all of that has to be reduced
to one force. Is that correct or not? FA.
Anyhow, so what should I put here? What you
see here, which is very simple. Minus 250,
minus 200i, and then minus 433, minus 300,
and minus 150j. You should not have any problem
with that. Although, a couple of people did.
Is that correct or not? Yes. So total becomes
equal to -- so you write it down and you add
up together. This becomes minus 450i because
it's 200 and 250. And these are all negative,
so it becomes minus 883j. And all of that
should be equal to FA. All of that should
be equal to that. Is that correct or not?
But many people, remember, write sigma if
X is equal to that [inaudible] is equal to
that, and you didn't specify. Some of you
said, "OK, this is sigma F at A." That's not
the case. Sigma F can be anywhere. Is that
correct or not? Sigma F doesn't have A, B,
C. Then you take the moment, you have to take
the moment about [inaudible] A, B, or C. Everybody
understand? Summation of the forces or summation
of the forces. Correct? So now, what's the
second part? This was condition what? Look
how simple it is. That's all I'm asking you
to do; add this until you get the right [inaudible].
Many people, of course, did, so I add a few
points, yeah. And then what's the second condition?
Summation of the moment of system one about
any point you choose, should be equal to summation
of the moment about the same point in system
two. But since we are reducing it at A, obviously,
this would be at A, and that should be at
A. Yes or no? Now, what is the moment of these
two? Now, here is the key. If I want to do
this, and these are four forces here, I have
to take the moment of all of that at what
point? A. But what is the moment of FA about
A? Zero. What else is left here? MA. So this
part is equal to? MA. Is that correct or not?
So let's write it down. The second part is
equal to MA. Let's put it in magnitude, because
all of the M's -- by the way, I'm sorry to
say, if you have not recognized that at this
time and you are putting for this moment IJK,
something is totally missing. Everybody understand
that? In the 2-D problem, moment is all about
the k-axis. [Inaudible] got that. But still,
there are a few people who have not done that
properly, and I am a little bit surprised.
Anyhow, all of the moment of these two are
either like this or like this that. We studied
this [inaudible] lecture one about whenever
you are talking about a moment. It either
goes like this or it goes like that. Is that
correct or not? So there is no IJK. It's much
simpler than that. You have to take the moment
of all of these forces and add this couple
to it about point A because that's where I'm
reducing it. Is that correct or not? Therefore,
of course, you write this number down which
is the [inaudible]. First of all, you have
400 there sitting as a positive moment. [Inaudible].
The unit is [inaudible]. You can put it in
front or you can put it for each number, so
[inaudible] for the entire equation. Then
you have there 300 times three. Look at that.
300 times three. You are taking moment [inaudible].
[Inaudible].
Oh, about point B?
Yeah, because it all starts [inaudible].
Oh, I'm sorry. OK. So I'll have to change
that. OK. I'm sorry, because this is for the
other class. So that's for point A, but we
are not doing that, so let's change that.
Notice this one has no reference to the point.
So it doesn't matter whether it's A, B, or
C, and D. This time, OK, let's choose then
point D and D. That's what we want to do.
Is that correct or not? But notice this time,
let's change that color to something else
because this time we put the red and then
we try to [inaudible] like this. This is FD.
I'm sorry, I just got this -- we are doing
it about A, so FD and FB. Therefore, we should
change this to MD. Because when you take the
moment about point D, we only have MD, and
FD doesn't have any moment there. So we could
do it from A or we could do it from D. Now,
this equation, you were [inaudible] about
this, so let's change that. This never changes,
so [inaudible] change that to D. Sorry for
that because I forgot that, didn't look here.
All right. Is that correct or not? Now, take
a moment about point D. So you have 300 times
three. OK. So that works there. And then we
have 400, 300. Then these two don't have any
moment. Then we have 250 and 433. So what's
the 250 times two? Look at it. 250 times two,
and it's going that way, therefore it is negative.
And finally, 433. Let me look here because
not looking here, I may make a mistake again.
433 times five. Correct. So 433 times five.
And this one is [inaudible] positive. So you
put positive there. And then you sum it up.
So the sum, it should equal to -- so MD in
magnitude becomes equal to 2,965 meter. Meter
in a vector form -- some of you wrote it [inaudible]
2,965k. [Inaudible]. Is that correct or not?
Yes. As I said, summation of the forces does
not have to be about point A, B, C. That's
why we get to the next part. Now, what are
doing in this part that many people missed
it? What are doing? We're reducing it to one.
But you have to use the same idea. That's
the problem. Many people just forgot about
the idea, just they took the moment about
point D and the force about D without realizing
what is the background, what's the idea behind
it. Still, the same equation applies. But
this time, this is your system. So you have
to draw that, but I mean, if you don't do
it, it still is OK, but as long as you understand
what you are doing. Is that correct or not?
This time, we want to review set two. One
force only. Where? On which line? OK. Let's
look at it. Line AB or CD? AB. OK. I want
to review set to this somewhere here. Is that
correct or not? What is the minimum force
I can have? Notice minimum force is this.
Is that correct or not? I cannot reduce it
to less than one, that's all I have here.
Is that correct? I already reduced it to once
force, so that one force will be equal to
450 negative meter. And what was the other
one? 883 meter. Is that correct or not? Yes?
Yes.
Now, what is the first condition? Now apply
that condition. I don't understand why you
cannot do it. So that shows me that part A
was a little bit on shaky ground. Everybody
understand what I'm saying there? What did
we use here? Summation of the forces in system
one should be equal to summation of forces
in system two; should be equal to summation
of forces in system three. Are they the same?
Yes. Minus 450i, minus 880, 3j which is here,
where is there. So that part is good. What
is missing now?
The moment.
Moment. Moment about which part? Now, here
is the key. Why don't you do a moment, but
you apply your technique rather than trying
to remember, and then emulate that [inaudible].
Try to remember what was involved. Many of
you wrote for [inaudible] sigma M equal to
sigma M. But I said where -- no question there.
You have to choose either A, B, C, or D, or
any point. Everybody understand that? That's
why I was thinking about A. So let's take
a moment at what point?
A.
A. Who said A? B? C? Any of them will do it.
Is that correct or not? If I choose moment
about point A, the only thing [inaudible]
is this distance. Is that correct or not?
Yes. That distance is one, or D, or whatever
you want to call it. Because the location
of that point is not in my hand, I have to
locate that point. And that's what I did in
class for you too. I'm surprised because this
example was exactly the same example like
I did on board for you. Go back, take a look.
It has four forces and it has a moment. You
remember that? Did you look at your notes?
No.
[ Laughter ]
So that's the problem. Now, what is here?
Now, remember that. I do not need to go back
to that one. Why I do not need to go to it?
It's more work to go here. I can't use it.
I can't use moment of all of that about point
A equal to moment of all of that about point
A. A couple of people tried that one. But
that's the longer version of it. What is the
shorter version of it? Isn't that reduced
to this? Is these two are the same? So why
not using this instead of the one? Is that
correct or not? Yes? So summation of this
system, red system, with respect to A should
be equal to summation of the moment of this
system with respect to A. Yes or no?
Yes.
So please write down. This is the solution
to that. Or any other point of your choice.
So summation of the moment of about point
A of system two, equal to summation of the
moment about the same point A of system three
now. Now, I have three systems here: system
one, system two, and system three. In other
words, all of that reduced to one force. Everybody
understand what happened? All of that in system
one, it was in done in two steps. First, I
reduced it to a force and a couple at D, and
then I reduced it to one single force, which
is better than going in one step. You can
do that. Now, what's the moment of this one?
Look at this one. This is the green one. What's
the moment of this one about point A? Does
883 have any moment about A?
No.
Does this one have any moment? Everybody can
do that. This is the whole idea. Everybody
can take the moment, but the idea behind it
was missing. So it is 450 times?
Y.
Y. OK. 450 times Y, equal to what?
The moment of --
Of all of that. Let's go back, put it in,
rather than put it [inaudible], this is the
way I did it, that this was in [inaudible].
Yes. But I know, I know what FD is. What's
FD? I can change that to actual FD. FD is
how much? This should have been -- remember
I made a mistake there. This should have been
FD. FD was equal to minus 450 minus 883. So
actually this is 450. And that is 883. Is
that correct or not? [Inaudible]. Now, does
this one have any moment about A? Yes. Does
this one have any moment about A? Yes. Is
this a couple? Yes. I have to add all of them
to get it. So what I should write here, which
I write in red, belongs to system two, is
this. First, I have 2,965 moment there. That's
a couple. OK. I'll put that there. FD is this
one, so I don't need that anymore. And then
I have 450 times this distance? What's that
distance? 450 times five. Yes or no? So it's
plus or minus? Plus. Plus 450 times five.
And then what do I have? I have 883 times
this distance. That distance is also equal
to five. Is that correct or not? Minus 883
times five. And these two hasn't to be equal.
You calculate Y. Y ends up to be equal to
1.78
[ Silence ]
Now, many of you, I don't know why [inaudible].
Many of you [inaudible] write M equal to FD.
No clue why. The whole point is just there.
You got the first part right for some reason.
Everybody understand that? Notice what I'm
saying, that both part A and part B are using
the same principle. Everybody got what I'm
saying there? Everybody understood what I'm
saying there now for future? Yes? Now, what
was missing is this. Probably some of you
saw what I was doing, that what we do here
is this. One more time, repeat. If I want
to reduce it to A, which I thought at the
beginning that's we are supposed to do, the
force is always reduced to what? It doesn't
matter about the point. But where I take the
moment about A will be a different number.
Is that correct? Here, I just put the [inaudible],
I asked you to reduce to one force and couple
at point D. So that's your force. That's your
M. There is one [inaudible] the magnitude
of that came out to 2,965. So I accept that
from you any way you wrote it down. Which
I had a little doubt, remember that, whether
you understood the first part or not. Then
I asked you the second question, which shows
exactly what you have to do. The second question,
reduce it to one force. Again, you may have
taken moment about here, or you can take moment
about this point. Everybody understand that?
This is what you see in the book. Now, I see
some people have looked at the solution in
the manual without understanding what's going
on in that. What they do in the solution manual,
probably, I do it this way for you to get
the idea straight. I don't care whether it
takes a little bit longer or shorter. What
they do in the solution manual, they use this
point as a reference point. Is that correct
or not? So they are saying the summation of
the moment of system two about point D must
be equal to summation of the moment about
point D. I understand what they are doing
here. Then they say here, OK, what's the moment
here? What's the moment of 450 about D? Zero.
What's the moment 800 about D? It's zero.
What moment do I have? 2,925. So the moment
of these two about D also must be equal to
2,965. Everybody understand? Then they use
this distance and this [inaudible]. They are
not using that. They are taking the moment
of that about this. Everybody understand?
So you get [inaudible]. Y and that distance
x. Everybody understand that? You saw that
in the solution manual. I hope you understood
it. But that's the danger, I told you before.
When you're looking at the solution manual,
I can see that you all have some great idea
-- a couple of you -- that have looked at
the solution manual trying to emulate that,
and you are writing it without realizing what
you are doing. Is that correct or not? Because
that's the simplest solution of all -- is
not to take a moment about A as I did. Look,
about A. I have to write all of this. Everybody
understand that? But about D, the moment of
all of that about D is equal to that one.
Yes or no? Yes. So the moment of these two
also about D should be the same. Is that correct
or not? That's why it comes distance x and
distance Y. Is that correct or not? Everybody
now understood what you did wrong? But that's
the whole idea. Before quiz; smiling. After
quiz; frowning afterwards. Yeah. OK. Now,
next time when there is a quiz, let's start
off smiling before and after. Yes or no? Which
gets me to the point -- see how much explanation
I did here? That makes me change your quiz
this time to Tuesday rather than Thursday
because I'm afraid I will not be able to finish
all of the lecture today for you to be able
to do your homework. So next quiz will be
not this Thursday, it will next be Tuesday.
And your homework will be due next Tuesday.
Everybody understand that? So that gives you
still another week to finish the subject which
is very important.
[ Inaudible Audience Comment ]
All right. That means that the subject on
hand is very important and there are more
coming. So please be careful.
[ Background Conversations ]
All right, guys. Now, let's start with -- first
of all, let's start with what we are understanding
from there; reaction. Any time you constrain
an object from moving in a certain direction,
you have to put a reaction there. Is that
correct or not? That depends on the type of
connection that you have. If it's a 2-D problem,
you only have horizontal motion, vertical
motion, and one location. This -- we'll discuss
it. Even somebody looking at the reaction
that we put, we always put it either negative
or positive from your side. Is that correct
or not? But when it gets to 3-D, I explained
that to you last time, any point in the space
can move this way, can move this way, it can
move this way. Then we have three constraints.
If we are preventing this from motion, we're
putting this in the 3-D against the wall,
and it is a ball and socket. Therefore, we
have to put AX, AY, and when we separate it.
But however, ball and socket can do this,
is that if it is fixed, actually all six of
them become unknown. AX, AY and AZ, MX, MY,
and MZ. So one [inaudible] takes care of all
of your six equations of equilibrium. Yes
or no? And now here in this building for future
reference. Again, I always do that not to
scare you, to help prepare you, that's what
it is. In this building, which is five-stories
high, I believe it's five. It's either five
or six, I don't know. Five. Is that correct?
Let's say there is ten rooms this way and
four rooms that way. So ten times four is
40. 40 times 5 is 200. Each corner is, when
the [inaudible] and columns come together,
it's six unknown. So 200 times six, it is
1200. Not kidding. 1200 unknown in the frame
of this building.
[ Laughter ]
And you want to solve it. It's six equation
of equilibrium. That's a little bit too much.
Is that correct or not? But you will learn
a way when you come to ME 218 and 219 in the
future courses, CE courses, ME course. Little
by little, you'll learn how to handle those
kinds of problems which you have more unknown
[inaudible] equilibrium. In static, we are
very simple. Your static is all about equilibrium.
[ Laughter ]
Why you are laughing? The simple case. Is
that correct? Because we are basically talking
about all of the determinant system. Determinant
system means -- by the way, this is not in
the book, you want to write it down. Determinant
system means the number of the unknown is
equal to number of equation, static equation.
Write it down for future reference, it doesn't
hurt. Any time the number of unknown is exactly
equal or less than number of equation, the
system is determinant. So all of your problems
probably would be determinant. This system
is called the system of forces, and moments
are called determinant. If number of the unknown
exceed number of equation of equilibrium,
then the system is called indeterminate. You
would get indeterminate problem in future,
not in this class. Is that understood? Yes?
[Inaudible] is a determinant, we'll get to
that. The [inaudible] are not fixed. When
it was fixed, we'll get to that, I'll tell
you what happened to the [inaudible]. Yes.
Correct. That looks like many unknown, but
it is not. Let's get to the next subject which
is the two force member. Anyhow, before I
go to the two force member, let's do couple
of more examples. Because what is important
here for you guys is the three-body diagram,
and understanding what's going on. That's
why I extended the time for you to really
work harder on this type of question. So let's
do one more example with number and then we'll
go forward with -- these are all very simple.
Many of you already have the answer. This
is problem 34 from your book, which is a bracket
like this pinned here to the wall. So this
is pinned to the wall. And there is a rope
here going like that around the [inaudible]
coming like that and attached to here. And
so roughly, this is the scenario, guys. So
this distance is five inch. And this distance
is seven inch. Five inches or seven inches.
And this height from here to here to there
is given also. 12. Just add 60 like that.
Oh, 12 is that one. 12 is total. So this is
six. Oh, this is not [inaudible], how can
it be? [Inaudible] shown here. Sorry. This
is another notch on the scale, so please write
it down. The way they've done it here is like,
they put A here and then they call A equal
to six. If this is 12 inches and this is six
inches, this would be half of that. So it's
wrong, so let's move that down a little bit
just to match what they are saying here. So
here is the pinned location. So the location
of the pin is six inches, and six inches from
top and bottom. So that's what we have here.
Still [inaudible] scale, so let's [inaudible].
I have to draw that because of the explanation.
We have this part [inaudible]. At this part,
this point B, this is point A, and C, D, and
E. Obviously, again, we have to draw the three-body
diagram of the bracket -- this bracket. Is
that correct? Disconnecting that from all
of its connections. This part is very simple,
you have done it. So basically you make a
cut here, you make a cut here. You don't care
about that one. You are looking at this bracket.
And you cut this one. So you make a cut here.
You make a cut here. And you make a cut there.
Is that correct or not? Yes? So four connections
we had and this becomes the three-body diagram.
So the three-body [inaudible], this is the
most important part of this analysis. This
was point D at the middle. So six inches and
six inches. And then here is A, let's put
it in red. So when I cut it here, I have here
a tension T. Is that correct or not? Yes?
But this is the same cable, therefore I have
here tension T. [Inaudible]. And here at D,
I have DX. Now, somebody asked me [inaudible]
all the reactions are always positive. Of
course not. The reaction could be positive
or it could be negative. So what I'm doing
here is this. Whenever you draw your three-body
diagram, all the unknown always --write it
down in your notes -- all the unknown always
should be in the positive direction. So if
the number comes positive, it's positive.
If it comes negative, it becomes negative.
But if you put it in the negative direction
and it comes negative, it should be -- it
doesn't sound right. Is that correct or not?
So please, make sure that you write this down.
I don't care what the forces are there and
et cetera, et cetera. All I am saying is that
-- oops, I forgot to put the extra force here.
There is a force P without that. We don't
have anything there, so we have a force P.
And the force P was given equal to 90 [inaudible].
OK. So that's the three-body diagram. This
is the low. This is the actual [inaudible].
The red one is the reaction. Notice usually
it's a good idea to show all of the reactions
in red. But notice, this is T and that is
T. We have only three unknown. And we have
three equations. Yes or no? Now it's your
choice which one to use first. You can write
sigma FX, sigma FY and sigma M. But usually
it's better to write sigma M first if you
do it right way. Notice, this is point A,
this is point C, and this was point D. And
this was point E. Obviously, when you want
to [inaudible] moment, you take the moment
about point B because you are eliminating
DX and DY. That's what I was using in the
quiz in a way. If you go to the D in the quiz,
you are eliminating the low. Is that correct?
That's why it makes it simpler. That's what
they use in the solution manual, and that
misled you. Some of you, not all of you. Everybody
understand that, yes? But the idea should
be established there. Therefore, summation
of the moment. At or about D equal to zero
gives us the following. Now I need this angle
or I need this slope. Look at that slope.
That slope is five, 12, and 13. So this is
five and this is 12. Notice that's why the
total was 12. Look at the picture there. 12
and five. And the other one is horizontal,
so no problem there. So therefore when I take
a moment [inaudible] about point D -- let's
do this one first. First become 19 times six.
Notice no cross product here. Now, here, two
or three people in the quiz, they tried to
use cross product. For 2-D, if you are using
cross product, yeah, you get the answer, but
that's not the point. Everybody understand
that? I tried to make the 3-D into 2-D. You
make a 2-D as a cross product format, that
means you are avoiding your right-hand rule.
That means you have not got the idea of moment
yet established in your head, which is not
good, because that's all you need for future;
finding the moment. This here is nine times
six positive. Yes or no? Then three times
six positive. Plus T times six positive because
T is your unknown. Then here, you don't have
to show that -- there is two of them, horizontal
component and vertical. The reason I did that,
some people forget one or the other. Both
these two components have moment about point
D. Again, you don't go D and the shortest
distance. What is the shortest distance? I
don't know where that is. Is that correct
or not? Still there are a couple of people
using that. You make it into horizontal component
and vertical component, and then take the
moment about point B. Yes or no? This one
is vertical component, yes? So it is [inaudible]
like that. You just go the same format as
I am going. 12 over 13 of T. Is that correct
or not? 12 over 13 of T which is vertical
times this distance which is?
Six.
No.
Seven.
That distance is?
Seven.
Seven. That's correct. That's seven. That
could be an error, a mathematical error, but
it's not essential. And then we go with the
sign of it. Which way is correct? Negative.
Then you add that one. Not because this is
plus or minus, because the sign [inaudible].
And that's it. And then we go to this one
because it's going this way. So that one is
five over 13 of T. And it's going that way,
it still is negative. Multiple the distance.
That distance from here to here is six. And
then we should have it there all complete.
And then [inaudible] equal to zero. [Inaudible]
this because [inaudible] these two are [inaudible].
You calculate T, T becomes equal to magnitude
of T, become 195 pounds, yeah, because that's
[inaudible]. Is that correct? After I do that,
I'm not going to do the rest of this problem.
After you do that, then you write sigma FX
equals to zero and sigma to [inaudible] DX
and DY. Is that correct? Yes. Notice this
part, sigma M and sigma FX, sigma FY, is going
to be repeated for every problem from now
on in this class, next class, next class,
next class. Everybody understand that? This
is essential part of every system in existence.
Every system is in equilibrium, if it is not
in dynamic state. In the static state, it
means it's not moving or it is constant velocity.
That's what it's going to be. Is that correct
or not? Yes? However, what is important here
is three-body diagram. Let's look at some
of your problems in your book. And by doing
that, I am going to go over every part of
the list that I gave you in order for you
to understand very simple technique of doing.
The next problem I want you to consider is
this. There is a wall like that. This is problem
number -- there is a wooden piece touching
the wall here and there. And then there is
a [inaudible] weight. This is problem number
36 in your book. This is point C, and there
is a weight hanging there. Of course, this
thing may slide down, especially if it is
frictionless. They are saying that this wall
and this wall is frictionless. If it is frictionless,
it means this is going to come down. Is that
correct or not? Yes? Correct? And then if
it's going to come down, then I have to hold
it up. So they are putting a cable here. They
call this point, point E. They call this point,
point B. They give you all of the dimensions.
I'm not going to do that because the point
is I want you to show the three-body diagram.
I'm not going to go through the sigma FX equal
to zero, sigma FY equal to zero, and sigma
M, because that is a routine that you have
repeatedly know on your homework. Is that
correct or not? Yes? The key, again, is three-body.
Notice, if I draw here a wrong three-body
diagram, nobody looks at this because all
of the forces are going to be wrong. Is that
correct or not? Yes? So be careful here. So
how do you [inaudible]? Notice this is a solid
wall. This is contact and contact; yes or
no? Again, you have to do this, everybody,
diagram like this; correct? Of course, you
change the 20 kilograms into meters because
you want to have some weight [inaudible] W.
W becomes 196. So 196 meters given. Right.
Notice I disconnect this wall from all its
connections. Now, here -- by the way, they
have given -- let's put it this way, it's
interesting. They have given this distance
equal to 75 millimeters and lots of other
distance. And they have given this distance
from here, from top to bottom. Here equal
to 200 millimeter. They have given us some
other distances, but I don't put it there.
Then we have here a cable being cut, this
wall being cut, this wall being cut, this
wall being cut. Correct? Yes. So therefore,
if the cable is being cut at B, this was point
C. At point B -- let's put here point AB as
well. A and B. So here we go at D. Of course,
we know that part, it's a tension T. So in
other words, the tension in that cable becomes
unknown. And I have to put forces here now.
This is no friction. The force is going distance,
pushes against the wall. Wall pushes back
that way. OK. So that's AX. And this one the
same way become DX. How many unknown do you
have?
Three.
Three. How many equation you have?
Three.
Three. Can you find a clever way of solving
this problem without taking moment about point
A, B, C, D?
[Inaudible].
What? Don't be afraid.
[Inaudible].
What?
Summation of distance.
Yeah. Don't you see these two force are -- we
have only two vertical force, two horizontal
force. Yes or no? This must be?
[Inaudible].
This must be?
[Inaudible].
So I still have this 196. This one is 196.
You see, this is a clever way of learning
the material. This is a couple. This is a
couple. The sum of this couple must be equal
to?
[Inaudible].
So it is short. Everybody understand that?
Did you get it?
No.
You are not smiling.
[ Laughter ]
And if you are not smiling, I have to explain
to you one more time. This is a couple. Yes
or no? The magnitude of the -- first of all,
this is two vertical. Agreed or not? There
is no two. So this must be hundred.
96.
96 [inaudible]. But I don't know yet that
two. Is that correct? Because whatever distance,
the other one should be how much? I don't
know how much to put. All right. This is a
couple; yes or no? That couple MC equal to
what? Magnitude what? It's 196 times what?
That's why I put [inaudible]. Times 75 millimeters.
Yes or no? Correct? Times 75. So 196, using
a couple idea, times 75 millimeters. So the
answer would be in [inaudible] millimeters.
Is that correct or not? However, some of the
horizontal force must be equal to zero, so
this must be equal at opposite. Some of the
moment must be equal to zero, yes? This is
one moment. This one, this is also a couple;
yes? That couple is positive or negative?
Let's put it this way. This one is?
Negative.
Negative. OK. The other one is?
Positive.
Positive what? D or DX, doesn't matter. Times
what? Times 200. I'm taking moment over here.
Is that correct or not? Times 200 equal to
zero. You find DX [inaudible]. So what I'm
saying is that from now on, and when we get
especially to 3-D, don't go with the [inaudible]
product. Find a way that all of those homework,
last line of your homework, rather than writing
one or two pages of math, write three, four,
five lines and you are done. I'm going to
show you all of that in a minute when we get
there. Is that correct or not? Yes? Yeah.
If the walls are a rough surface, would that
[inaudible]?
No, because [inaudible] if the [inaudible]
is the same [inaudible] the company-efficient
of [inaudible]. Yeah, and so on. If they are
the same, again, [inaudible]. But that is
if you cannot find [inaudible]. But then you
can't do that, because you get to the friction
next chapter. Next chapter [inaudible], not
this chapter, after next week will be friction,
we'll talk about it.
[Inaudible].
Yes, it wouldn't be three couples, but [inaudible]
not simply, [inaudible]. Still we solve it.
Because the F equals [inaudible]. We'll get
to that. Anyhow, that was [inaudible], that's
the shortcut of it. I mean, if you want to
take [inaudible] moment about A or B or C,
you can do that, but that's [inaudible]. Let's
go to the next problem. Next problem is problem
number -- I thought this problem are very
interesting, and I'll also show you some of
the techniques that you are supposed to use
in solving this problem. This is number 41
in your book. OK. Problem number 41 in your
book is the following. It is a sort of bracket
cut like this.
[ Silence ]
At point D, we have a force P applied there.
Then here, there is a little cut there, point
A. And there is a [inaudible] here, point
A. At this point, there is a force 30 pound.
P is given equal to 15 pounds is given. There
is a force hanging here at 30, right? And
then there is another cut here. And there
is a [inaudible] here, they call it point
B. And all of the dimensions that they given,
I don't care about dimensions, as I said.
Then there is a roller here attached to the
wall. Because what happened here, when you
pull it down this way, this side wants to
go up and hit the roller, and roller prevents
that from going up. Everybody understand?
Therefore, I have to put the force going down.
Is that correct or not? Yes? So the object
is given, all of the dimension is given. There
is another force. No, there is a 30-pound
force here. There is a 15. All the dimension
is either this, this, that [inaudible], and
we don't care about it because all I am doing
is a three-body --
Diagram.
-- diagram. Not solving the problem for you.
Is that correct or not? All right. So here,
if we want to three-body diagram, we have
to do a few things. First of all, do everybody
understand what's happening here? What's the
difference between here and here? Is this
[inaudible] attached? I mean, it is part of
this assembly or is something in the back?
Something in the back.
Say this is attached to some other item and
this is like that. Is that correct or not?
[Inaudible] that that can be moved away. Is
that correct or not? So when we do the three-body
diagram, we are doing the three-body [inaudible]
this problem. That's what you should do. You
should erase this and take that out because
that's part of the other object. Is that correct
or not? Yes? But I leave it there just for
you to see what happened. Therefore, I have
here a force going like that. That's [inaudible].
Is that correct or not? What kind of connection
is this? Now, remember, I explained that to
you last time. I hope you all remember. It
is in your handout. There's a cut here. There
is a pin which is attached to that. This pin
can move horizontally, but it cannot move
vertically. Therefore, the vertical movement
of this object at this point, because this
pin attached to the other piece. So this front
object cannot move up and down. Is that correct
or not? Therefore, I should put here only
AY, no AX. Is that understood? Yes? Somebody
came to me again and said, "What's the reaction?"
I said, "Well, I explained that reaction to
you." This was from the other class. Because
I have a little behind in the other class
as far as the table is concerned. For you
are going through the whole table. Everybody
understand about that? That reaction, any
time you constrain something from motion,
you put a reaction there. OK. That's one [inaudible].
Is there any constraint here? And a freedom
here. This side is freedom. This side is unknown.
So I have to put here in force. Of course
the angle will be given to you. That's the
direction of that. [Inaudible] 30 degrees
or whatever. Yeah. 30 degrees. Yeah. They
have given this angle equal to 30 degrees
which makes this vertical angle equal to 30
degrees. Here is 30 degrees. This is 30 degrees.
Is that correct or not? Yes? Now, how many
unknown do we have? This, you have to call
it D. No DX. That D has two components: DX
and D1. The rest of them are one component.
Either one component or the I component or
J component. Is that correct or not? However,
still, the principle applies. How many unknown
do I have?
Three.
Three. And I have three equations. I should
be able to solve it. Again, no problem. This
one I'm going to do with [inaudible]. This
is problem number 46. This one has a little
important point in it, so I'm going to do
it. This is similar to one of your homework
as well. This is a key here for understanding,
again, the equilibrium. This is like a crane.
Some of you have already done that one as
well. So let me get my [inaudible] here. This
is similar, it's a smaller [inaudible], but
the idea is there. So if this happened to
be a crane which has this part and this part
-- they have put at this side, they have put
750-meter force there. And they have given
you this dimension equal to 500 millimeter
from here to there. This is A, they call this
one B. They call this one C. They call that
one D. Let's call this one E. At point E,
there is a force of 450 meters. The distances
all are given. So this distance is given,
equal to 400 millimeter. And notice a few
things that I'm going to explain to you. This
very much, except this load looks like a crane.
Is that correct? Have you seen the crane outside?
They [inaudible] the weight hanging here.
And this crane was to go that way. Now, this
could be on the truck. If it is on the truck,
so the [inaudible] has to hold this -- should
not move. Everybody, should not top, I mean,
this can't be -- top of this -- when you have
heard many times that some of these cranes
are going to fall down and people get hurt.
That actually happened at one of the [inaudible]
last few days. And then sometimes this is
fixed if this is -- you're at the Port of
Los Angeles or [inaudible], they are much
larger and they are fixed to the ground. This
one is fixed to ground. Is that correct or
not? Yes? So therefore, it's being put in
the concrete and moved it down or whatever.
So that means this is fixed support. Is that
correct or not? This by itself can't work,
but you have more load here, you have to prevent
this from moving that way. And if you look,
this is not the case here. If you look at
many cranes, next time look right inn here.
They put some concrete slab. Have you seen
that little concrete block there? Of course
you can see that. They are balancing this
moment with that moment in order to make this
stable. Is that correct or not? However, except
this one, if it is mobile. But since it is
to the ground, so they have put here a cable
and attached that cable to the ground here.
Is that correct or not? So there is a cable
attached from point B to point D. The height.
This distance. From here to here is 250 millimeter.
This is 150 millimeters. And the height is
600 millimeters.
[ Silence ]
I have to disconnect this object from all
its connections. OK, guys. For me it is easy.
For you, I don't know what you are going to
do.
[ Laughter ]
I'm going to erase it. Is that correct or
not? But don't put it on top of that. Everybody
understand? What we are putting here replaces
the connection. Everybody understand that?
And also, this is being disconnected. Yes
or no? So I'm drawing now the three-body diagram
of this crane, or frame in this case, which
is like that. Is that correct or not? Yes?
So what should we have at point C? Now, remember
what I said last time? This is at?
[Inaudible].
It's not a [inaudible] support is like that.
Fixed support is like this. Is that correct?
Remember that. So I should have what?
CX [Inaudible].
CX. Very good. CY. Notice, if you get that
idea and moment C. Yes. Right. And what should
I put in the cable? We know that from lecture
one, or from physics. You have to put here
tension in the cable. How many unknown is
that?
Four.
Three.
Seven.
Why three? You said three. Do you agree with
three [inaudible] or not? Now, I know what
you are afraid -- you are afraid to say four
because four become indeterminate. Yes or
-- but yes, it is indeterminate. If I give
it like that, then how many unknown is there?
Four.
Four. CX, CY, MC, and T. Is that correct?
That's what I call indeterminate. Is that
correct or not? However, but you cannot solve
that problem, remember that. So what all they
have done in this problem for you? They gave
you the tension in the?
Cable.
Cable. So if I give you the tension in the
cable, then we make it into determinant. So
yes, don't be afraid. What you see and what
you've written, if you are sure of that, admit
to that. This is one, two, three, and --
Four.
-- four. Therefore, yes, you have written
that, you're afraid to say. Notice, nobody
answered me because you were afraid to say
four because four is indeterminate; yes or
no? However, that's the case. So four unknown
and therefore indeterminate, so I have to
do something. So this is what we changed that
to a known quantity. That's purpose of this
calisthenics. So therefore, they have said
the tension in the cable is how much? 1300
meters. So they put here 1300 meters. And
[inaudible] is, this is five and this is 12.
250 and 600 is the ratio of five and 12. You
don't see it first, but it is like that. Five
times 50 is 250. 12 times 50 is 600. Yes or
no? Correct?
No.
[ Laughter ]
All right. Five -- this is 250 [inaudible]
600 which is 5:12. OK. Now, how many unknown
do we have?
Three.
Now, why I'm saying that is very important
about C. Maybe that was reason you were sort
of debating. First, you write sigma FX equal
to zero. Yes or no? So that's simple. Let's
write it down. Sigma FX equal to zero. I said
I'd finish this in totality. You get there
-- nothing there. Here you get five, 1300.
Of course, that's 500. Is that correct or
not? Yes, or no? Five over 13 of 1300. And
it's going this way, therefore, it is plus.
Yes or no? Then we come here. All we have
is CX. Is that correct or not? Plus CX, equal
to zero. Therefore CX becomes minus 500 meters.
So we calculate CX.
[Inaudible].
What else? Oh. I'm sorry. So let's then change
that. Yes, of course we have that. Minus 450.
So that changes. That's [inaudible] minus
500, so let's see what that is. That is minus
900. Oh, no. No, that's right. That's plus
and that's minus. So this goes to the other
side because minus 50. We're subtracting.
Correct. Minus 50. Sigma [inaudible] Y equal
to zero.
Why didn't we do the moment first? Why didn't
we do [inaudible]?
It doesn't matter. I can get with those [inaudible].
[ Laughter ]
Right. Yes. You usually do that. Yes. But
I leave that for the last because remember,
there is only one horizontal unknown and one
vertical unknown. So we do that when there
are several horizontal and several unknown.
That's a good question. Remember in the previous
problem, I took the moment about point D first.
Why? Because I had several horizontal forces
and several vertical forces. Here, I only
have one horizontal and one. So sigma FX gives
me the first one. Sigma FY gives me the second
one. So it's not necessary. Yes. So if that's
what your question was, that's really what
I'm doing here. Is that correct? See, Y should
be balanced all the vertical ones. Is that
correct or not? Which is easy to see. Nevertheless,
you have minus 750 going there, nothing there.
Then you have minus 12 over 13 of 1300 there.
And then we have plus CY and equal to zero.
So CY is going -- hopefully is going upward.
I mean, it's not downward. The truck cannot
go up there. It has to be pushing down. Is
that correct or not? When it's pushing down,
the reaction is upward. And that's exactly
what they are showing here, 1,950 meters.
Correct? Yes?
Yes.
And finally, moment about point. Which one
should I take the moment about?
[Inaudible].
Summation of the moment at C equal to zero.
Now, I want everybody to translate this for
themselves. This doesn't mean MC equal to
zero. Although I have said in this class,
and in future classes, people will do that.
That is nothing to do with that MC. That MC
could be hundred millimeter or 2,000 millimeter.
Either plus or minus. We don't know yet. Is
that correct? What is this? What did I write
here? Read it for me. What is that?
Sum of all the [inaudible].
Exactly. As if this point is here or point
is there. Everybody understand that? Use this
application, then you will be fine. Some of
all the moment, everything that I see here,
red or black. Everybody understand that? About
point C must be equal to zero. That's why
I put here [inaudible]. In the book, you don't
see that. Some people even don't put that,
you see. Some of you are a little worse than
that. You don't put summation there. MC equal
to zero. This is MC equal to zero. That's
not the case. Everybody see what I'm talking
about? How do you do that? That is wrong.
It should be summation. And the book write
that. I add even. At or about point C must
be equal to zero to get you into that position
that you want not to forget about that MC.
Is that correct or not? Then if you do that,
then start from here. So what's the moment
of that about point C? 750 times 500. So you
can write it in meter if you want. 700 in
this case, it's better use to millimeter rather
than [inaudible] meter because the number
is less. Is --
Positive.
-- positive. Because that's the one [inaudible]
this. This one could be [inaudible] and that's
the result of [inaudible] or some horizontal
forces, which is 450 times what? 400. [Inaudible].
Is that correct or not? Plus 400 times 24
meters. And that going that way. This one,
force is balancing force, because it should
reduce this. These forces try to [inaudible]
tower. This force try to help us. So the moment
is [inaudible]. Is that correct or not? So
we already calculate that this was 500 and
this was 1200. Yes or no? 1200 times what?
1200, 50; yes or no? This is 1200 times 150.
So minus 1200 times .15. And then this one
is five over 13 which is 500. 500 times the
entire height, which is 600 millimeters. And
that is also negative. Now, what should I
put that? Should we put that equal to MC?
No.
No. Notice, that's the mistake that some people
made. First of all, some of them, they keep
this as MC, and they say MC equal to that.
That's not what we said. Summation of all
the moment about point C equal to zero. I
took care of that one, I took care of that
one, I took care of that one. Now we are coming
here. CX doesn't have any moment about C,
correct? CY doesn't have -- but we have a
couple sitting here [inaudible] sitting here.
If it was here, would you add this to the
system?
Yes.
Yes. So add it the same system. And the way
we wrote it, it's positive. Is that correct
or not? Plus MC equal to zero. So this is
what we are saying, that all of the actions
plus the reaction must be equal to?
Zero.
That's what you read for me. You remember
what you said? You said the summation of all
of the moments about point C [inaudible].
Please do not do this any other way. Many
people in future classes, they put this equal
to MC. And they come [inaudible] say, "Oh,
I got all of the numbers, but all number is
in reverse." I say, "OK. Because you did not
apply the proper understanding of the formula."
Is that understood? OK. So that was the only
point, that MC ends up to be very little.
MC ends up to be minus 75 millimeters because
especially if this was a mobile. Notice, they
did this as a mobile frame. Because notice
the horizontal force is very little, so the
truck is not moving back and forth. The vertical
is OK because the truck can take it, because
the truck was pushed. But look, the moment
is very little too, so there is balance there.
Is that correct or not? However, if suddenly
this load becomes much larger or go further
by mistake, the operator extend this year,
then you may couple the crane. Is that correct
or not? Yes? Which takes us for the next two
minutes to go through the two-force member
and three-force member. Question number 67
and 69 that many of you asked me about. Is
that correct? Yes. OK. So therefore, let's
talk about -- now, this is all of the 2-D,
applies to 3-D as well, so the 3-D should
be very simple. But I'll give you another
half an hour to finish that. Which we start
on Thursday, hopefully, and finish it. That's
why I extend your quiz. Is that correct or
not? I gave you a little bit more detail here
because these are very, very important in
the future of your work. OK. Now, what is
a two-force member and three-force member?
So let's say I have here an object here. Please
write it down. We are talking first about
two-force member. Very special scenario we
have there to help you to understand that.
This is a two-force member. Let's say that
I have here a force FA applied at point A,
and I have here FB applied at point B. Obviously,
this is a two-force member. Yes or no? Now,
let's say that I add another here, F2, FA2
here. And F2B here. These are all vector,
of course. This is a two-force member or four-force
member, or three-force member. What do you
call this one?
Two-force [inaudible].
Two-force member. Very good. So you are ahead
of me. Still is considered two-force member.
Why?
[Inaudible].
You can find the result. Is that correct or
not? So really, the two-force member not means
only two-force FX and FY. Everybody understand
that? So let's make it general. If it put
another F3 here and another F4 here, or I
can add that; this still would be considered
two-force member. And you correctly answered
that. The reason is all of the forces at point
A can be -- because this is concurrent at
this point. So you can find the result [inaudible]
of that one, yes? And result [inaudible] of
this one, put it here and here. So two-force
member doesn't mean only two force. It means
two points of application. So please write
it down. So all of the forces are applied
at two points of application. Of course, if
there is another point C here and there is
another force there, that would not be a two-force
member. Is that correct? Understood [inaudible]?
So therefore, this object ends up to be like
that I have a point A here and I have a point
B here, like I showed that. Then I put here
RA, and I put there RB, in arbitrary format.
Remember, this system is supposed to be in
equilibrium. Otherwise, we are not in this
chapter. From now on, actually, every object
would be in -- everything that's not moving
is in equilibrium. Every object in the world,
as long as it's not in motion, would be in
equilibrium. So that's the case; two-force
member. Equilibrium is there, don't forget
about that. Since I only have RA and RB, obviously
it is two-force, must be equal at opposite.
In other words, sigma FX equal to zero, and
sigma FY equal to zero means X component of
that must be equal to X component. If this
is going forward, this must be going back.
If this is plus five, this must be minus.
There is no other force there. Everybody understand
that. The same thing, you said sigma FY. Correct.
If this is going up, this must be going down.
So the way I have done it is incorrect. Is
that correct? In order this system to be in
equilibrium, according to equilibrium condition
[inaudible] equation, these two forces must
be equal at opposite. This is RB. Now I can
see if I draw that exactly parallel to that,
then it is condition one is satisfied, condition
two is satisfied. But is it in equilibrium?
[Inaudible].
What is the third condition? Remember that?
We have three conditions. One, two, and three.
[Inaudible] here. Where is the third -- is
this moment of this equal to zero or is it
just a couple? Two force equal at?
Opposite.
Is that a couple?
Yes.
Is this going to be something like that; yes
or no?
Yes.
So this object is going to rotate like that.
Therefore, if this is the case, it cannot
be in equilibrium. So what I can do is this:
the moment of RB with respect to B equal to
zero. Yes or no? The moment of RA with respect
to B, sum must equal to zero. Yes or no? It
cannot be. I have this distance. I have to
change it until what? Until RB become like
this. Is that correct? Now, if I put RA in
the direction of AB, is the moment of RA about
B equal to zero? Yes or no?
Yes.
Is the moment of RB about B equal to zero;
yes or no? However, these two are supposed
to be equal and opposite. Therefore, the only
possibility is when I connect A and B together
and two forces equal at opposite. So four
unknown become only?
One.
One. Because these two are equal and opposite.
Look, that's what you said. Now you are all
so proud [inaudible]. Yes? Let's go number,
guys. Let's say there's a bracket here. Let's
make it, this is five inch and this is five
inch. If I put here 20 pounds, I should put
here 20 pounds. So this is the point of application,
this is the point of application. Yes or no?
Correct? I'm saying the same thing in number,
but you understand because suddenly, you went
all quiet. I noticed that. OK. What I said
there, suddenly quiet. [Inaudible]. When that
quiet comes, that means that OK, hold up.
If I put here an R, and R here at point A.
This is point A, this is B. Yes or no? You
said if this goes horizontal 20, this should
go horizontal 20. Yes or no? Equal and opposite.
If I put here, let's say 30 going up, this
would be 30 going down. Anyhow, this becomes
two-force equal and opposite. Which conditions
one and two only satisfied. Condition three
is not satisfied. Yes? So if I make a resulter
of this force which becomes something like
that, and the resulter of that force becomes
something like that. Now, they are at distance
from each other. Notice, the moment of this
force about this point B equal to zero. But
moment of A about this point B is not equal
to zero because it is at the distance D. Is
that correct or not? Which creates a couple.
The only way I can change that, this is not
possible. The only possibility is that I connect
these two and put FA there. Is that correct
or not? Now the moment of FA about B is zero.
This one also should be zero. So some of the
moment is zero. However, since these two are
equal, so this is the result. You connect
AB together and you put two forces equal and
opposite. If you put it this way, the item
is opening up, it will be in tension, or you
will put it this way, it would be pushing
down, it would be in compression. So [inaudible],
you had here AX, AY, here you had DX and DY.
You said that this equation, AX must be equal
to DX. Is that correct? Equal and opposite.
With this equation, DY must be equal to AY
equal at opposite. With the third equation,
summation of the moment with respect to any
point equal to zero, that's the only possibility,
all three conditions now is satisfied. So
that's the only way it's possible. Is that
correct or not? So from now on, when you see
a two-force member, you replace it with -- now,
look at the handout problem quickly, it takes
another few minutes. So this idea would be
applied to one of the handout problem number
two or three. Let's see which one it is. So
let's go back to the handout problem. Because
I don't want to do your homework for you.
Many of you asked about your homework. But
this is what you have there, which is a handout,
all sort of [inaudible] system that you have.
I just briefly draw it here, so you see it
like this. Look. There is a force here of
400 meters. Let's not go through the number.
Let's do that. This is a force here. There
is a pin here. This is what you see in your
handout. And then there is a pin connection
here, pin connection here, pin connection
at D. And these are given .2 meter or 200
millimeter, .2 meter. And these distances
are given like that. .2 meters, .1 meter.
There is a pin here. There is a pin there.
Yes or no? Correct?
Yes.
If I do not recognize a two-force member,
that's what you have, some of you had problem
with number 67. Is that correct or not? Yes?
Yes.
If I do not recognize a two-force member here
from now on, very important factor, I have
to erase this and erase that. Look what happened,
guys. So I'm probably going to repeat that
later next time. I'm going to put here AX
and AY. And here I have to put DX and DY.
How many unknown do I have?
Four.
Four. Couple of you showed me your homework,
67, and you end up with four; you should.
Because you didn't know about the two-force
member. Now, here is the scenario. Do you
see a two-force member?
Yes.
Which one? Is it ABD, or is it AB -- what's
the other one? ABC or is it BD?
BD is exactly what I draw that. So if this
is BD, it's like that, so I can't separate
this. This is all it is. Go like this. Is
that correct or not? You can redraw that.
But of course, here, I said five inch. And
five inch, here you have .2 and point -- not
necessarily always because this could be eight
inch, this would be four inch. But you get
the angle. Is that correct or not? What you
put there -- two ways you can do that. You
can keep this connected. But since this is
a two-force member, you only put one force
here going up or going up or going down. Everybody
understand? In the direction of D. Look at
this. If this part is connected to the body,
I have not disconnect that. If I did not disconnect
that, I disconnect that one. But this is a
two-force member, the force is going like
this. Is that correct or not? So how many
unknown do I have? Three. But maybe you understand
this one better. Another option, you draw
this and you put here, your force is there,
400. Is that correct or not? Yes? Here you
put AX and AY. Now, this is two separate members.
Is that correct or not? And this one should
be like this. Is that correct? So what should
I put here? The force equal and -- oh, not
that way. The force equal and opposite. So
it should be like that. Is that correct or
not? Yes. And since this is 45 degree, that
angle is [inaudible]. Then it becomes solvable.
Which is it? This is option one which you
understand probably better. Or this is option
two. Both of them are acceptable. One is being
in two member, one is in one member. Is that
correct or not? Yes? One is being cut here
and there. One is being cut here, here, and
there. Everybody understands it? Yes? So both
of them solve the problem. So you should do
the same thing about problem number -- actually,
there are more one-force member, two-force
member coming, so we'll discuss that next
time. All right.
[ Music ]
