[ Music ]
If you want one-to-one, you have to make an
appointment.
You have to call [inaudible] send him an email,
or give him a call.
These are the times that he has [inaudible]
Wednesday for two hours, 11:00 to 1:00, which
is before your quiz, or 4:00 to 6:00 [inaudible]
what I gave you last time.
Everybody remember that.
Don't take a picture.
I want to add to it [inaudible].
Or if you want one-to-one privately, question-and-answer,
again, this is the time.
We'll be every week.
Now, aside from that, this is another workshop
for two hours, so please write it down.
If this is tomorrow, 1/29, and will be 12:00
to 2:00, building 15, room 82, 182, which
is library.
And 2/1, which is Monday, 5:00 to 7:00, room
15.
I mean, building 15, room 2919.
And again, 2257, the same room.
And this could be repeated.
Any time they have time, they'll let me know,
and I'll put it on the board, or next time
put it on the Blackboard, or wherever.
There are several of you who may need still
a little bit extra help.
That will do it for you, okay?
And basically, that will be -- and also, I
gave you [inaudible] which is always available.
I don't know whether any of you have gone
to him or not [inaudible] okay.
And on top of that, of course, we have to
give you your homework, so these are your
homework assignments for next week, and those
problem number, question number questions,
number one and four on your handout, which,
I believe, is pages four and five.
One is on page four.
The other one is on page five.
You can go over that, because I'm going to
do one of them today in class, and I want
you to use the similar procedure for finding
the [inaudible] about x, y, and z.
And then also, the rest of it, which is new
subject 71, 74, 77, and 80, which are couples.
And then 82, 96, 102, 114, 124, and 128, which
are equivalent system, etcetera.
We're going to get there.
We will discuss the [inaudible] after today's
lecture, probably you should be able to do
the first two rows of homework.
The last two will remain until I explain it
to you.
There are, again, 12 problems there, okay,
and the first two rows, you should be able
to do it this weekend.
Any questions before I start?
Yeah.
[ Inaudible Question ]
The same chapter.
Chapter four is about equilibrium of rigid
body [inaudible].
By the end of this chapter, basically, application
of all the principles of [inaudible].
Then after that, we are going to use to solve
equilibrium of the rigid body, and then some
[inaudible] etcetera.
Okay, now, there is an easy way of doing some
of the questions that you answer in this homework,
and many of you came to the office hours today,
and we were discussing this.
Some of you are still not showing up.
I hope you go there.
Wherever you go, all of you, many of you need
to discuss this question with somebody, and
get the right answer from them.
Everybody understand that.
If you are not doing it, you are going to
fall behind or something is not right, because
these are not easy questions to be answered.
So the best student still have some questions
here and there.
But there are easy ways of doing it, and there
are difficult ways of doing it.
Just to show you what I said last time, which
is the summary, that the two methods that
I show you, this is repeat.
There's nothing new here.
Let's say that we have here a force.
This is just simply showing what we have done
in the past, which has a vector representation
of 2i minus 3j plus k [inaudible] for example.
And let's say this is applied to a Point A.
It doesn't matter where Point A is, so we
can put it anywhere you want to.
This is the A. This is force, f.
And Point A has the coordinate of 4, 3, 5
meter [inaudible].
And let's say that one way of doing that,
I said if you want to calculate moment about
-- oh, everybody knows how to do that.
Is that correct or not?
All these [inaudible] cross product of f.
is that correct or not?
Let's do that, just to repeat.
All of you have done it.
There's nothing new there, but I'm doing it
for you to see that mo becomes equal to roa,
cross product of f, if you are using cross
product, which, for this problem, simply becomes
-- because this is zero, zero, zero.
So it becomes, of course, i, j, k, and then
becomes 4 minus 0, 3.
I just made it very simple because I want
you to get the idea.
The numbers is not important.
And then the component of the four, which
is two and, as you see, minus 3 and 1.
Of course everybody knows how to expand this.
When you expand this, this becomes equal to
18i plus 6j minus 18k.
And the answer would be [inaudible] which
has a magnitude.
This is [inaudible].
Here is what we talk about [inaudible].
This is -- this right now, what I have done
here, this is vector Mo become f.
Vector applied at Point O, yes or no?
That's Mo.
This is Mo.
Now, it has three projections.
Mx, My and [inaudible] and minus 80.
Now you can use this method if you want, as
I said before, to calculate all your [inaudible].
Now, if somebody says but this axis still
is parallel to x, let's say this is the axis
[inaudible].
As long as that is parallel to x, you say,
"Okay, who cares.
I can put o, or I can put here o prime, and
then put here x prime, y prime, and z prime
there, and take the [inaudible]."
Then you have, again, [inaudible] x prime
axis, y prime axis, and [inaudible].
You should have done it this way.
You should have relocated the o for the tenth
problem, which was problem, I don't know,
149.
Because the distance between the two [inaudible]
were not given.
I don't know how you did that one, but that's
[inaudible].
Now, the way that it is in the solution manual
is this.
You have to pick up, if I want to take the
[inaudible] I have to pick up a point on the
x axis, yes or no?
That could be this point, this point, that
point, or it could be o.
And that's what they did.
In the solution manual, you see this.
Please write it down.
Mx equal to -- although I explained that before.
But now that you have your homework, maybe
you are understanding it better.
Is that correct or not, for everybody.
So they write it like that.
The lambda of this line, of course, is i.
Is that correct or not?
So they use this formula [inaudible] product
of m, which is a moment about a line on the
line, which should be o or any other point.
[inaudible] o is the best point, because it
has the coordinates of 0, 0 and 0.
So therefore, Mo.
And then, of course, this is the magnitude
and this is the direction.
Yes?
But if I write this one, again, in the [inaudible]
format, this is i.
This is triple product.
I don't have to put ij here, but this dot
product [inaudible], yes or no?
You have one, zero, and zero.
Then Mo is sitting there.
I have to put there 4, 3, 5.
And then I have to put 2 minus 3, 1.
So you have this option versus this option.
What do you think the answer is here?
Look, this is zero at zero.
Instead of i, you have one.
[inaudible].
So the answer is, of course, you have to put
i there, because this is a magnitude [inaudible]
direction.
So here, they come up with 18i again.
Similarly, you can change i to j.
And the k, you get the moment about x axis,
y axis, and z axis.
The choice is yours, but if you are using
the solution manual, at least understand that
this is the method that they are using, which
I did not recommend.
I said use this method.
Everybody understands it, yes?
However, if the line is there, and the force
is there, this has nothing to do with your
x, y, z [inaudible] problem.
I don't know what problem in your homework
you had.
One of the problems was like that [inaudible]
to find the moment about i, then you do that.
Is that correct or not?
Now, problem -- there is three problems there,
which I have a lot to talk about those three
problems, which I wanted you to do first.
You should have done it first.
Now, let's see whether there is an easier
way than even this one or that one, to get
to this solution to solve out that problem.
One of them is there [inaudible] that you
want to turn it around.
The other one was the part of a boat that
was hanging there.
Remember, everybody?
Yeah.
Now, to do those problems, you have to understand
this, that first of all, when we have -- please
write it down.
This is a technique that everybody uses, but
I'm summarizing for you, to see that in many
of these problems, we don't need to use any
cross product or dot product.
It's not required.
Many 3D problems turn into a --
2D.
-- 2D problem by using force distance.
That's all you need.
However, you have to understand what the scenario
is as far as the x, y, z axes.
If we have a set of coordinates like that,
x, y, z, let's first understand that what
we are talking about is this.
Let's assume I have here a plate sitting here,
in the plane of x, z.
Is that correct or not, yes?
Displacement, there is a load applied on it.
It's going to be rotated in three ways.
One, what I'm talking about.
Let's say there are hinges here, and this
plate goes like this.
This is the rotation about the x axis.
This is the --
y.
-- rotation about the y axis, and this is
the rotation about the --
z.
-- z axis.
That's what we are talking about, Mx, My,
and Mz.
It could go positive, positive, positive,
which I showed you last time.
Is that correct or -- or go negative, negative,
negative.
Is that understood?
Then let's say -- look at this now.
I have here this plane, and this is, let's
say, 100 pounds.
This is equal to ten inches, and this is equal
to six inches.
And I want to find out the moment about x,
y, z axis.
This is a much simpler case than we were talking
in office hours.
Can you find the moment of this 100 pounds
about x, y, z axis without using the cross
product?
Everybody see what I'm talking about?
[inaudible] put your hand here, everybody,
around this [inaudible].
Just go around the x axis.
So create a moment about x axis for rotation
about axis.
And it is going to be clockwise, as soon as
you are standing there.
Therefore, it is negative.
So Mx [inaudible] or by force distance method
must be equal to 100 times, what, 100 times
6?
Okay.
And it should be negative, but that is -- remember,
that is [inaudible] of the Mo.
You are only finding the magnitude of that,
the sign of it is negative.
Is that correct or not?
But that's it.
[inaudible].
Then does this force cause this plate to rotate
about the y axis?
No.
No, because they are parallel.
See, that's the rule.
If the two forces and the axes, if the forces
and axes are parallel, there are no reason
their axes try to rotate.
Is that correct or not?
But does it rotate, this plate rotate about
z axis?
Yes.
Yes, so it will be what?
100 times.
Put your hand again this time.
Go around the axis.
That's what you do.
Use your right hand rule.
If this is the first time you are doing it,
use the right hand rule, and then use this
way.
Is that correct or not?
So that had Mz equal to, what, 100 times 10,
and it is -- look at your [inaudible].
It is positive plus 100 times 10.
Obviously, this becomes minus 600.
Let's put a unit there.
Pound-inch.
And then this one becomes equal to 1,000 pound-inch.
Without using any cross product, I end up
with Mo to be equal to minus 600i plus 0j
plus 100k pound-inch at moment.
Notice the same goal achieved.
Is that correct or not, yes?
Of course, not always the problem is that
simple.
The problem is a little bit more sophisticated.
Open your note, handout.
Open your handout.
There are five problems.
Number one, two, three, four, five.
Number one and four is your homework.
Go to page four and five, of course.
You see?
These are actually [inaudible] and if you
see extra information, don't worry about that.
This problem, the same format.
Look at this dot here [inaudible] to calculate
the [inaudible] and how this behaves structurally.
Everybody understand that?
[inaudible].
So you see x, y information.
There is a point there.
There is a point there.
Don't worry about it.
All you have to do in this problem, find the
moment of these forces about x, y, z [inaudible].
Everybody understand that?
Without using any --
Cross product.
-- cross product.
[inaudible].
Now, in order to show you how this works,
I am going to do problem number three for
you.
So go to the next page, page five, and I'm
going to do that, or we are going to do it
together.
I ask the question from you, and you should
be able to answer that.
But we need to find [inaudible].
Where are they?
[inaudible].
Oh, there it is.
I put it there.
Okay.
So therefore, I need to draw this on the board
here for you all to see.
Of course, you have it in your hand.
Okay, the problem is as follows.
So problem number -- which one is -- oh, here
it is.
Can I see one of your handouts please?
I have extra.
I didn't bring it with me.
[inaudible].
Here it is.
We have here a rod sitting in the direction
of the x axis, and then there is connected
to another rod in the direction of the z axis,
so this is going to be [inaudible].
And this is y as usual.
And this road has nine inches in this direction,
nine inches here, for the total of 18 inches.
And this leg is 18 inches.
And there is a force here, 40 pounds in the
direction of x.
There is a force here, 30 pounds in the direction
of negative y.
There is a force here in the z direction,
75 pounds.
And there is a force here in the y direction,
equal to 80 pounds.
And let's call this one point o.
The way they call it here, abc.
Correct?
Thank you.
The idea, is you want to find the moment of
all these forces, about point o, one method
that you all should be now familiar after
doing all your homework is the cross product
method, because this is a 3D problem, yes
or no?
So cross product method is lengthy.
By the way, you don't get much out of that,
because what you see there -- this was -- how
much was that?
That was 75, 75 pounds.
So it should be parallel.
And then all you have to do, which I have
it here, Mo equal to.
So this is what you write.
Mo, moment about, or if you are using cross
product, which I am not asking you to do.
But if you cannot do it in y [inaudible],
do it this way, check your answer.
Everybody -- because most of you are familiar
with this, is that correct?
Mo is equal to r.
You are taking roc, cross product of fc, which
has one component there, plus rob, cross product
of force at b, which has two components.
One along the x, one along the y.
You have done that in your homework, so [inaudible]
you are familiar.
Plus roa, cross product of fa.
Then we say determinant, determinant, determinant.
Then you get ijk, ijk, ijk.
It takes a few minutes.
Is that correct?
Then you add all the i's together, all the
j's, k's.
So you have the moment about all three axes,
yes or no?
That's one way.
Now, this method is good.
But basically, if this becomes a design issue,
and we want to find out what's happening to
this rod [inaudible] for future reference.
You don't know it yet.
If this rod rotates this way, this has become
a torque.
Somebody asked me what is the torque, the
difference the torque and the moment.
One component of the moment, become a torque,
you don't have to do it now, but that item
is going to rotate like this.
Is that correct or not?
So that's the torsion.
What do you put another moment like that?
That becomes a bending.
Everybody understand.
But that comes for future classes, not yet.
For time being, I want to that.
I want to use r cross product, which come
up with some number without realizing it.
I want to see how this assembly behaving.
Is it turning around the x axis?
If you are turning it this way or that way.
Does it turn about the y axis?
If it turns around the y axis, all right,
this is attached to the wall.
There are some bolting system.
I can see which way the bolt is being pulled
out, which way the bolt is going in by finding
the direction of the moment.
You're getting what I'm saying?
This fill-in is important for an engineer,
not r cross product.
r cross product is a math.
You put it in your calculator [inaudible].
You get the answer.
That's what I said, don't use it.
Whenever you do that, you don't learn anything.
Is that correct or not?
Whenever you copy the solution manual, or
you put it in front of you, you change the
number, "Okay, they did it this way.
I do it that way," you don't learn anything.
The time that you learn this material, when
you, independently, you read the question,
the connect the concept to the idea, the ideas
to the problem, and then you solve the problem.
Now, you do 50% of it, 60% of it, that's much
better than the other method.
Everybody understand that?
However, when you get stuck somewhere [inaudible]
or whatever, you solve that.
You ask somebody else to help you out.
Is that correct or not?
Now, let's do this.
Now, you can organize or put this in a table,
if you want, or the other way around.
So let's put it in the table, see whether
that makes it simpler.
[inaudible].
You put forces here, then you write Mx's here
for all the forces.
My there, if you want.
But notice, this is only for 3D problems.
You don't do this for 2D problems.
2D problems is one-dimensional.
Is that correct or not, yes?
All right.
Now, let's look at each force -- look at this
force at a.
So put that force here, 75 pounds.
Look at a.
Now, look at this one.
If you have a force here applied to this member,
and pushing it this way, is this going to
rotate about the x axis?
No.
This force causes the whole assembly to rotate
about the x axis or not?
No.
No.
Why?
This is the reason.
If the axis is here, and the force goes through
it.
See, it crosses there.
There is no rotation.
The rotations has to be at distance.
You don't have a distance.
It goes right through that.
It doesn't rotate.
So write it down in your notes.
Where the force and the axel intercept, there
is no moment about that axel.
Of course, earlier, I mentioned it.
You wrote it down.
If the axel and the force are parallel, you
all answer, it does not cause any rotation.
[inaudible] it is the rotation is there when
you get into this scenario look, this is the
axel, and I put here [inaudible].
Everybody understand?
If I put the [inaudible] this is going, this
is what you answer me there.
Is that correct?
Everybody saw this, but this is what they
gave.
This is the force.
There is the distance.
There is a rotation this way.
If I push it down, the rotation changes.
Is that correct or not?
That is exactly what you need, nothing else.
None of this is [inaudible].
Does this instrument rotate because of this?
I'm not talking about the other [inaudible]
force.
Only this force.
Does it rotate this instrument about the x
axis?
No.
Does it rotate about the z axis?
Notice they are parallel.
But does it rotate about the y axis?
Yes.
Yes.
How much is that?
75 times --
18.
-- 18.
That's one thing you want to know.
Oh, which one was it? y axis.
Zero, 75 times 18.
The only question is whether it is plus or
it is minus.
Look at it.
It is going this way, so it is going like
that.
Is it plus or minus?
Plus.
If it's going like that, if it's going that
way, it is plus.
It's going like this, or it's going like that,
which way?
That way.
So therefore, it is minus.
What you just did here is this.
Which one is easier?
Or which one is better understandable?
There, you are engineer.
You know exactly what you are doing.
Is that correct or not?
This is just math.
If you do this, sit down at home.
That's the answer.
As simple as that.
You are laughing.
That's right, you [inaudible].
All your problems, 53, 149, I assure you that
problem 53 or the other one can be done with
this method, can be done in 10 to 15 seconds.
Everybody understand that?
But learn it first.
Okay. [inaudible].
[laughter] I don't want to do that.
Okay, let's talk about this 30.
All right, 30 pound force going down at Point
B. What's the action [inaudible]?
Does this one rotate this about the z axis?
Actually, this is a 2D problem.
Look at it.
Is that correct or not, yes?
30 times, what, 18?
That's about the z axis.
Yes or no?
Yes.
Yes.
30 times 18.
Is it moving this -- which way?
This way or that way?
This way or that way?
Negative.
Negative, that's right.
Does this one cause this instrument to rotate
about the y axis?
No.
[inaudible].
Does it cause it to rotate about the x axis?
No.
Okay, that's part of this, not the whole thing,
because there are two forces there.
Is that correct or not?
And finally [inaudible].
So 40, which is simpler than that.
This 40 pound force -- this is 30 pounds,
I'm sorry.
40 pound force, which is at Point B. Okay.
This one is easier than all of them.
Does this got any rotation about the x axis?
No.
y axis?
No.
z axis?
Yeah.
Oh, wait.
No.
No, nothing.
Actually, this is zero, zero, zero.
Look how simple this is.
These two together is the second cross product.
The first row is this one.
Row two and three is the result of that one.
You can check it at home.
Everybody -- and finally, we get to the [inaudible]
at C. At Point C, we have 80 pounds going
like that.
Does that one create any moment about x axis?
Yes.
You see, it is turning it like that.
Is that correct or not?
It's turning it like that.
What is it?
It is plus.
80 times what --
9.
9, very good.
80 times 9.
Any rotation about the y axis?
No.
No.
You know why you are answering all that so
nicely?
Because you already got the fill-in about
the moment about x axis, y axis, and z axis,
many, many times.
Is that correct or not?
And finally, does it create any moment about
z axis?
Be careful.
Yes.
You see that this turns to the z axis.
How much does this turn to the z axis is --
18.
18, and it is going --
Positive.
Plus, okay.
So 80 times 18, and that's plus.
And then this is the answer.
And this last row is this cross product.
There's nothing wrong with that, but be sure
that you learn this, because this is what
you are going to do -- yes.
What you are going to do in future for all
your ME 218, 219, CE 3, 4, 5, 6, whatever
you're going to do.
Any design, that's what you do.
This way, not that way.
You never see that again in your design courses,
until you go to graduate school.
Is that correct or not?
Of course, you see that in lots of other dynamic
or math courses.
[inaudible] if I add this all together, this
is moment about, what?
x axis.
x axis.
These are the moments about the --
y.
That's the moment.
That's [inaudible].
So the answer for this problem is you can
check it at home, so you will see that.
That's the same way you do the problem number
one and four, with the same method.
This becomes -- you just put the summation
here.
You put 720, and this becomes minus 1,350,
and this becomes 900, because this is plus
and minus.
Of course, the unit, you can put it on top,
is pound-inch, or [inaudible] meter or whatever,
which applies to all the numbers.
And you don't have to do it this way.
As long as you [inaudible] calculate moment
about x, y, z, you don't need a table either.
Is that correct or not?
Which takes us to the next question, and that's
question number 53.
Is that correct or not, yes?
Okay, let's use this method for problem number
53, not the one in the solution manual, okay.
[laughter] Or not the one -- oh, yeah, that's
the one, probably, most likely you look at
it, didn't you?
How many of you did look?
I'm not looking at you, so you don't have
to tell me.
Well, that's [inaudible].
Somebody left his pencil in my office.
Okay, this is the question that was raised
[inaudible] one of your homework is this.
There is an x.
I'm just explaining to you.
I don't have to use any number.
There is a wall [inaudible] this wall was
in x and y axis, yes or no, and it was tilted.
Okay, this is exactly what I was telling you.
Yeah, it was tilted backward, and you want
to bring it forward.
In order to do that, somebody must go back
there, and push it, not one person.
If this is a big wall, you cannot do it by
yourself.
Is that correct or not?
Everybody understand?
You have to bring it forward.
In order to bring it forward, this is the
rotation about the --
x axis.
x axis, yes or no.
they have given you that number.
They said that we need the -- what number
did they give you?
They said Mx, it is required.
In order to tilt this wall forward, we need
a moment of 4,728 pound, foot, okay.
So you have that much moment going this way
about the x axis, yes or no.
Now, however, the way they have given to you,
there was a line here, a line there.
They call this one ac, for example, and this
one was d and f.
And then, of course, when you put this cable
on the wall, attach it to the wall, and then
you put the machine or arrange that you pull
it, you'll tighten the belt.
The belt creates -- I mean, the cable -- there
is a tension, and the tension causes the wall
to come forward.
Is that correct or not?
And the way you solve your problem, which
is absolutely correct, was like that.
You end up with two forces there, yes or no?
And one force was here, and one force was
there [inaudible].
Is that correct or not, yes?
Correct?
Now, these forces have [inaudible], so one
way of doing it is this.
Take your moment about point, o.
So use r cross product, r cross product, etcetera,
etcetera.
Or use the moment about x axis.
Either way, all of them, either use this method
or the other method.
That gets to the answer, yes or no?
however, when you use this inspection method,
this is simpler than [inaudible].
See what happens here, you have to write ab
[inaudible] tab.
Let's write it down.
This is what you should have end up.
tab end up to be, because it was given, end
up to 15i minus 180j plus 180k pounds.
Is that correct or not?
Because the magnitude of that, it was not
given.
tad was -- tdf was -- this was ac, I'm sorry.
b was here.
That's the same thing.
tde was not given, but that was equal to its
magnitude times -- this part has nothing to
do with the moment.
That's part of a static.
So this is one and half minus 14, 12 divided
by 18 and half.
I believe [inaudible] foot or inches.
I think it's foot.
Is that correct or -- yes, foot.
It doesn't matter, because as I said [inaudible]
foot over foot.
So that's the two forces, right?
Now look, you need the moment about the x
axis, yes or no?
So let's look at this differently.
This force has a component like this.
Does it have any moment, they y [inaudible]?
Does it have any moment about the x axis?
No.
Does this one have any moment about x axis?
No.
Only one that has moment about the x axis
is --
z.
z component.
So that's all I need is this [inaudible] z
component of that force.
So the moment Mx, which should be equal to
4,728 equal to 180 times this height.
That height was given equal to, how much?
Twelve.
Twelve.
Then the moment of this one, but this is the
same format.
x doesn't have any moment.
y doesn't have any moment.
Only z has the moment.
The z portion of that one is this.
12 over 18 and half.
Is that correct or not, yes?
Times tde [inaudible].
Multiplied by -- this height was given differently.
That one half was equal to, how much?
Fourteen feet, the sum of these two [inaudible].
That's the problem.
That's the solution to that problem.
Can you find td, yes or no?
With one line.
Yes?
Yeah.
Good.
Is that how you used it?
No.
No, okay, good.
[laughter] Okay, now use it from now on.
Now let's go to problem number 152 , the boat
that you are struggling so much about it.
[inaudible].
Are you all surprised?
Yeah.
Why?
Did you get it?
Yeah.
Good, that's what I'm saying then.
It is much simpler than what you think.
When you use this one, there's nothing there.
No feeling about it.
This, you know exactly how the force is acting,
which direction it is.
Now let's go to the boat problem.
So in boat problem, this is the same scenario
again.
I'm going to erase that, and put that one
here.
That one, some of you have a problem with
it, but some of you didn't ask the question.
First of all, let's put it this way.
This was -- I think there is the pulley here.
Is that correct or not, yes?
There was a man standing here.
There was a rope here, attached from this
point, going to this pulley, going down to
the boat, and going up.
And here is the boat, the weight of the boat
hanging there.
Yes or no?
The magnitude of this tension, let's say -- it
doesn't matter what is given, what is not.
They turn around, they can give you three
problems with the same idea in it.
What is important here is how you calculate
moments about x axis, y axis, and z axis,
because that was the idea they're presenting
to you.
Let's go, for example, about the z axis.
Let's say that we want to find out the moment
Mz.
This is the question.
I don't know.
There are two, three problems.
They all are the same thing.
Now, first thing first that you have to understand
this, the tension in the cable is teased.
Is that correct or not?
You are drawing this free body diagram, yes
or no?
So you should erase everything, draw your
free body diagram.
This was a question that many people came
to my office about it, because they did not
understand the nature of the problem.
This is the same rope going around the pulley,
going back to the pulley.
We discussed that a long time ago.
The tension remains the same.
Is that correct or not?
[inaudible] the tension was given equal to
82 pounds.
Is that correct or not, yes?
However, there are three of them now.
Look, this is the tension, this is the tension,
this is the tension, because you are cutting
it there, yes or no?
These two are vertical, yes?
So this is 82j, or total, for total of 164j
going down.
Is that correct or not, yes?
Then you have to write this one as a vector,
because this is not the magnitude of that
t.
But it was t -- what was it?
I forgot.
What was t?
What was the letter format for that one?
Where is that point called?
Is it a or b?
Let's call it here.
[inaudible].
That's it.
So there for this, you have to write the tension
tab.
Is that correct or not?
But the magnitude, that still is 82, correct?
If the question is, what is the moment about
the z axis, look, this one doesn't have any
moment about z axis.
This one doesn't have any moment about the
axis.
x component of that has moment about the x
axis, yes or no?
y component doesn't have this one, even [inaudible].
All you need is x component of that force.
Everybody understand that?
That's as simple as that.
The moment Mz is equal to tx.
You find that one, of course, when you expand
this one.
You have done it.
Is that correct or not?
Multiply this height.
That height was how much?
I forgot.
That height was six, eight?
What was it?
Come on, guys.
You did your homework.
You don't remember it?
Open the book.
I don't remember it.
You did the homework.
[inaudible] How much?
The height is 7.75.
Ten point five?
Seven point 75.
[inaudible].
Okay.
This height was 7.75.
That's it.
That's the answer, yes?
I'm sure that you now understand what your
preference from now on should be, yes or no?
Yes?
All the 3D problems, in future, when even
you go to next chapter, I'll give you a 3D
problem.
Try to use this method.
If it is too complex, look here.
The advantage is this.
Let's put it this way.
These forces are all parallel to x, y, and
z, and many times, that's the case.
Is that correct?
But if there are three or four forces going
into different directions, and then you want
to use cross product, and that gives you the
advantage, use it.
I'm not saying that.
That would give you the answer.
However, learn this method as well.
Is that correct or not?
Because especially when you are talking about
moment about x axis, y axis, and z axis, many
components do not have any moment about certain
axis.
There are lots of zeroes there, so don't waste
your time by using cross product.
Make sense?
So these two are now solved.
Can you do that by yourself now, those two?
Of course, you do it.
If you have any questions, you would come
to my office.
Is that correct or not?
But one thing when you go to those workshops,
please do not ask them to do the homework
for you, or help.
When they do the homework for you, the advantage
of that homework already disappeared, because
I don't want you just -- you act like a calculator.
This is the problem.
I have seen a student in the past come to
me and say, "Well, I did all the homework.
I got [inaudible] on the homework."
Then I said, "Okay, sit down.
Let's talk about it.
How many times did you refer for each problem
to here, to me, to somebody else?"
"Many times."
Okay, okay.
So then, you didn't do the homework.
That's not the way you do it.
Is that correct or not?
The homework, when you do it is when it comes
to you first.
Is that correct?
Your ideas come to you.
Of course, this is a homework.
You are a student.
You should do this.
You look at the problem.
You should somehow relate that problem to
the concept that was discussed in the class,
and then you go look at my example, or a book
example, then you say, "Okay, this is what
route I should take."
So you go forward.
Bu still, there are questions that you want
to ask.
There was many of them in this set of homework.
Now let's move on.
The next subject is interesting subject, so
let's look at it.
This is total Mu, so you want to learn it.
So this is the subject of the second row of
the question there.
So before I tell you what those are, I want
you to do this for me.
There is a rod here, okay.
So let's say this is Point A, this is Point
B, this is Point C [inaudible].
This is Point D, and at Point E, I have here
a force of ten pounds, going up.
At Point D, I have ten pounds, going down.
And let's say from A to B is four inches,
from E to C is six inches.
Let's call this one C. From C to D is another
four inches.
This is a very simple problem, but it shows
you a fact that you are going to see [inaudible].
Take the moment of these two forces about
Point A, and tell me what number you will
get.
What's the moment of this force about Point
A?
It is 10 times, what, 14.
Plus or minus?
This is very simple.
You should do that.
This is 14 plus, yes or no?
what's the moment of this force about Point
A, going this way?
10 times 4, negative.
So please write it down, 10 times 4.
So the answer is 10 times 14, which is 140
minus 40, so 100 pounds each moment.
Yes?
Now let's take the moment about Point B. Here,
I'll change the point.
Yes or no?
What's the moment of this ten pounds about
Point B?
Zero.
Zero.
What's the moment of that point about Point
B?
Ten.
10 times 10.
And it is [inaudible] 10 times 10 equals 100
pound-inch.
Now let's take the moment about Point C. What's
the moment of this one about Point C?
Everybody should be able to do that.
We are past this stage.
This is very [inaudible] yes or no?
Yes?
10 times 4.
Is it?
And this one, 10 times 6, also positive.
So 10.
The answer is 100 pound inch.
Okay.
Can you guess what is the moment about D is?
A hundred pound-inch.
I'm glad.
So 100, 100, 100.
Yes?
By the way, taking moment about this Point
E, which has five inches here, would be, again,
equal to 100.
Taking moment about here would be 100.
Taking moment -- what's happening here?
Why this is the case?
Is there anything special about these two
forces?
[inaudible].
No.
You said the right word, but we don't know
that yet, do we?
These are two forces equal and opposite.
When two forces are equal, this is important
to recognize.
When two forces are equal and opposite, we
call them a couple.
This is not for every two forces.
The two forces are not equal, this is not
happening.
The two forces must be equal and opposite.
So that's the meaning of the couple.
You just said it, correct.
Yes, good.
You went ahead, looked at the book and -- I
wrote it down here about the couple, yes or
no.
I said it to everybody.
I said this row is about the couple, is that
correct or not, yes?
Those are forces.
And what is the idea behind the couple?
See, that's the whole point.
The moment of that couple, about every point,
is the same.
It doesn't change.
Is that correct or not, yes?
Notice I just don't have to prove it.
I will prove it for you mathematically later,
but look what I did.
I took moment about A, B, C, D, E. All of
them end up to be equal to 100 pound-inch.
They are all equal.
So any other point I take, it would be the
same result [inaudible].
So the idea behind the couple is, when you
have a couple -- write it down.
Which are two forces equal, the opposite doesn't
have to be in this plane.
It could be in this plane.
This force going up, this force going down
with the same magnitude is a couple, yes or
no.
But this couple is about which axis?
Now, let's understand that you end up with
100 pound-inch of what, i, j, or k?
Look at it.
i, j, or k?
k.
Therefore, this couple, which from now on,
we show it like that, this is 100 pound-inch.
You will see that in your instruction from
now on.
You see lots of arrows like that.
Of course, if it goes counterclockwise, it
will be positive, because that's the net effect.
The two forces cancel each other.
Everybody see that?
However, if the two forces come on top of
each other, which [inaudible] equal to zero,
that is no couple.
Is that correct or not?
So that is how the system works.
In future, if you have a pipe system like
that, and you put here a force here, a force
here, these two forces are equal.
This system is not going to rotate.
Is that correct or not?
Everybody understand that, yes?
Because the moment of that one, and the moment
of that one about the center is -- one is
negative, the other one is positive.
So it's zero.
However, if I create this kind of action,
you see now this is going to be rotated if
there is an axel here at the center of this,
this pulley is going to rotate this way.
Is that correct or not?
And then about a moment, we can show it like
that.
This Mo is equal to t times radius, whatever
the radius is, times 2.
Because there is one here, one here, or become
equal to t times diameter, which is the same.
Is that correct or not, yes?
This is the -- now, don't get this wrong.
This is a one-dimensional couple, so if you
take a drill in your hand, and you're drilling
this way -- I'll show you that.
You are creating a couple about the z.
This is, in actuality, what we have done here.
This couple can be represented as a vector
m of a couple.
I'll put C, because that's the couple.
It doesn't have to be any special point.
It could be any point.
Must be equal to 100k pound-inch [inaudible].
However, if I take my drill, and go vertically,
the couple is about the y axis.
Similarly there, would be about the x axis.
However, if you take a drill and go through
this panel, which I hope you don't do, okay.
So you go this way.
Yes, everybody [inaudible].
There is a couple, but this couple is -- this
is the line of the couple.
Is that correct?
It has i, j, and --
k.
-- k components.
So the couple could be one-dimensional, or
it could be three-dimensional.
3D and 1D, or 2D.
It doesn't make any difference.
So you can have all of them.
Everybody understood that?
But the net result is the same.
That means if I have two forces here, going
one up, one -- equal and opposite, I'm creating
a couple, and the moment -- you should write
that in your notes, put it in a box.
The moment of couple, because this is the
one you have to use in this homework.
The moment of the couple does not make any
difference about any point.
Would be the same.
Is that correct or not?
For that reason, I'm going to do one example
quickly, that you get the idea how this works,
then you can do the rest of that through your
homework.
And of course, I cannot finish the whole subject,
but that would take care of the couple, at
least, for time being, which I said you can
do the first row and the second row.
Let me get my example somewhere from here.
So if I can find it.
Where did I put my notes?
Professor, we have a question.
Yes.
Can you repeat what you said about the moments
of the couples?
Because you said to write it down.
I said the moment of the couple is equal to
the same everywhere.
[inaudible] it's about everywhere.
Every point is the same, correct.
I'll show you in a minute what -- you see,
look at here.
That's what I said.
Look, the moment this couple about A was the
same, about B [inaudible].
What about C, and what about D?
And how about E?
It even goes here, and here, and here.
[inaudible].
So the moment is the same about any point.
So you have to pick up a point which is easier
then to -- I'm looking for an example here.
If I can find it [inaudible].
I had a problem.
Oh, here it is.
Oh, okay.
Here.
Now, this is a good example for you to sort
of get this question that you asked.
Okay, here we go.
So here it is.
This is 2D.
I'll just do it in 2D, but 3D will follow,
okay.
Let's say we have here a force f going that
way, and there is a force f going this way.
Of course, minus f.
So these two are equal and opposite.
That's the technique.
Let's say that this distance is one meter,
and this distance is 2 meters.
That's the location of that point B [inaudible]
point of application, at A and B. And then
let's say that this is not in the scale again.
This is half a meter, and this height is given
four meter.
The height is given from there to here, four
meter.
Is this a couple or not?
Yes.
Two forces equal and opposite.
Now, if I ask you what's the magnitude, first
of all, you don't care about the point.
If this couple has the magnitude, yes or no,
what is that couple?
Yes, like the other one.
Is it 100, 120, or 130?
Is it plus, or it is minus?
We don't know.
But we can do two, three things.
We can take the moment about Point O, because
moment of a couple, about every point, is
the same.
Is that correct or not?
Is that a good choice, yes or no?
Or do I have a better choice than that?
Notice what I just said.
The moment about this couple about every point
is the same.
Should I choose O, or do I have a better choice?
[inaudible].
How about Point A?
Because they're all the same.
But if I choose that, then the moment of that
force about A is --
Zero.
-- zero.
All I have to do, take the moment of that
one.
That's what I did here.
I don't know whether you noticed.
When I took the moment about D, the moment
of this one was zero.
I only have to do that.
This is the shortest one we wrote here.
Is that correct or not?
That's exactly what you should do.
So you don't have to go around, although the
moment of this couple about every point is
the same.
But the easy point will be either A or B.
Everybody understand that?
This is the moment [inaudible].
We need the magnitude.
This is a 2D, so we need the magnitude, the
direction.
Is it plus k or minus k, yes or no?
Because this is not a 3D problem.
This is a 2D problem, yes?
Correct?
It's like this one, except the forces are
in angle.
Now, what you can do now, you break those
forces into components.
So I should give you the magnitude of force.
So let's say the force f is equal to five
[inaudible].
Of course, this would be given.
And let's say that this angle is given, which
is the same as this angle.
It's given equal to 30 degrees.
So that angle, which is equal to that angle,
equal to 30 degrees, which makes these two
forces equal and --
Opposite.
-- opposite.
Yeah, angle should be the same, correct?
So make them a couple.
The moment of a couple about every point is
the same.
I can take the moment about O.
This is [inaudible].
O, A, or B. The best choice is either A or
B. So I'm taking Ma equal to Ma.
So let's put it this way.
Is that correct?
This is what you were asking, yes or no?
Equal to what?
All I have to do, break this force into two
components, which you show it either in dash
line that you don't want to [inaudible].
So anyhow, let's show it in blue and -- like
that.
This is 5 cosine 30 degrees, and this is 5
--
Sine.
-- sine degrees.
Is that correct or not, yes?
Okay, does this 5 cosine 30 have the moment
about Point A?
Yes or no?
This is simple.
They are far distance then.
Is that correct or not?
Distance is how much, two meters and half
a meter.
Therefore m equals to five.
I write it in blue, because that's what they
have there.
Cosine of 30 degrees.
Remember, I can break this in components.
That's the rule we learned previously.
Is that correct or not?
Yes.
So 5 cosine 30 degrees times what distance?
Two and half.
Is it plus or minus?
Plus.
Plus, very good.
And then we have this one.
Look at that one.
5 times -- because we have to take care of
that.
5 plus -- [inaudible] 5 times sine of 30 degrees.
Correct.
What distance are we talking about [inaudible]
is that the vertical difference from A to
B, which is 4 minus 3.
Which way does it go?
Put your hand there.
Which way does it go?
This, we should already learn it.
This, we have done it in previous homework,
yes or no?
And it also plus.
So 5 times 30 degrees time 4 minus 1, which
is 3 meters.
So these are meter, meter.
Those are pounds.
And that's also plus.
So when you add these two together, these
two together become equal to 18.3 positive,
183 [inaudible] meter couple.
What is the direction of it?
Of course it is plus, so it is plus k.
Now, you want to be sure what we talk about
here, you can go at home, take the moment
about Point B, take the moment about Point
O, take the moment.
It will all be 18.3, but this is the simplest
of all.
Is that correct or not?
Now I can show you something else too.
So if I break this one in two components as
well, do we two couple here?
I see one couple and two couples.
That's exactly what we did.
Is that correct or not, yes?
Yes.
The moment of the couple actually, again,
equal to four [inaudible] everybody, because
look, the moment of this couple is forces
times this distance between the two, yes or
no?
The moment -- because we are out of time,
I am going to give you a few more hints, and
I will do more examples next time about 2D
and 3D.
But you try your homework first.
Go as far as you can.
This idea should be there.
Need a little bit more elaboration when it
comes to 3D.
Is that correct or not?
However, if you have a force like this and
like that, this is f.
This is minus f, or the other way around.
All you need is [inaudible].
Is that correct, which is perpendicular to
the two, yes?
That will do it as well.
Actually, this way, you can find the shortest
distance between [inaudible].
Here, I have 18.3.
Can I find the shortest distance between these
two lines?
Yes.
Yes.
Eighteen point three equal to force times
distance.
Is that correct or not?
Again, you have that in your homework, D equal
to 18.3 divided by 5.
This is [inaudible].
Come on, guys.
[inaudible] listen to me.
This is part of your homework, okay.
You can find the shortest distance between
two lines as well.
Is that correct or not?
Yeah.
Yeah.
You don't have to go through geometry.
Okay, we are done with the lecture, so let's
go to the quiz.
So this quiz should be simple.
Section two or section three?
Section two.
Thank you.
[inaudible].
Thank you very much.
[inaudible].
Put everything aside, guys.
How many people are here?
One people left.
Only three?
All right.
Five.
One, two, three, four, five.
One, two, three, four.
Put your name down, and will explain the problem
to you.
One, two, three, four, five, six.
[inaudible].
Okay.
Five.
Oop, we are short.
[ Music ]
