[ Music ]
All done? Very good. [Inaudible] the pulley.
I haven't discussed that one yet, so therefore,
the one that we cannot do, we are going to
discuss the last [inaudible] homework that
is something about the pulley. Some of you
may already have the answer for that question.
We are going to discuss it. Besides that,
you had a question. Yes.
I was just wondering, how do you get the answers
to the like, class exercise problems? Do we
just come to your office hours to check?
No, I can put it on the board.
Oh, OK.
OK.
[ Laughter ]
You can always ask that question, I'll put
it on the board. And which problem do you
want answered? Which problem did I give you?
Which problem do you want? [Inaudible]. I
don't know how many I gave you.
All of them [inaudible].
You gave us quiz number two. It was the [inaudible]
packet.
Yes. The quiz number two, I can give you that
answer. That answer -- force F. This is for
quiz number two. The answer for that problem
was F [inaudible] is 141 pounds. That's one
question. The other question that you might
not have the answer [inaudible], I gave you
problem number two, [inaudible] from the handout.
Or that answer is in the handout. Yes.
[Inaudible].
Which one I did not give you, or which one
you don't have the answer? I don't remember.
Question number [inaudible].
What? Oh, those?
Yeah.
Those, I don't have it.
OK.
Those are extra. I thought you were asking
a homework problem.
Oh, no.
No. Those, you can ask me during the office
hours.
OK.
And then sometimes after it pass a week or
so, I do not bring those solutions with me,
so you have to --
Oh, OK.
So if you need, you can send me an email or
something, or you know, we can discuss it
in [inaudible]. Those are not required. You
need only to do the homework assignment.
OK.
All right.
I was just [inaudible].
Those are extra, so just for your [inaudible].
As long as you know how to set up those correctly,
don't worry about the answer, because usually
if your setup is correct, your three-body
diagram is correct, you should have no worry
about that.
OK, thank you.
OK. Are you OK with that quiz number two,
because you asked for the solution [inaudible].
Are you OK with quiz number two? Yes or no?
I have a question.
Yes.
I was having trouble finding the angle for
[inaudible].
That's quiz number two you are talking about?
Yes.
Yes. That's exactly my point. You have to
look at that problem. Some of you already
saw me during the office hours. Actually,
many of you were there, that's good. The more
people, the better it is. So what that means,
somehow you have to try it, and if you cannot
get it done, then I will help you to find
out what the answer to that question is. Nevertheless,
you have two [inaudible] and you need to draw
the three-body diagram of each one separately.
Yes or no?
Yes.
Then you draw the three-body diagram of the
top cylinder. This cylinder touches the wall.
There is a vertical wall here, yes or no?
Yes.
Therefore, there should be a normal force.
There [inaudible] perpendicular to the wall
cylinder which is towards the wall. Wall pushes
back towards the cylinder. So obviously the
normal force should be like that. So we call
it MW. Is that correct or not? Yes?
Yes.
Then, of course, you're assuming this is [inaudible].
Remember, I'm putting everything at the center
of all of the forces, intersect at the center
of this [inaudible]. Then there is a force
of -- I don't know what you called it. That's
cylinder A or cylinder B. I don't remember.
A.
A. OK. So that is WA, which is weight of cylinder.
You can show all of this in magnitude or you
can show it in a vector form. Choice is yours.
Is that correct or not?
Yes.
As you I asked you before, you can even write
at least four [inaudible] vector. [Inaudible],
but the essential part of the problem is a
[inaudible] idea. Is that correct or not?
Then there is a normal force [inaudible] here.
Notice when I do that, the benefit of this
homework is already gone. If I draw you the
three-body diagram and after that you are
submitting this homework, the effect of that
homework is already gone. Because after that,
you are only adding or subtracting, which
everybody can do. Everybody understand that?
This is the problem, it's been asked probably.
And the question was what is this angle? Is
that correct or not?
Yes.
Yes. So the question is very -- what is that
angle? That angle is not the other angle given.
But a student make a mistake that that angle
is not [inaudible] that is given for the other
surface. It's got nothing to do with that.
This angle is somehow to do with the size
of two cylinders. Yes or no?
Yes.
So actually, this normal force, if you look
at that, this is cylinder A, cylinder B sitting
here. This force N, is in the direction of
AB, correct? Yes. So cylinder B is here. So
all you need is to find the direction of AB.
To find the direction of AB, look at that
triangle that forms between A,B. It is in
your handout. And this horizontal force. This
is the radius of this cylinder which that
is that part. So it is like that because I
don't want to crowd that picture. This triangle
sitting like that there. Is that correct or
not? Yes? So that is six inches. This one
is -- I believe this is six inch because they
are touching here at this point. This is six
inch or four inch, something like that. To
the total is?
Ten.
Ten inch. So this one becomes eight inch,
because that's the ratio of three, four, five.
So [inaudible] nature of this, three, four,
five ratio. You can find that angle or the
other angle, which gives you that angle. Yeah.
Correct? Good. So that's the first one. Now,
for the second part, you want me to draw the
three-body diagram at the second part, or
can you do that? See, if I do that, then I'm
solving the problem for you. That's not good.
Everybody understand what I'm saying? This
lower part has actually four force. The lower
cylinder, the other one has four force. Might
as well give you some indication. You see,
that's the whole idea here. If you don't think
about the problem yourself and you do not
settle it yourself, you are not understanding
what the question is. Me doing it for you
doesn't teach you, because none of the other
problem looks like that. You have to dig in,
you have to really sort of put in all of your
effort into solving the problem, put the data
together. And you'll see they already probably
have done all of the three-body diagrams.
They have one question, and that's what [inaudible]
for that angle alpha, what was the value of
that angle alpha, which I showed them how
to do it during the office hours over here.
But this one, of course this normal force,
which is between the two cylinders, so you
can call N1 or NC. Some people call it NC.
That's why. NC means between the two cylinders.
Of course, that force comes here after you
solve this problem now. This problem has two
unknowns. It's MW and NC. Yes or no? Because
this is given angle, [inaudible], it's a very
simple problem. You come here, you put that
N there. This is NC. Again, equal and opposite.
That is now solved because this is [inaudible],
this is done. Then you go here and this time,
let's put all of the unknown in color, so
we know what the difference is between known
and unknown quantity. Then here, we have a
force F. And magnitude of that is unknown.
Of course, we have here a W, weight of cylinder
B, which again, I'll put it in magnitude.
And that is given. Now what else do we have?
There is a contact or not contact force. Let's
put it this way. The contact force is in the
form of the normal force and friction force.
However, we are assuming there is no friction
there, so we only have normal force. Yes or
no? But normal to what? Normal to the incline
surface. Is that correct? And that incline
surface has an angle of eight. OK. So that,
practically I solved the problem for you.
So this one is N [inaudible] and that angle
is [inaudible]. And I asked you to practice
that because when you practice that one, as
I said, you probably are going to get about
40 or 50 homework which has these two lined
perpendicular to each other. When we get to
the next chapter, we're talking about the
moment. Many times, we have to find these
two directions, so you need to practice what
is this angle and how you can [inaudible]
that angle, which is a very simple case. But
you should do it once or twice in order not
to think about it, get it more quickly during
quizzes. Yes or no? All right. OK. That was
the question you raised, so that's fine. The
rest of the problem is straightforward, and
you have many, many questions. Some of you
had that answer during the office hours. Let's
go to handout problem number five, which has
something to do with the [inaudible]. And
some of you already know the answer to that
question as well. So look at the problem number
five in the first handout. And let's see what
is the problem or how do we solve this problem.
This is the scenario. You have a post here.
And then there is a little cylinder here attached
to that. There is a ring here. There is a
cable here. There is a cable there. And there
is a weight, W, here. And there is force here,
F. Of course, I do not have to draw that [inaudible]
where they are. This is point A, this is point
C, and that' is point B. This distance, CB,
is given because we want to have [inaudible]
from here to here, [inaudible] given equal
to one [inaudible]. And this distance from
C to A is [inaudible] meter. W is given equal
to or the mass, M, for this box is given equal
to 20 kilograms. The question is what's the
value of force F, or what is the magnitude
of force F, and what is the tension in cable?
And after you have done all of your how many,
this should not be a big deal. Is that correct
or not? Yes? So everybody knows that you have
to draw the three-body diagram of [inaudible]
A, correct? But look what happened here. If
I draw a three-body diagram of [inaudible]
A and I put the W here, let's calculate now
the W. W, of course, becomes equal to 20 kilograms
times G which is 9.81. And you get 196, going
only three digits. You don't have to go to
the fourth digit. 196 meters. So the weight
is given. Force F is unknown. Force F is unknown.
Notice, the system is in such -- because of
force F, the system is in equilibrium in this
position. You [inaudible] force [inaudible],
the whole system is going to go back against
the column. Is that correct or not? That's
the whole idea here. Anyhow, and then, like
before, we see two cables here, yes or no?
This is the first look. So there is a cable
here. This cable shows the tension. This cable
is the tension. These tensions are not in
the same direction, so they are different.
Yes or no? Now, let's draw both of them here.
So one going in that direction, which I showed
you before how we do it. One is in the direction
of AC, and the other one roughly is in the
direction of AB. Is that correct or not? Yes?
So let's call this one TAC. And this is magnitude.
And this is TAB. Or you can call it T1 and
T2. How many unknown do I have?
Three.
How many equations do I have?
Two.
Two equations. Sigma FX and sigma [inaudible].
Can I solve this?
Not yet.
Not yet. So something is missing. If I have
three unknown and only have two equations,
sigma FX equals to zero, sigma FY [inaudible].
Obviously I cannot solve it. Is that correct
or not? So what did we miss?
The moment?
The moment. Why the moment? We haven't talked
about moment. Have you taken this course before
or have you heard about it before? We don't
know yet anything about moment. Yes or no?
Yeah.
If I have two equations, I cannot have more
than two unknown. That's a fact, correct or
not?
Yeah.
So what's wrong with this problem? What did
they miss? That's the whole idea behind your
next two homework, or the last two homework,
which are the pulleys homework. Read the question
one more time. You see, it says something
there in the question. You have to pay attention
to the -- is this two different rope, or is
it the same rope?
Same rope.
You see the difference between the idea here?
If you have -- this is you have something
to remember for future. If I have a ring here,
and this ring, there is a rope attached to
here and there is rope attached -- [inaudible].
I could have two different tensions. Is this
the same rope?
Yes.
Actually, it's not the same. If you look at
it, I did it purposely like that. Notice,
the rope comes through the ring and goes out
of the ring. That is the same rope. Therefore,
the tension, if it is the same rope and we
have no friction there, the tension must be.
[Inaudible].
Listen, don't get me wrong. This is the vector
in different direction, but that's the vector
[inaudible]. the magnitude of these two vectors
must be the same. Is that correct or not?
Same distance. Write it down in your notes.
Same distance. This is the same rope. This
is the idea you have to use in the last two,
three homework that you have. If the same
rope and you are not considering the friction,
tension must be?
The same.
The same. Is that correct or not? Yes? Therefore,
this whole solution [inaudible] problem is
that TAC, magnitude must be equal to TAB magnitude
[inaudible]. Let's call them T, just one T,
because there's only [inaudible]. Now, how
many unknown do you have?
Two.
Two. Although there are three forces there,
but in effect, you have only two unknown and
you have two equations to solve. You can go
by ratio or you can go by the angle. This
time I'll got by angle to see that both of
them -- this angle, you can solve it from
this equation. That angle -- tension of that
angle equal to one over two. Yes or no? So
let's call that angle equal to alpha, and
let's call that angle equal to [inaudible].
So this is [inaudible] problem. First, you
set up this, because that shows you that this
is solvable, otherwise you will not be able
to solve it. And then you calculate alpha
equal to tension minus one [inaudible] equal
to what we have there. One meter over two
meters, or alpha becomes equal to 26.6 degree.
And [inaudible] become [inaudible] minus one.
Or you can use this [inaudible] if possible.
Either way, it's correct. Actually, the other
one is [inaudible] three, four, five. But
nevertheless, that is equal to two and a half
meters over two meters, or equal to 51.3 degrees.
And as you see, the rest of the problem is
simple. You [inaudible] sigma FX equal to
zero to set up one equation. And sigma FY
to calculate the tension. Actually, let's
write sigma FY first. So write sigma FY first,
because depending on the type of equation,
you can write the one that helps you the most.
So in this problem, sigma FY must be equal
to zero. Notice, again, this is the vector
equation, you don't need to put J down. I
saw one or two students come and put a JJJ,
III there. That's not required. This is not
a vector equation, everybody see that. Anyhow,
you get T times sine of 26.6 degrees because
this was 26 sine of this one. But you are
vertical. This is cosine, that's sine. And
then plus T, sine of -- as you see, the reset
is very simple. 50 -- how much was it? 51.3,
.50 times .3 degrees, then minus 196 equal
to zero. Obviously, you can solve for T. T
becomes equal to 160 meters. And then immediately,
you put into sigma FY. I'm not going to do
the rest of it, it's very simple. Is that
correct or not? [Inaudible] sigma FY, you
calculate [inaudible]. Correct? So that is
not essential here, but I did need to ask
that anyway. So I try to avoid as much math
as possible in class because it is a simple
problem, I [inaudible]. The key was just this
one, that's it. That's [inaudible]. All right.
Now, some of you have tried your pulleys problem.
It was just a couple of problems that I have
given you. This rope is the same, so you put
the same tension. Correct? Now let's go to
the handout problem that I have given you,
the one that goes to the next page. [Inaudible],
but there is a handout, there is a pulley
problem I have given to you. I don't know
what page it is. Here it is. Go to page ten.
This is out of your book.
[ Silence ]
Let's assume the weight of that block, that
[inaudible] there, there are five different
scenarios of pulley [inaudible] and pulley
system. Holding a box, and you want to find
the tension in each cable. Let's start from
system one. Assume W equal to 600 pounds or
[inaudible]. What did I give you? Did I give
you a number? No? 300. OK. Let's change it
to 300 pounds. OK. The weight of the box is
given 300 pounds. And the scenario is like
that. But I cannot draw that because this
is not [inaudible] like this, this is the
rope. And this is the 300 pounds. This is
T. What's the value of T? Commit yourself.
What's the value of T?
300.
300. Yes?
Yeah.
Sorry to say, you're wrong. Do you see mistake
you are making, guys? That's exactly my point
about this problem. If you do not pay attention
to what has been said, [inaudible] you think
this tension in this cable is 300. It is a
wrong answer. Now, what you didn't do and
you should have done there, you should have
drawn a three-body diagram of this pulley.
Yes or no? OK. Let's do that. They purposely
did it this way because they want to in effect
make you think about the problem and come
up with the right answer. Let me put it this
way. If you have here a rope and here is the
box, and the box is 300 pounds, is the tension
[inaudible]?
Yes.
Is this the same?
No.
You see that? They are not showing the rest
of this. This could have gone up here and
attach to there. Everybody see that? So if
you draw the three-body diagram of this cylinder,
I mean, this pulley, as you see there, you
have here, of course the [inaudible] is the
one that attach to the box. This one, you
are cutting it twice. Yes or no? So actually,
this is the three-body diagram. [Inaudible]
write it down in your notes, underneath the
same page, you can put it the same page under
[inaudible]. So the three-body diagram is
the tension going up, and 300 going down.
Obviously, the answer is 150. Notice, what
you are doing. What's the difference between
the two? In this scenario, if this length
is two meters, you are losing two meter cables.
In this scenario, you are losing four meter.
Everybody understand that? Or [inaudible]
four feet versus two feet. Everybody understand
what I'm saying? So it's twice as much, yes?
So you are paying the penalty by using more
cable, but then the tension in the cable is
half the other one. Everybody understand that?
Now, if I do more of this, I should count
to the less and less number, to the point
that you will see in the factory the guy like
me sitting there. [Inaudible], like me, older
guy. OK. Sitting there putting a rope, and
then there is a weight of [inaudible] that
is going up and down. Isn't that interesting?
Yes or no? Yes? Why? Because they are using
some pulley and rope system to achieve that.
Having that in mind, go to [inaudible] two,
or scenario 2B. Tell me what's the tension
in the cable. Don't make the same mistake
you made here. This is the [inaudible]. Draw
the three-body diagram of the pulley in question,
which is the lower pulley. I don't care about
the upper pulley because that's on the wall,
we don't care about that. Yes or no? And tell
me what the tension in the cable should be.
How many cable, in other words, do you cut
when you go draw a little circle? Come on,
guys.
[Inaudible].
Where is your handle? There. No, you can [inaudible]
there, no problem there. But do it. That's
the whole [inaudible]. Draw a little circle.
Yeah, you can do it there. That's not your
book, you can draw that. So what should be
answer?
[Inaudible].
So you are afraid to say it. You are thinking
either 150 or 100. Which one is correct?
100.
150.
150, that's right. Because, you see, that's
the whole point. Because, unfortunately for
the benefit of the camera, we don't have that
on the camera. I don't know whether you can
take the picture and show it there. But however,
this is the center view there. So let me look
at one of your handout [inaudible]. OK. Here
is the second one, which second one is like
this. So this is the ceiling. So the rope
comes past through the pulley, goes up, goes
through the other pulley, which is attached
to the ceiling. But then this one comes down
and they come before T. Remember, again, this
one is not attached to here. This one can
go anywhere. Is that correct or not? This
one can go all the way to something else.
Yes or no? But this is the weight. This is
your three-body diagram. So how many cable
is being cut?
Two.
Two. What's the answer? It's 150. So this
would be W equal to 300 [inaudible]. And then
[inaudible] three-body diagram becomes exactly
like that. Is that correct or not? Yes. Sorry
for [inaudible]. Now, look at the third one.
What's the third one? Now, this time, because
careful. See, all you have to do, this is
what I'm asking you to do. Draw a three-body
diagram for the lower pulley and see how many
cable is there. So what will be the answer
for this one?
100.
100. See, now, everybody getting it correct.
This is the whole idea of learning process.
The first time, maybe two-third of students
had the question or more. Now, [inaudible]
to 60, 40, 30, 20, we go down. Now everybody
has the answer. Yes or no? It's 100. What's
the next one?
[Inaudible].
The fourth one. 100. Yes. Both correct. Now,
here is again, a little challenge. What's
the last one?
150.
100 or what?
[Inaudible].
100 or 75. What do you think? No answer? Come
on, guys.
[ Background Conversations ]
I gave you a hint. One of the two is correct,
100 or 75. Which one do you think it is?
75.
75. All right. This is the scenario, guys.
Again, I need [inaudible] your picture because
I don't have mine with me. But this is a book
anyway. So here is the scenario. Actually,
this is a very nice problem. Actually, this
is it. So this is the ceiling, and then there
is a pulley there, it goes like that. OK.
And then the rope comes here, that's the lower
pulley. This is the box. OK. And then there
is a link here. And this goes around here
as the tension. Goes around here, goes there,
goes around here, goes around here, goes back
there, something like that, if I have the
picture [inaudible]. Notice, when you draw
this three-body diagram, this is the tension
here. You have actually, then you draw the
three-body diagram of this part. This is how
to show this part. Yes or no? Yes? So you
have one, two, three, and four. So thank you
for this one. My picture is very ugly, but
you have the right picture in front of you.
Yes or no? So therefore, it is 75. Now, can
you tell me what is the force in this link?
Look at this link there. There's a little
piece of metal there. Yes? Connect those two
together. How much do you think, first of
all, that link is in pulley mode, which we
call it later on, tension. And how much force
do you think that link has?
75.
75 or 150?
150.
Because there are two of them [inaudible]
on top of that. Look exactly like that. You
want to do that. Let me show you the three-body
diagram. If you want to show the three-body
diagram of that one, it's like that. When
I cut it here, I can cut it up there, which
is one, two, and three, and four. Or I can
go a little bit lower. Everybody understand?
If I go lower, then I see one, two, and a
link. Yes? OK. So let's go that way. So here
it is. One tension here. One tension here.
And I have this link, the piece of metal,
which I cut it there. Is that correct or not?
Yes? But that piece of metal, notice, it's
connected to the same cable, but it passes
twice. Is that correct or not? Therefore,
I have here in effect T and D. Is that correct
or not? Yes? That's why you came up with 40.
And now, a person -- what's the purpose of
this assembly? Of course, if I want to make
that, I will go to the last one because the
last, I have 75-pound force applied to the
rope. Yes or no? At the end. But the weight
that goes up and down is 300. Is that [inaudible]
application? Of course it is. Yes? Everybody
saw that? Now, using this concept in some
of your homework would resolve your issue,
but remember that you have the people that's
hanging from the cable, but some cables are
the same. If they are the same, it doesn't
matter. The tension is the same. Magnitude
of tension the same. But they are in two different
directions. So you can use that concept, I'm
sure you can do your homework. Is that correct
or not? Those of you who have looked at your
last two homework, that was the question that
you raised. Is that correct or not? Now, the
next question, don't worry about it. If you
can do that question, you are at top of the
class, I'll tell you that. If you don't do
it, don't panic, because that is a little
bit more difficult than the other one, but
it is not required. Everybody understand that?
It's just to test yourself for the fun. If
you can do the other one, then you can come
to my office, or next time I can give you
the answer. But it's not required from you.
The next problem is a more challenging problem.
I do not expect you to be able to do that.
However, if you want to challenge yourself,
go ahead and do it, and we'll discuss in class
again. But that's not part of the curriculum.
Is that understood? Yes? Any questions? We
are done with this part of the book. Any question
there? Now you are ready to handle all of
your homework, correct? Or answer all of the
little questions that's left out, right? Let's
move on to 3-D, yes? OK, guys. So you have
now to be careful a little bit of what we
are doing. Now, before we go to the vector
in the space, which we already have, by the
way. I have to explane something to you which
is related to 3-D problem, but I will explane
it in the 2-D format. [Inaudible] those are
direction angles.
[ Silence ]
And direction cosine. Or more likely lambda.
Lambda is the unit -- this comes very handy
for 3-D problem. However, I have to explane
it to you in 2-D which is much simpler to
understand. The idea is this, that something
that you already used in your homework many
times, and that idea is the following. So
you have here -- I'm going to put the idea
here. So here you have XY system of coordinate
in 2-D. Then you have here a force F going
like that. And that angle usually give you
equal to [inaudible]. Yes or no? We talked
about this in the past. F [inaudible], you
can use it two, three different ways, you
can break it into two components, FX, FY.
But eventually, F ends up to be equal to F
cosine theta I plus sine theta J. Either pound
or meter, or whatever [inaudible]. Is that
correct or not? This is the representation
of a vector in 2-D, correct? Which many of
you used. However, if the force goes this
way, one becomes positive, X component becomes
positive -- I'm sorry. theta X is positive,
Y is positive. If it goes here, X component
becomes negative. Y goes positive. You have
used it in the past and you have used it in
your homework. I'm going to change that a
little bit, for the reason of, not because
of the 2-D problem, because of the 3-D problem.
I'm going to call that angle equal to theta
X. What is that angle? Why did I call it theta
X? That angle, please write it down, is the
angle between positive direction of the force.
Please write it down. I see that some of you
are not writing that. Because this is not
in your book, in this format at least. theta
X is the angle between positive direction
of the force and positive direction of the
x-axis. Similarly, this would be [inaudible].
Usually we don't need that because I use cosine
and sine. Everybody understand that? But I'm
changing that. Instead of sine of that angle,
I'm using cosine of that one, which is the
same. Is that correct or not? But I'm going
to call that angle [inaudible]. So theta Y,
again, with similar scenario is the angle
between positive direction of the [inaudible]
and the y-axis. Yes or no? Now my equation
turns into this equation, F equal to -- rather
than having cosine sine, I have both cosine
now. Write it in terms of cosine and cosine.
Cosine theta XI plus cosine of theta YJ. Again,
unit the same. Is that correct or not? Yes.
So all sine goes away, becomes all cosine.
Now remember, cosine theta X was -- because
remember what we said last time. I don't know,
I hope that you remember. Each force equal
to its magnitude times a unit vector. Yes
or no? Each force equal to a magnitude times
a unit vector. So this is lambda X and this
lambda Y. Yes or no? The same as cosine and
sine. We discussed that last class, correct?
Yes.
Now what happened here is this. If you use
this one -- by the way, in 3-D, that's why
I have this here. In 3-D, if this happened
to be x-axis, this happened to be y-axis,
that happened to be z-axis, we don't have
this scenario. How many angles do we have?
We have theta X, theta Y, and --
[Inaudible].
That's what I'm getting at. Everybody understand
that? I am using this technique to get to
that one, especially because of having three
angles, not two angles right here. Now, right
in general, we have a theta X or theta Y for
2-D problem. In 3-D problem, you have theta
X, theta Y, and theta Z. These angles are
called directional angles. That is the name,
which are defining the lambda in other words.
Is that correct or not? Yes? So this is in
2-D. So in 3-D, then F becomes equal to F
times cosine theta XI plus cosine theta YJ
plus cosine theta -- all I'm adding is the
[inaudible] component. Yes or no? And you
have seen that in the past, this is the same
part. But notice, since I have three angles,
no longer I can use sine and cosine, because
sine and cosine was for 2-D problem. Yes or
no? So I'm changing it all to cosine. So there
is two definitions. The theta X, write it
down. Direction, angles actually are theta
X, theta Y, and theta Z. By definition, there
are -- the angle is in positive direction
of the force and positive direction of x-axis,
y-axis, and z-axis. Yes or no? Correct? OK.
Why did we do that? What's the advantage of
that? What this means? It means this. If I
use this technique or that technique -- please
write it down. If I use all cosine in this
format with this definition, this should be
in a box, and put a star in front of it. If
I use this format or all cosine format with
definition was given to you, you do not need
to use any sine, the sine automatically will
appear. So you don't care which way the vector
goes. Automatically, you get all of the pluses
and all of the minuses. Yeah, I know you are
surprised. Now I'll show you in a minute how
it works. Is that correct or not? Yes?
Yes.
Do you want to see what I told you?
Yes.
OK. Let's practice it in 2-D. Always a good
way to do it in 2-D. Look. If there's no problem,
go this way. Now let's go this way. Let's
say this is a force F, and I state magnitude
of F is given equal to 1,000 pounds, for example.
And let's say that this angle is equal to
30 [inaudible]. Just purposely I give you
that angle, it won't be 30. I don't want -- everybody
knows how to write this in the old form. The
old form is this is cosine 30 degree negative,
that is sine of 30 degree. It still will do
it for 2-D, don't get me wrong. If you are
2-D, still you are using the same technique.
Don't change anything. But understand what
I'm talking about the 3-D problem. Yes or
no? OK. In 3-D problem -- now, I want to use
all cosine. So tell me what is theta X equal
to?
[Inaudible].
How much?
[Inaudible].
You see, you're thinking now. Is it 30 degree
or is it 150 degree?
[Inaudible].
Read what I told you to write down. Read it.
What is the theta X? Read the definition of
theta X.
[Inaudible].
Yeah, that is the key. It's the angle between
positive direction of the force and positive
direction of -- isn't that what you have written
down? Apply that. You see, this is the whole
trick here. So when I said theta X is positive
direction of X and positive direction the
force, the angle is not 30 degree, the angle
150 degree. Do you see that? So in that respect,
[inaudible] take X, which is direction angle
of this one, is not 30. If it was here and
this was angle, then it would have been 30.
Is that correct or not? So it is 150 degree.
And theta Y is how much? Use the same thing,
the positive direction of the force and positive
direction of the what? So the angle is?
60 degrees.
You see that? Now everybody got [inaudible].
That's the whole scenario here. Theta Y is
60 degree. Now I don't need this. All I am
saying is that F is equal to -- F times cosine
of [inaudible]. Notice I'm not using sine
anymore. That's what I said. Do not use any
sine. Actually, if you use the sine, you ruin
the whole scenario. Is that correct or not?
I don't know where the force is. This is given
to me. Everybody understand what I'm saying
there? Cosine have no sine here. I said do
no use any sine. Cosine of 150 degree I plus
cosine of 60 degree J. Either pound, or yes,
the magnitude was pound, so it is equal to
pound. So let's find out what the answer is,
the F equal to -- get your calculator. This
is 1,000 pounds. And what is cosine of 150
degree? Use your calculator. You will see
the answer automatically [inaudible].
[Inaudible].
.866?
Negative.
Yes? Negative or plus?
Negative.
Negative. You see, the sine came by itself.
Did you see that? Yes? Lady in front?
Yes.
OK, good.
[ Laughter ]
.866i plus cosine 60 degree which is?
[Inaudible].
.5 or J. So the answer would be minus 866i
plus 500j pounds. I did that for 2-D, but
believe me, with 2-D, we don't need to go
through it. Again, for the second or third
time. For 2-D, you still can use cosine and
sine. This was for the purpose of proving
that if I use this, because I have a force
here, guys, like that, force F, is it coming
towards us or going towards the other room;
can you tell me? I don't care. As long as
you give me theta X, theta Y, theta Z. I can
write my force in the vector form. Yes or
no? The only question, how we find the [inaudible].
So this would be in 3-D. If I used this technique,
but first of all, as I showed you here, that's
why I have this again, one more time. So I
have here a force here, and a vector here.
Again, from now on, remember this in this
class, because I'm not going to repeat that.
This line is our x-axis, this line is our
Y, and this line is our z-axis. And that follows
the right-hand rule. So please write it -- everybody
knows what the right-hand rule is, yes or
no?
Yes.
OK. Good. All right. So define it for me.
What's the right-hand rule? Who said yes?
I'm kidding. What is the right-hand rule in
other words? Because some of you didn't say
that. I just want to. Because we need that
for the next chapter a lot, especially when
we to the moment. How do you define -- like,
this is x and y-axis. z-axis coming toward
us or going towards that one, it's like this.
You put your finger in the direction of the
X. Everybody do that because we need that
to understand, everybody. And then you turn
it around 90 degree. This finger should be
in the direction of the Y. Then the thumb
is in the direction of the Z. That's the right-hand
rule. One, two, three. Is that correct or
not? X, Y, and Z. If you are left-handed,
sorry, you still have to use right hand. That's
the rule of the game. So therefore, this would
be the following. So that is the rule you
are using. And this theta X, and theta Y.
Therefore, if I'm using this corner of the
table, this is X, this is Y. And that is Z.
And a force like that, this is now theta one.
Agreed or not? Yes? This is theta -- this
angle is?
X.
Theta X. And that angle is?
[Inaudible].
Which is very difficult to measure. Now, we
want to come up with the component of the
force in 3-D. How do we do that? This is the
question we want to answer, then we go to
adding and subtracting, et cetera. So now,
remember, I said theta X, theta Y, and theta
Z. What is cosine? Direction cosine. Actually,
direction cosine are these three numbers.
Write it down in your notes. Cosine theta
X, cosine theta Y, and cosine theta Z, which
is the same thing. [Inaudible] cosine theta
X is lambda X. Cosine theta Y is?
[Inaudible].
So please write it down. So because of this
rule, the three of them are -- this is very
important, again. This is the right-hand rule.
And again, write it like that just for future.
FX equal -- we had that. We had F times cosine
theta X. We had FY equal to F times cosine
theta Y, and FZ. All I'm adding is the third
component [inaudible] F times cosine theta
Z. As I said it earlier, theta X, theta Y,
theta Z are the direction angle. Direction
cosine are lambda altogether, which are these
three numbers. But cosine theta X, remember
that, cosine theta X is lambda X. Cosine theta
Y is lambda Y. Cosine theta Z is lambda Z.
As the result, cosine theta X squared plus
cosine theta Y squared plus cosine theta Z
squared should be equal to one. Because that
is the unit vector. Everybody understand?
Because that's very important [inaudible].
Sometimes I'll use that. In one of your homework,
they may give you two angle. The third angle
you don't need it. You can use this rule,
because the square of all of them together
should be equal to one. So please write it
down The rule is the are direction cosine,
those are direction angle. And then because
of this being lambda X, lambda Y, lambda Z,
everybody knows that all three components
squared added together should come to magnitude
of vector which is one. So that was your answer.
Correct. Theta X squared plus cosine of theta
Y squared plus cosine of theta Z squared should
be equal to one. Which you can use it in some
of your [inaudible]. Now comes to the application
of this system. So in the next question we
want to answer, this was extra information
on top of what we want to do. And now we want
to write. Two things we want to do today:
write the forces in the vector form. Yes or
no? Right, in 3-D. In 3-D, OK. Write forces
in vector form. In other words, come up with
FX, FY, and FZ. Two methods we are going to
do. Method one, we are going to use projection,
which should be simple, probably you already
have done it in physics. Method two, perhaps
you haven't done that yet, is by position
method. Using position method. And after we
do that, the fact is very simple. Like the
previous chapter or previous discussion, if
I want to add several forces, all I have to
do is write R equal to F1 plus F2 plus F3,
and the same symbol can be used. Now, let's
do that first part. Now, what's the projection?
Let's do it through an example. So I'm doing
it through an example, but the formula is
the same way. You do not need to remember
the formula at all. You need to just use this
procedure. OK. Let's say that we have here
a force -- let's do it this way -- a force
OA, which represents the force F. I believe
I assume the force F magnitude -- this is
the idea we want to discuss. This is an example.
Magnitude of F equal to 1,000 pounds is given
to us. Notice, in order to come up with this,
I need magnitude or two angle, or I need all
three angle or something here. I have to be
able to come up with always of the answers
like that. So this is done in different format.
Let's say that this angle is given equal to
60 degree. So what I'm saying to you one more
time, if this is the [inaudible] -- -- and
this is the vector going from there, this
angle is?
[Inaudible].
Don't look at the picture because picture
might not be as clear as here, what I'm doing
here. Is this your vector? The black [inaudible]
is a vector. This is y-axis. This is x-axis.
This is a very simple procedure. I'm sure
that you already have done it. But drop a
point from here down here. Yes or no?
Yes.
The projection of this force over horizontal
plane will be a line here. Yes or no? However,
if I draw another line between here to here
-- everybody see what I'm talking about? A
line from here to here gives me this line.
What is that? That is?
[Inaudible].
FY, isn't it? Yes? That's the projection of
this force. In other words, I have to draw
one line parallel to this plane, yes? Which
we are going to do that here. But remember,
this is [inaudible]. In other words, we are
going to project that here on the horizontal
plane, and I get this vector here, which I'm
going to call it FH. So this is on the horizontal
plane, remember that. One more time. This
is the vector. I'm drawing a line here. That
line FH is here. Is that correct or not? Yes?
That is not FX, it is not FZ. Remember, this
is x-axis, as we said earlier. This is your
y-axis. And this is your z-axis. Now, I have
to draw this line here, assuming this is perpendicular
here. You don't see it here, but that's 90
degree because that is this line. This is
going [inaudible] horizontally. Yes or no?
This is 90 degree. This height would be FY;
correct? Because it is F times cosine of theta
one. That's what's written there. Is that
correct or not? So if this is 1,000 and this
is -- now I show it in color -- from [inaudible]
to B, that is your Y component. Yes? Agreed
or not?
Yes.
OK. So what's Y component equal to? So FY
equal to F times cosine of theta Y was written
there. So it is 1,000 times -- let's do it
underneath that. But I do not want [inaudible]
picture. It is 1,000 pounds time cosine of
60 degree. We already gave you cosine of 60
degree is .5. Yes or no?
Yes.
Yes. .5. Therefore, it's 500 pounds. So this
length, which we call it FY in the other vector,
or as a magnitude. It's 500 pounds, but it
is in the Y direction. [Inaudible] 500 [inaudible].
Correct?
Yes.
Now, what's FH equal to? Can I calculate FH?
See, I cannot yet calculate FH and FY, but
I can calculate this. If this is 60 degree,
what's this [inaudible]?
30.
Go back here to this picture. If this is 60
degree, what's that angle? 30 degree. Or you
can use the sine of 60 degree or cosine of
30 degree to come over this length, which
should be a little bit shorter than 1,000.
This is 1,000 and you are projecting it here.
Is that correct or not? The line on this wall,
yes? Which is line, oh, let's call it C. Yes?
It is on horizontal plane. But the magnitude
of that is not anything that I want yet. FH
equal to 1,000 times cosine of 30 degree,
or sine of 60 degree. Is that correct or not?
Yes? And that is .866 times 1,000, so it becomes
866 pounds.
Can you show again why it's 30 degrees? I
didn't see.
No, he's working here. You can't [inaudible].
This angle, total angle is 90 degree. Yes
or no? If this angle, I gave you 60 degree,
what's that angle?
30. Yeah.
30 degree. Then I have one projection over
the y-axis, the other projection over the
horizontal plane or called this length from
here to here FH. Which this ends up to be
866. And the vertical one ends up to --
1,000.
In other words, if I show you this plane,
I didn't mean to complicate this. If I show
you that line in a 2-D system, we are not
in the XY plane, we are in the Y at this plane.
Everybody understand? We are in this plane.
In this plane, this is what you have. This
is the y-axis, agreed?
Yes.
This line is not X. That line is going along
OC, yes or no? That's the OC axis somewhere
here or in that direction. Now, the force
was something like that. This was the question.
You know that's what you asked, that question.
This angle was 60 degree. [Inaudible] like
that, similar to that is [inaudible], I cannot
show it in this scale. Therefore, this was
F1, yes or no?
Yes.
Yes? But then when I project that here over
this axis, let's do a dash line because that
is -- let's call it horizontal line, that's
better. So this is the line horizontal there
going back OH, not that way. So that was the
force. This was point A, this was point B,
and this is point C on the h-axis. Yes or
no? Right. So I said, of course, this is such
a simple guess. In other words, what you are
doing, you are changing a 3-D problem into
a 2-D problem. Yes or no? Correct? So that's
one plane. So you calculate first FY, and
this line. That line ended up to be 866. So
everybody agree to that, yes? So this was
500, this is 866, because this vector, I call
it FH, which is neither FX or FZ. So what
do I need to come up with FX and FZ? I need
now this angle. Yes or no? Right. So if I
give you that angle equal to 40, be equal
to [inaudible]. Now, again, you probably want
to see it here, yes or no? Correct?
Yes.
One more time. For the last time, guys. Otherwise
we cannot go any further. This is 60 degree.
That is 30 degree. That is [inaudible] projection.
This is the method of projection, right? That
[inaudible] on the floor. This angle is how
much? Now I have a 2-D [inaudible] of the
X and Z. This angle is given equal to 40 degree.
The other angle would be of course, 50 degree.
Is that correct or not? In other words, if
I go now to the plane of X and Z, this is
the second plane. I hope that you can see
that. So if this is the plane of X and this
is the plane of Z, you are looking from top.
This is the top view. And this line was line
OH. We end up to point C, and that was 866
pounds. Yes or no? Correct? In order to find
FX and FZ, all I need to do, like what you
are doing in the last, [inaudible]. Is that
correct or not? You need another angle. And
that angle, the way I did it to you [inaudible].
I gave you this angle equal to 40 degree.
So this would be your FX, and this will be
your F?
Y.
No, not FY. That's not Y anymore. That is
Z. Is that correct or not? Yes? So therefore,
first you calculate FH, then finishing this
problem, you draw a line parallel to the Z,
you draw a line parallel to X. That's what
I did there. Then you find FX and FZ. So FX
becomes equal to -- this is the goal [inaudible].
FX equal to -- now, 866 times cosine of -- how
much cosine of?
[Inaudible].
860 times cosine of 40 degree. Very good.
Cosine of 40 degree. You can get your calculator
and check my answer. I'm sure it's correct,
but I want you to make a -- so that becomes
equal to 663 pounds. And of course, FZ becomes
866 cosine of 50 degree or sine of 40 degree,
which is the same. And that becomes equal
to 557. Just be able to find FX, FY, and FZ
of this vector. Yes, by doing [inaudible]
projection. Everybody understand? Now, here
is the answer, guys. I don't need that. So
that is the answer given to us. Now our vector
becomes equal to. Now look at this. F now
becomes equal to how much was FX we end up
with? It's 663i plus 500j plus 577k pounds.
We wrote our force in, a vector force. Yes
or no? Correct?
[Inaudible].
557? What? Yeah. 557. Yep. Correct. [Inaudible].
557. Yes? Now, do you want to check to see
whether this correct or not?
[Inaudible].
Now hold on, we haven't done the angle yet.
That's another question we're going to answer.
There are a lot of homework like that. Now,
first thing, you want to get your calculator.
The square of that, the square of that, the
square of that. Add it together, you should
get?
One.
Not one. 1,000.
[ Laughter ]
All right. That's 1,000, yes? Correct? Are
you getting it? Come on, guys. Don't be lazy.
In order not to go to sleep, some of you think
this is too easy, or I don't know [inaudible].
When you start working, then you are [inaudible]
your calculator. Maybe from now on, I'll put
some mistake here and let you check it. OK.
To get you all involved. What did you get?
Of course you don't get 1,000. You get 999.99.
That's correct. Because I said I'm going to
[inaudible]. Correct? Because of the round-up
error. Now, can you see this is the whole
thing? Now I can find my theta X, theta Y,
and theta. Actually, theta Y, it already was
given. Can I find theta X? Remember now, you
said this is theta X. It's so difficult to
measure this. Is that correct or not? This
is theta Z. Actually, you are right. We are
going to calculate now our theta X, theta
Y, and theta Z. Because many of your homework
is like that. Find the direction and [inaudible]
lambda X, lambda Y, and lambda Z. OK. To do
that, all I have to do, divide this by its
magnitude. Remember, lambda equal to what?
Lambda equal to -- what is lambda or any vector
equal to? Each vector -- this is the way we
said. Yeah. We said lambda, it was each vector
divider by its?
Magnitude.
Isn't that one of the equations we had much
earlier, yes or no? Or you can grab it from
here. Look. Cosine theta X equal to FX over
E. Cosine theta Y is FY over F. Or you can
use that technique which is the same way.
Lambda equal to 663 divided by 1,000. What
do you get? .663. Yes or no? Yes? So anyhow,
let's write it down [inaudible]. It's 663i
plus 500j plus 557k pounds divided by 1,000
pounds. Pound and pound. Therefore I don't
have to do with this much detail, but then
you can do it yourself. So it becomes .663i
plus .500j plus .557k. Do I need to use a
unit here? Of course not. Because what are
these? These are cosine theta X, cosine theta
Y, and cosine theta Z, which are lambda X,
lambda Y, and lambda Z. Yes or no? And hopefully
all of them would be less than one. If you
end up with the answer 1.25, you are in trouble.
Yes or no?
Yes.
Yes, the cosine cannot be larger than one,
yes? However, some of them could be positive,
some of them could be negative. If the number
comes to minus, it means the angle is larger
than 90 degrees. Remember that. So please
be careful here. So therefore, this technique,
we are going to use, obviously cosine theta
X must be equal to .663. No unit because pound
over pound, because cosine anyway. And theta
X becomes equal to 48 and a half degree. Do
I need to calculate theta Y?
No.
Of course not. Theta Y already [inaudible]
already equal to 60 degree. I use that part
of my system. But nevertheless, theta Y. Write
it down here. Equal to 60 degree, because
that was part of the data I used. Is that
correct or not? Yes? So [inaudible] cosine
theta Z is equal to .557. And theta Z becomes
equal to 56.1. Notice I'm going to three digits
in all of [inaudible]. OK. Yes? Now, there
are a couple of homework which I [inaudible].
Now, what I'm talking about today, guys, is
not about this quiz and this homework. These
are for next week. But I'm ahead of you in
order of you again to be able -- when I give
you a homework assignment on Thursday, which
is a bunch of this and bunch of new stuff,
then our lecture today, Thursday, you will
do your homework again. Tuesday, I'll make
some examples so you are polishing up all
of your questions hopefully. And some of you
saw how crowed my office was this morning.
I don't know. Yes. So there 15 people were
waiting. Usually Tuesday is for [inaudible].
If you can avoid Tuesdays, it's a little bit
better because they don't have a chance like
you guys to see me during the day in the office.
You know, that the Thursday office is after
their quiz. My classes for them is 7:45 in
the morning and 9:00. So you have that advantage
to see me both Tuesday and Thursday. So if
I give a little bit priority to them, don't
get angry with me, OK? It's because they are
more anxious than you guys are. OK. But that
is the routine we are going to work on. Now,
this was one method. Now you can go to the
several problems like that. There is a tower
here. There is a livewire going down there.
They will give you some angle. But the point
is this, you do it by projection. Either this
plane or that. It doesn't have to be in this
plane. You can project it in some other plane.
The format would be the same. My two projections,
you end up with all of the information you
find as far as the lambda, direction, vector,
[inaudible]. The second method, position vector,
is a little bit different that we go over
that and then hopefully we do some example
on Tuesday, at least principle [inaudible]
position vector, what a position vector is.
my presentation is a little bit different
from the book as far as the notation. But
all is the same [inaudible]. Again, just to
understand it, I'm going to do it into the
first and then expand it into 3-D. So this
is the position vector idea. So let's say
that we have here again, a 2-D problem rather
than a 3-D problem. And let's say there is
a point here given equal to -- this is X,
this is point A, has a coordinate of XA and
YA. Is that correct or not? Then this point
is connected to point B, which has a coordinate
of XB and YB, which is [inaudible]. So this
is YB and that is XB [inaudible]. In physics,
but for you it doesn't make any difference,
it you say line AD, in math it makes a difference
if I say AB or BA. Everybody understand that?
The direction of it is totally opposite, but
it's the same now. So the position vector,
from now on, if I want to show you a position
vector starting from A, ending at B, that
position vector would be like this. I call
[inaudible]. This is just by habit. RAB. This
[inaudible] show you where it start and where
it ends. RAB started at A, finishes at B.
And we call that a position vector. OK. And
that has a vector representation here. RAB
equal to what? Can you see that? What is this
length? What's X component of this length?
[Inaudible].
YB minus [inaudible]. XB minus XA. Obviously.
XB minus XA or the difference between the
[inaudible] of the two points multiplied by
[inaudible]. Because that is a longer [inaudible].
Yes. Correct? And then what is -- you see,
this is what I need, a calculator. This length,
which we already did, and this length. Yes.
Which is YB minus YA. So plus YB minus YA
times J. Correct? This is a position vector
in 2-D. What's the unit of that? Depend what
coordinate. If I'm using meter, meter, meter,
meter, it would be in meters. In other words,
if I find the magnitude of this vector, that's
the length AB. Is that correct? If this happened
to be five meter, that is four meter, this
is three meter. This length would be five
meter. Yes or no? So the unit would be in
the length. They are position vector from
point A to B. Obviously RAB is minus RBA.
Yes? Now, expand this for 3-D. If you going
that now, [inaudible]. All I have to do, add
the third coordinate to it. Yes or no? Which
would be what? I'll put it in red. Plus what?
Plus ZE minus ZA. Times what? K. The unit
will be in length. The length would be inch,
millimeter, and et cetera, et cetera. We have
five minutes to do one of the examples in
the handout again. This is [inaudible] more
advanced. Let's go to handout problem number
-- we are in TB now scenario. So the same
handout page, go forward. [Inaudible] page
ten finished, page nine finished, ten, eleven.
Go a little bit further. Now here we go. Notice
I did page 12, probably number six, question
number six.
[ Silence ]
So everybody got [inaudible] position vector.
This is the definition of position vector
and the length. However, how do we use it
comes next. And by the way, this procedure
that I'm going to show you, by the end of
this class, you are going to use it to two,
three, 400 times. So learn it now in the next
five minutes, or three minutes. Is that correct
or not? Let's see how simple that is. OK.
Now, it is really that simple. So here it
is. Look at the handout. I have to put it
on the board because some of you may not have
the handout. And then for the benefit of the
recording as well, I need to do that. Notice
what is given there. Unfortunately, there
is a cable going from point A to B, and there
is a force going here. And that force is along
AB. Remember that. The force is not length
of AB. Length of AB could be ten meters, five
meters, six meters. The force -- this is a
cable from A to B. There is a tension of F
or T in the cable. They show it by F, so I'm
going to show F. Is that correct or not? Yes?
Yes.
Now, what's the coordinate of A? Coordinate
of A, I'm going to put it here. A has a coordinate
of 200 at minus 100 millimeters. So that's
where A is located. In other words, X is positive,
Y is positive, but Z is, in other words, it's
in back of us. Is that correct or not? Location
[inaudible]. Yes? Z is negative. What's coordinate
of the B? In picture in your head, these [inaudible]
all [inaudible] short. X is 800. Y is 500.
And Z is minus?
[Inaudible].
Again, it's going backwards. So this point
is somewhere in the other room and going more
further back. Is that correct or not? But
2-D doesn't show it. This is all 3-D. Is that
correct or not? Yes? The question is, this
is the second method, what position? [Inaudible]
the force [inaudible] vector for what position?
Vector. OK. Now, this is again, [inaudible].
The question is find the force F in vector
[inaudible]. Now, it is not projection. This
is a position vector scenario. Yes or no?
Any solution, guys? Any idea about what I'm
trying to do? Each force equal to its magnitude
times its direction. Yes or no?
Yes.
OK. Is the magnitude of the forces given?
I forgot to put it there. The magnitude of
force is given equal to?
[Inaudible].
Write it down. The magnitude of force is given
140. 140 what, meters?
Meters.
Yes?
Yes.
OK. I'm purposely asking you to pay attention
to your notes, OK? Now, we know that fact.
I used that many times. Each vector equal
to its magnitude times lambda. Yes or no?
Therefore, vector F equal to its magnitude
times lambda. But I need lambda F. Yes or
no? But is lambda F equal to lambda AB? Yes
or no?
Yes.
Answer. Quickly, guys.
Yeah.
F is in the cable AB, yes or no?
Yes.
Yes. So is lambda the same?
Yes.
Come on. Faster. So it is the same. So if
I find lambda AB, that's the lambda force,
yes?
Yes.
OK. What's lambda AB? Now, in order to find
lambda AB, I'm going one or two steps backwards.
So first I write what? Position vector AB,
yes or no?
Yes.
Position vector is the length of B which is
the length of the cable in a space [inaudible]
in 3-D. Correct?
Yes.
So what's that one? I erased the formula.
You have it in your notes. It was X of B minus
X of A, which is 800 minus 200, therefore,
600i. Very good. You can write it immediately
if you want, or you can leave it at that [inaudible].
Is that correct or not? Then YB minus YA which
is 300. Very good. Then ZB minus Z. So [inaudible]
minus 300. Minus times minus become plus.
Plus 100 minus 200. Very good. Minus 200k.
What's the unit here? Be careful.
Millimeters.
What?
Millimeters.
Millimeters. Very good. That's the unit. Correct?
Now, so far so good. However, can I find lambda
AB?
Yes.
[Inaudible]. Look, it's right there in red
marker. Look, it is there on the board. Lambda
equal to each vector divided by its?
Magnitude.
Didn't I say it last week that we are going
to use this at least hundred, 200 times? Look
how many times I'm using it so far. Lambda
equal to vector divided by its magnitude.
What's the magnitude of RAB? Get your calculator.
Go a little bit faster. Magnitude of RAB is
600. I'll write it here, but I know the answer
here. I want you to work on it. Actually,
this is -- I have seen people doing that.
This is minus 200 to the power of two become
plus. I don't want you to put it in here minus.
That's a no, no. Everybody understand that?
Yes? OK. That's wrong. Minus 200 squared always
become positive, correct? So what is it, 700?
700.
Yeah. 700 what?
Millimeters.
Millimeters. That's right. The length is 700
millimeters. Look there. Look what happened.
After that, all I need is lambda AB, which
is supposed to be equal to lambda F because
they are all on the same line. Is that correct
or not? If I calculate the AB, that's like
lambda A. And lambda AB is lambda F. And that
is its vector divided by its magnitude. Since
we are running out of time, this 600 over?
700.
700. So we are writing it right. 6 over 7i,
300 over 700 plus 3 over 7j. You don't need
to put it in the decimal point, this is better.
And minus 2 over 7k. Do I need to use a unit
there? No, it is millimeter over [inaudible].
Is that correct or not? Yes, because these
are cosine theta X, cosine theta Y, cosine
theta Z. [Inaudible]. Now, what is the force
equal to? Force equal to 140 times lambda
AB. That's why many times this happen in the
book too. When it becomes a round number,
do not make it in decimal point, because you
[inaudible] 140 divided by seven is how much?
140 divided by seven [inaudible]. You don't
need calculator. It's equal to 20. 20 times
six is how much, is 120. We don't need calculator,
do we? OK. All right. 20i plus 60j minus 40k.
What, what's answer? [Inaudible]. Not [inaudible].
Did we break this into components? What method
did we use? Position method. So any time -- this
is the homework that I gave you. If the coordinate
of the -- if you have here a tower, one of
the television towers [inaudible] the cable
here, there is a cable here. If I give you
coordinate here, coordinate here, coordinate
here, coordinate here, then you use the position
vector. You find the forces. You can add,
you can subtract, which I'm going to do [inaudible]
next time. But [inaudible] the rest of the
problem exactly what we did in the past. Yes
or no?
Yes.
Except the equilibrium and that's also very
simple. So we will finish this subject and
we'll move on next class. But remember, Thursday,
your homework is [inaudible] about those [inaudible].
[ Music ]
