[ Music Playing ]
Not that tray. Usually in the first quiz Abbie
should be around 14, 15 I'll talk to Anthony
or about 70, 75% if was lower than that, much
lower, not much lower but it was 60% that's
good. But it does going to be curved at the
end your grade is going to be curved so we
continue in this system. Now couple of questions.
One, you may show this as a box or you may
show this, your free body diagram, as a particle
which is a point. But when you put it in the
box at least put your forces in the appropriate
places, everybody understand that? Where you
put the box don't put the force here because
this box is not being pulled away. Mistake
number one. Mistake number two, this end and
this TA tension in the cable are not perpendicular
to each other. Some of you made that assumption
and therefore you put here A and B in that
format. So that is not the case either, mistake.
Mistake number three which was common is this.
If the box is sitting on the slope or in this
case sitting on the surface, the forces are
equal and
Opposite.
Opposite, what was the idea behind the free
body diagram? You wrote it down. Read it.
To disconnect the item from all its
Connections
Connection including the touching the and
colloid surface. Some of you not drawing the
free, this is the free body that actually
means free of everything, is that correct?
This is not actually this end is the effect
of ground on their box, is that correct or
not? The box pushes toward the ground, the
ground pushes against the box to hold that
box in that position. So those were many common
mistake in your free body diagram. Basically
I did not subtract too many point but be careful.
That is the idea behind the free body diagram.
This is your homework. It is alphabetic so
you will pick up your homework and pass it
on. So, your homework average-wise was okay.
They look at it but we don't look at it for
the you know, accuracy of your whatever, your
homework. But we do it to see what you have
done and how much. Anyhow this is the free
body down the left side is the free body diagram
of box A. The right side is the free body
diagram of box B. Of course when you get here,
WA was given. WA was given in magnitude. You
can put again another thing. You can put your
forces in the vector form or you can put it
in the magnitude. You cannot mix it together,
they're all, they show, some of you show one
force that the vector does one. And if you
showing vector is all vector you have to have
a dash line on top of that. Do not make it
such a jumbo mumbo, just make it a little
draw your free body diagram a little bit more
accurate. You are bound not to make any mistake.
If you not draw your free body diagram correctly,
you may make a mistake. So therefore let me
based on that I should have put here put all
the angle there. Many people make again I
warn you about that. Many people make this
angle equal to 40 degree. They call the other
angle equal to 50 degree. Wrong. I ask you
to practice there. It was in your homework,
a couple of your homework, yes or no? You
did mistake there.
No, I put those [Inaudible].
You are the one who didn't draw the free body.
You did it in your head, I subtract 3 point
for you to just don't forget that, okay? Otherwise,
and some of you made a mathematical error
at the beginning you lost 1 or 2 point because
your procedure was correct, all your number
is, I mean wrong, but then your procedure
is correct. So you got credit. Anyhow, you
can see what I hear is very simple scenario.
WA was given that' what I1. WA was given equal
to? 100 pounds so immediately all you have
to do, draw this one first, that draw that
second, the biggest mistake is few people
did it this class the other class, I don't
remember. They add sigma FX of this one to
sigma FX of that one. That is a no-no. That
is a very bad mistake. Okay? By the way, the
way you can read my handwriting sometime I
write lots of thing very quickly so it is
not, no, that grade high but you can't get
the idea. Some of them, if you get one question
mark that is a mistake that everybody can
make. Two question mark means it is not a
good mistake. Three question mark forget about
it. It's not correct. You understand that
means you did a very bad mistake. So I don't
know how many question mark you guys got.
Nevertheless, this is the free body diagram
and this is all you have to write. You write
"sigma FX" beside on your exam, why, which
is the standard form here and there. I didn't
draw it there so here your X and Y. So you
don't need to change the axis because only
W is vertical. The other one or not. So therefore,
you write sigma FX this is the idea. Sigma
FX which is equal to. You look at that. This
one has X component, this one have X component
so you write T, oh by the way. TAB is a vector
going that way. TBA is going to vector that
way. However, the magnitude of those two are
the same. So you want to write here, this
is all is required. TAD magnitude equal to
TBA magnitude. Let's say for simplicity we
call it "Ted," a tension in the cable, T,
right? If you are showing originally as a
magnitude then you can call this one T and
they try call it T. Everybody understand that.
You cannot call this vector T and this is
vector T. They are not the same. They are
going in opposite direction. That I made it
clear. Of course these are minor mistake but
you should get used to doing these correctly.
Nevertheless, you looked at TA, T has it,
or is that our component and this has a horizontal
component. So this is all the math you need
to know. Some of you were a little bit, what,
it was difficult for you to solve this problem,
but I gave you at least 10 homework that you
should have done the same thing. So should
get used to the idea. So that's T cosign of
30 degree and this one is sign of 50 degree
or cosign of 40 degree, however you want to
write it. So that is magnitude of MA time
sign of 50 degree equal to zero. This is point
866, this is a number, you get it from your
calculator, so one, you find one in terms
of the other one less calculate in terms of
the other one. T become equal to point 766
which is this divided by cosign 30 degree
866. And times MA or equal to point 885 MA.
So the point is you will find one in terms
of the other one. Is that correct or not?
Yes? Then you go to sigma FY and sigma FY
you look at your picture again,so from the
first one you have TAD, or T I call it T,
so might as well use T. T sign of 30 degree
which is half 45. Then you get plus MA cosign
of 50 degree. Minus 100 pound equal to zero.
You already have T in terms of MA, you bring
that here. You simplify this, is that correct
or not? Yes? And my direct, do not use your
calculator. You need to show me some step
how to be fair to other people, you need to
show how you solved your equation. Couple
of you, again, use your calculator to solve
this for this point part of quiz part. I need
to see the detail. Anyhow, from this if you
substitute this into here, this equation become
all in terms of MA. And MA become equal to,
after you solve it, become 92 point 2 pound.
Obviously after you find MA and you get, you
can find T because I asked you not even to
find the tension, I asked you to find all
the other forces. [Inaudible] some of you
misunderstood that. It says find the tension
in the cord and the forces acting on each
block. That means contact forces. Okay, [Inaudible]
become equal to point 885. That which is 81
point 6 pound. So you put the answer in the
box. That's this much math. That much free
body diagram. Then we go to the second part
of course. Now we have the tension even W
is given the only two unknown is P and N which
is very simple. Actually these two are vertical
so you write sigma FX equal to minus here
it is, how much was it, tension, it's called
to 81 point 6 time cosign of 30 degree which
is that equal plus P equal to zero or magnitude
of P or P become equal to 70 point 6 pound.
Of course again I ask you to calculate NB
as well so you write sigma FY everybody see
that? Yes? And B become sigma Y which is simple
to do. It's 141 pound. You put that 2 in the
box and you are done. Alright your homework
is coming. Your, you see the result of this
quiz, next quiz should be easier because your
homework was easier and you should be able
to do it. And is there any question before
I start? Any question about your homework?
About the homework, the very first problem
that was assigned I think it was like 2 point
7, I was, is it on the plane?
Okay, let me, I don't know exactly, let me
take a look to see which problem we have here
so I verify it, oop, alright, now before 272
you're talking about? 272 force the 600 it
is coming toward us or it is like the one
I did problem B. It is all in positive direction.
Yes? But it's 450 the other force. The other
force is going, look at it, X negative and
Z positive, Y positive because on the second
port. It's like this, let me, that's for the
sake of everybody so if this is your X and
YZ diagram but this is your X. X going like
that. This is positive X, this is negative.
So if your force going like that here, so
if X become negative it is in that direction.
That's all.
It just looked,
Yeah, the picture is yeah, 3D is difficult
to show but that's where it is here. So it's
not that big. Now few of you did very badly,
I'm telling you, those of you who got less
than 10 it's not good. Those of you who got
less than 5, something drastically is wrong.
You have to move on. This static is the core
course. You'll need to know this as anybody
who have taken 219, 218, or Spain [phonetic]
or other significant quarters or mechanical
engineering courses, ask them how important
some of you were in my office today when you
saw the student coming and helping how important
static is and everybody should know material,
static material perfectly. However, a few
of you are not showing that sign that you
are yet there. I don't know what happened,
it was a mistake or it was something you didn't
pay attention. For example, many people put
MA equal to 100. I don't understand that.
That's a no-no. How can NA become 100? Because
you are used in physic that this is force,
this is a box sitting on the floor. Obviously
if this is the case, this is W and this is
the N, and these two are?
Equal.
Equal. How can they W become equal to N, I
have no clue that N going this way because
there are other forces on that. That's one
mistake. You have to get out of physic classes
at elementary classes come to the static level,
everybody understand that? Today you will
see we are going to talk about lots of math
and we are going to expand our math into something
that a little bit more sophisticated than
what you have seen in the past. Is that correct
or no? Another thing is this. You have [Inaudible]
in physic, at least 5 people in this class
or the other class they use this idea. This
is a box sitting on a slope. And nothing else.
If nothing else and there is no friction,
the box is going to slide down, yes or no?
If there's friction the box may stay there,
is that correct or not? In that scenario you
have seen that. In that scenario if there
is a body with no other attachment, no cable,
nothing, and then there is a weight here.
Of course the weight going down there. Then
you must have an end and you must have an
N. Then N become W cosign of that angle and
F become something similar to that. Everybody
then, you cannot use like that idea and several
people used that idea for their equilibrium.
You have to get to the equilibrium idea. I
don't care you have 3 courses or 4 courses
or for simplicity I only gave you three forces
here and four forces there. But the idea is
some of the forces must be equal to zero,
then sigma FX equal to zero and sigma FY equal
to. And that is a principle and that's the
one that you have to use. Okay? At least this
average was little bit low, so I'm suggesting
this. You should need tutoring, okay? So here
is that your tutoring schedule if you want
to. First of all go to Room 9167 or which
is this building, or 169 which is the first
floor in front of the elevator. Monday 1,
2, 3 pm, Tuesday 3, 5 pm, Wednesday 3 to 5
pm, Thursday 3 to 5 pm, Friday it's 12 to
2 pm. This is, this office hours is one of
my oldest student who has been working with
me for 2, 3 years. He has taken all my classes
and he is knows all the little tricks and
everything that I talk about in the static
and the strength. So he can answer all your
question, he is very good. So that's his tutoring
hours. There are other people who also have
beside this time in the same room. They are
tutor. You should check with them, check their
schedule, go there, talk to them, and if you
need any help if I'm not available and you
need more help, you need to go there. Some
of you really need to, quickly, remedy the
mistake that you make in the first quiz. Is
that understood, yes? Just to help you out.
If you need them. Some of you may not need
it because you are doing okay and you don't
need these or you come to my office and I
think you feel that's sufficient that's so
that day. But if you extra help but the idea
is that you go there don't ask them to do
the homework for you. If somebody solve the
homework for you, you're ruining that problem
for yourself. The idea is that for you to
sit down like this problem, to find out what
principle is being asked, what concept is
being used, and when you do it your homework,
you practice in your homework, when it come
to the quiz you quickly will be able to be
able to finish it. Other words, you are going
to struggle with it. Is that correct? But
not some mistake for mathematical error which
is okay, but there were many principle mistake
too, as far as the [Inaudible] et cetera,
et cetera, et cetera. Is there any question?
No? Okay, let's move on then [laughter]. No,
I'm just, I'm saving time. I was hoping, but
I can answer the question one or two, but
the point is we have so much work to do in
so little time we have to finish all the material.
I was just going to ask for you to close the
handout 2 and 5 because I think I did the
wrong handout. If you could just close the,
The handout problem that,
2 and 5.
To put it there?
Yeah, the homework.
How can, okay.
Because you said you would [Inaudible] if
we ask.
This time I did not ask you any, for the homework
number one?
Yeah, because I did I think the wrong problem.
Oh, I see. You want the answer you made there?
Yes, I want the answer.
Right now? [Laughter] Alright, so I put the
answer there. If there are some question that
the answer in the back is not in the back
of the book, you can ask me, I can put it
on the board [Inaudible] just to help everybody
else. Now what we want to do today is we want
to go in the idea of the moment. And moment
vector. But before we do that, we need to
understand the, let me backtrack a little.
Let's take a quick look because one of you
are homework is so let's look at the previous
handout there is one more problem left. I
want to discuss that one as well. There's
one more problem I believe is problem number
7 in your previous handout, which is a 3D
problem, which I am going to put it on the
board for you guys because there is a little
discussion involved there. Few people ask
me question about this type of problem, this
is the question that is being asked there.
So here it is. X this is XYZ then it is, I
try to do my best by drawing this picture
like this. So you don't see that, you don't
see there, okay, so alright. There is an object
sitting at the middle of this, this is problem
I don't, it's problem number 70 your handout,
yes or no? There is an object sitting at point
A which at the middle of this sort of slope
inclined surface. This length, this is not
in scale, it's 10 feet. This length is 8 feet.
And this height is 2 feet. And then since
this is no friction there is no friction,
the object is on the inclined surface then
we have attach that to one cable here and
one cable here. The point B is in on the Y
axis, the point C is somewhere there, apparently
on the XY axis. C has the coordinate of 8
4 0 feet Z is part of zero there [Inaudible]
must be XY, XY plane. And this one is on the
Y axis which has coordinate of 6. A zero 6
zero means the height is from there from the
center of coordinate is 6 feet, these are
again feet, and the question being asked find
the tension in the cable at forces acting
at A for equilib. It has a little similarity
with one of your homework. In part of your
homework you're supposed to calculate the
tension in an object which is sliding or is
standing or on this inclined surface. There
you are going to do only one tension. Later
on I don't know whether I asked you to do
the equilibrium as well or not, I don't know
whether I asked that question or not. You
know [Inaudible] Is that correct or not? Yes?
Alright. Now, how do we solve this problem?
First, if I draw the free body diagram of
A, what force is acting on that one? Couple
of them you know. There is a tension here
and a tension there. Is that correct or not?
Yes? We discuss that in the past. Therefore,
graphically direction it's given there but
I'm going to put here a tension, TAC, this
is the discussion about how to do this problem.
Then there is a tension TAB and those direction
is given which we have done in the past many
time. You did it in your homework, correct?
What else is that? Of course it's the weight.
The weight of the object whatever it is. Which
goes downward, yes? So that is the weight
which I'm going to draw it black because it's
given. Yes? All in vector form. What else?
Normal force.
Normal force, that's right. Now the normal
force if I draw it there which somewhere I
can do it this way like that. Normal force,
I'm showing it as a vector as well, that correct
or not? Normal to what? [Multiple speakers]
Normal to that inclined surface that's exactly
the problem. Now how many are known can I
have? Is this a 2D problem or 3D problem?
Of course it is a 3D problem. How many equation
do we have? Three. How many unknown can I
have?
Three.
Three. How many unknown do you see there?
Of course I should draw in in red as well,
I'm sorry, let's change that to red because
that's also unknown. Yes? Correct? Knowing
that N is perpendicular to the incline surface,
it must be according to the rule [Inaudible].
Now, tension in cable 8C I can write this
one as we had done in the past. TAC this is,
I'm just writing explaining the problem. I'm
not doing the problem because that's your
homework. Is that correct or not? TAC you
write it in the magnitude of 10 C have landa
of Ac which we use position record divided
by is magnitude. Yes or no? Because we have
done it, because the direction is given, therefore
that is one unknown, yes or no? The second
unknown would be, this is what? The second
unknown these are the vector format, this
is TAB which is the magnitude of TAB which
is the question being raised time landa of
AB again using position [Background Sound]
idea. So that's two unknown. Now what do you
know about N? In general, N is NX, NY, and
NZ should it should be three unknown and two
unknown become five unknown. I cannot solve
it in that way. If I don't know anything about
the N, N is the vector in a space, any vector
in a space has straight component. XY and
Z I have to have that. So that is not solvable.
Is that correct or not? I asked you to solve
it. So N must be only one unknown. How can
I write it exactly like that? Is that correct
or not? Yes? So let's see whether we can find
it through the geometry of this picture so
I have to write, if I write N equal to a magnitude
of N, magnitude of N is a number, times what?
Times?
Landa.
Landa N. Your eyebrow went together again,
see, whenever you do that that I can see your
faces guys, see I always checking on you to
see whether you are following me or going
there, she's hum, what is he saying again?
Yes, landa can be written any force, doesn't
matter, any force. This is what I said here.
Any force when it written in this format and
you already know landa it's one unknown. When
you write it in the TX TY TZ that is three
unknown. Yes or no? That's what I said. I
already have one unknown, I have the second
unknown, I cannot have more than, one unknown.
I decided one unknown to be magnitude of N,
then therefore I should somehow from geometry
I should find the landa of A, yes or no? Okay,
what's landa of A? You said it's perpendicular
to that incline surface didn't you say that?
Yes? Let's explore that. So each landa, this
is the whole point, see we are going back
each landa doesn't matter with the land N
or landa other, equal to cosign data S cosign.
Tater Y and cosign of tater Z. Is that correct
or not? The cosign data is the direction cosign
where it defining the landa of each direction,
yes or no? Therefore, let's write it like
that. cosign tater X I of that line, cosign
tate YJ plus cosign of tater ZK, this being
landa X landa Y and landas. Yes? If we know
tater X, tater 1, tater Z, then their problem
are solved, is that correct or not? Then I
only have three unknown. This and then I write
sigma of X, sigma of Y, sigma of Z then my
problem's resolved. Is that correct or no?
Now to do that you have to look at the picture
like that. So look at the picture of Y, this
is Z and Y. Then you see this. There. This
is 2 feet and this is 8 feet. See in the picture
you have it there. Look at that surface if
this shadowed on the side, is that correct
or not? Yes? Is that what you see in the plane
of YZ? Yes or no?
Yes.
Correct? So what is this angle? Can you calculate
that angle? That angle a tangent of that angle
is 2 over 8 so you can get that angle. I don't
know how much it is that, but that is we find
let's see whether I have it here. You just
calculate, somebody calculate and give it
that angle. How many degree?
14.
40?
22.
14. Exact? Because we go three digit, remember?
Yes? Some of you are not following the three
digit rule either [Inaudible] some of you
writing with pen [Inaudible] point some of
you would be call me two digit. Three digit
is minimal beneath, is that correct or not?
14 point 02 is 14. Is that correct or no?
[Laughter] Use three digit. Okay, so if that's
the case this is the direction of the N, yes
or no?
Yes.
Yes? Okay, what's this angle? What's that
angle? If this is 14, this is 90 degree. What
is that angle? That is 76. Yes? What is that
angle? Positive direction of N and positive
direction of the Z. Is that tater Z? What
was tater Z? You wrote it in your book. You
have to understand what you, wasn't tater
X, tater Y, tater Z the angle with positive
direction of a force and positive direction
on XYZ axis? Yes or no? If this is direction
of the N that you have chosen, N is perpendicular
to this line, yes or no? Correct? Therefore
this is positive N and that's positive Z,
this angle must be taterZ. Yes or no? This
angle which is 14 degree must be positive
Y and positive that angle is tater ?
Y.
Y. So you have only tater Y equal to 14 degree,
tater Z equal to?
6.
What is tater X?
Zero?
Not zero, 9 [Inaudible] cosign of 90 degree
is zero. Is that correct or not? Yes? The
angle is 90 degree. This line that's perpendicular
to here when you draw it here, has no direction
with the, let's see, that go that way, so
you come with the per, actually parallel to
this plane so I believe that that's what you
want. Is that correct or not? Therefore, just
making a simple just become equal to, so this
become we write it here so that become equal
to magnitude of N time zero I plus cosign
of you have it there, cosign of tater Y was
14 degree, J plus cosign of 76 degree K and
so you have three unknown. One, two, and three
when you write your three equation you can
solve for that. This was just in case you
get a problem like this so in other words
the N can be defined by geometry. Is that
correct? Now before we go to the moment, a
moment vector, we need the dot product and
cross product. Yes?
Can you review real quick how you found the
data [Inaudible] degrees?
Tater X is look where that is. Tater, this
is going perpendicular [Inaudible] tater you
have to find the direction of the normal to
this surface, let's see then. It goes, now,
it goes like that. So the line will be like
this, perpendicular. It goes that way. It
doesn't show this way, I have to go like that,
perpendicular. It go like that.
Okay, I see now.
It goes like that then it become perpendicular.
I showed its, that is the wrong, I'm not going
to do anything else, but it goes like that.
So it goes to that line X and it pick up 90
degree. Anyhow, now before I go any further,
we have to just briefly,
Don't have to erase, wait [laughter].
Alright. Can I erase this? Yes?
Yes.
Okay. I need the board. There are couple of
homeworks, it's couple of homework that you
have to do same night, just wanted to give
you some idea of how to work it out, then
detail of it you have to get up yourself,
is that correct? Yes? Anyhow, now before we
go to the moment, we need this idea. First
of all let's go quickly because you are supposed
to know and you know everything but I want
to review it real quick. Let's say A is a
vector equal to AXI plus AYJ plus AZK we need
that and we need 3 vector equal to BXI plus
BYJ plus BZK, vector usually have the unit
where we don't care about here because we
are talking about math part of it. By definition,
we have two product. Dot product and cross
product. Dot product or we called it the sailor
project. Yes or no? Because the result is
[Inaudible] not a vector, you already know
that so. A dot product of B equal to what?
By definition. You know that. Not from here
now. Equal to what? Magnitude of first vector
time magnitude of second vector time what?
Time sign of a now it's very important to
understand what is sign of a angle, tater,
which if this is A and this V, this is the
specifics directions start from A goes toward
B. It must be that order, yes? Everybody agree
that? Do you remember that from your physic
classes or math classes? More or less? This
is, we have to know that. At the result it's
not a vector. It's magnitude magnu side tater
[Inaudible] one distinction. When I have a
dot product and there I want to do E dot product
of A, then I have AB sign of minus tater,
yes or no? Yes? A, B, sign of because if this
is tater now I'm going that way. Is that correct?
Everybody knows that when the sign goes counter
clockwise it is plus or it go counter clockwise,
if it goes clockwise it's negative, if it
goes counter clockwise it is positive. Then
multiply by sign of minus tater but however
sign of minus tater and sign of tater, let's
put it this way it's better, sign of are the
same. Yes or no? Therefore A dot product equal
to B, I mean A dot product B equal to B dot
product of A. You remember all of that, yes?
Because that's not true for the cross product.
So A dot product of B equal to B dot product
of A because these two are the same. Agree?
Not shaking your head because you'll use then
your calculator. The way I see it is this
they're all a circle, this is plus tater,
this is minus tater, they're all the same
angle. Notice both of them having the same
sign, yes is that correct or not? Yes?
Yes.
However they have a different cosign everybody
see what I'm talking about. The star cosign
one is bigger.
It's the other way around.
Other way around. cosign is the same side,
okay, so wait a minute. This is sign and this
is the cosign. Therefore this should be
cosign.
cosign. Right. Okay, now we got it right.
Okay. We find that our mistake. Correct? Alright.
Right? Okay. Therefore, since cosign is the
same therefore these two are the same, therefore
this is correct. I made it be, yes? Now, the
second definition, so we going through that
one. Now we go through the expansion of that.
Now we want to expand A dot product of B equal
to AXI plus AYJ plus AZK dot product of BXI
plus BYJ plus EZK. Notice if I dot product
these two vector together, I have three terms
here, three term there, so I should get nine
temp if I do it in expanded form. Yes or no?
However as all of you know that's not true
when I do that the first term is AX VX I dot
product of I. Is that correct or not? Yes?
What's I dot product of I?
1.
1. Because cosign of zero is
1.
1, is that correct? Actually this is the case
sitting here. cosign of zero S1. Is that correct
or not? Therefore this become 1. The second
there is AX BY, yes or no? So you have AX
BY but this time I have I dot product of J,
obviously these two are perpendicular to each
other, cosign of 90 degree is?
Zero.
Zero. Yes or no? Correct? So all the term
IJJKKI all those term become equal to zero.
The only term that remaining become one is
IIJJ and KK. So let's put a few more dot up
here representing the rest of it, but the
answer is this. AS DX plus AY BY plus AZ BZ
now, I have two definition here. One is returning
this form. So please put this in the box,
box 1. And also put this in box 2. You can
use it in this format or you can use it on
the other format, both of them together. However,
if it's just the two of them are the same
I can find if somebody gives me a vector A
and somebody give me a vector B and I don't
have this tater can I find cosign tater with
these two operation? Yes? This is a rule that
it is given in one of your homework which
will come this week. Is that correct or not?
There is a cable going this direction, there's
a cable going that direction, you need the
angle between the two cable. You have to find
these two cable and dot product here in this
form. Put a question mark equal to equation
two, is that correct or not? A question mark
which is this one. So cosign tater become
equal cosign tater become equal to A dot product,
B which can write this [Inaudible] divided
by magnitude of A and magnitude of?
B.
B which is scalar numbers so you find a tater,
is that correct? This is something we are
going to use a few time here and there to
calculate the angle between two lines. Okay?
As a matter of fact, you 
go back again to that three that I found the
sum of the two forces if there was a cable
going this way, a cable going that way, now
we can find, did I do that for you yet or
not?
No.
Not yet. Okay, we did couple of problem which
have cable going in different direction. Yes,
sir, you can do that by using this tape. There
is another application that I am going to
leave it later on when we get to the moment
about axle then I talk about the second application.
This is one of the application of dot product.
The second application come later on. Now
this was a dot product, now we want to do
the same thing for cross product. So this
time I was ahead of myself because I always
feel with moment so that's what we do in the
moment, we use a cross product, dot product,
always called sign not cosign. [Inaudible]
sign there. Okay, now this time the vector
is A. This vector is A and B this time we
want to find A cross product B. So we show
it in this format. By definition you have
heard this before I'm sure know the answer.
The answer should be a vector. Yes or no?
Now what's the result, I mean what's the definition
again? It is A magnitude of A magnitude of
B this time of course sign tater, yes or no,
correct? Right? Is that it? Is this complete?
Is it a vector the way I wrote it?
No.
So where is the vector? This is a magnitude,
yes? So something missing here. You said it
correctly but I didn't put it here correctly
not yet. Everybody understand it. So here
is actually magnitude of that vector but you
know A cross product B should be a vector
yes or no? Let's call that vector [Inaudible]
C is that correct or not? Yes? So the results
should be a vector here what I have is another
number, 20 time 10 time point 5 or point 6.
Is that correct or not? Therefore I need here
landa that's exactly what I gave you the other
problem. Landa in the direction of N. So you
have to put landa there which is a unit vector
in the direction of plane of A and B. So please
write it down. This is the vector C is perpendicular
to the plane of A and B. So this is vector
C. This is the cross product result which
is very important for today's lecture, the
rest of the lecture. So if I take a vector
A cross product with the B the angle between
the two from A to B is tater, I should get
vector C, but vector C is perpendicular to
the plane of A at B. Is this complete or not
yet complete? See I draw it that way. Why
not going this way? Where I say normal I just
say going up or going down. Is that correct
or not? To every surface there is a for for
going up and down. It is not yet complete.
Write it down in your note. It is vector C
is perpendicular to the plane of A and B and
follows the right hand, now you have the direction.
See, if you miss any of this part in math
classes, you don't have it complete. And that
we cannot do here because here we have to
be precise. Is that correct or not? Yes? So
therefore, one more time, now what happened
here you put your hand in the direction of
the A such a way to turn into the B so A and
B are in this plane, actually C coming again
we cannot show it exact, C is coming in our
direction. Is that correct or not? Or if A
is here, then there as you see it here for
example. Here we go. If this is A, this is
BAB, look it coming this direction. But BA
is going upward it sign of tater is as you
just saw it, I make a mistake, the sign of
tater and sign of minus tater are opposite,
yes or not? Therefore with that definition
A, write it down, this is the definition mark,
so what that right hand rule, add that, this
in the direction of the right hand rule. We
tell that is complete. And therefore now it's
become a vector A cross product [Inaudible]
become a vector C that's the vector C, this
is the magnitude, this is the direction, and
it's going that way. Is that correct or no?
Yes? Now, remember that. The other one A cross
product B is not equal to B cross product
A as I just show you there. Is that, one more
time, A cross product B going this way B cross
product A you going that way. Is that correct
or not? Therefore it is minus. And then we
go back to that expansion. I don't have to
do that again but we do it the same way. This
time see it's easy for me to erase it if you
have to rewrite it so I give you a little
time. This is you want to rewrite this but
you want to change it from dot product to
cross product. So you change that, you rewrite
it please in your note, so this time we have
cross product and cross product, yes or no?
Again, we should get nine term in journal
but do we get nine term or this time three
term? [Inaudible] everything we get this time.
Three.
Three? Not at all. Actually is a little bit
more than, actually we get six. Let's see,
find out. Here it is. The first thing it is
AX DX I cross product of I. What's I cross
product of I? Sign of zero is zero. Actually
the term that we had before now it become
equal to? Zero. Then the second term we get
plus AX BY what? I cross product of J. What's
I cross product of J? 1? No. First of all,
cross product of two vector has a magnitude
of that is 1, yes, correct. But the cross
product of two vector should be another vector,
yes or no? Come on guys, I should teach you
math or we're in a static class. So where
are we now? K. Thank you. So I cross product
J become equal to K. Why J cross product of
I become minus K. And it goes [Inaudible]
on you have to work it out, depends whether
they are positive or they are negative, so
I'm not going to do that. So you get actually
3, 6, 3 term become zero. IIJJKK this time
RR zero. But the other six term remaining.
What's the best way to present that one? There
is a several way of doing it. One is in the
book so I'm going to put it here. You put
a circle there, you put IJK here, if you are
going this way it's all positive. If you are
going this way it's all become negative. In
other words, I cross product J is K. K cross
product of I is J. However, J cross product
of I is minus K. K product of J is minus.
Do we need that? No. What do we need what
we are going to use is this guys. We put all
the term here. All of them become a determinant.
Remember that? Now you smiling because you
have remember some of these. Something come
to your mind. It become a determinant but
in that order. The first rule is what? The
first rule is IJK and of course everybody
knows how to expand a determinant. Yes or
no? Correct? And then the second rule is AX
AY in that order. The first vector come in
the second row. AZ and finally the last one
become equal to BX BY and BZ. Now you can
expand it look what happen. I become AY BZ
minus AZ BY remember that technique, yes or
no? Everybody should remember that. I asked
you to study this over the weekend. I don't
know whether you did it or not, however we
are going to make enough practice of that
by the end of this class. You are going to
use this determinant another two at a time,
so you are going to learn in, whether you
want it or not. Is that correct or not? Yes?
Yes.
But did you see the difference between the
two. What is the scalar the other one is and
the direction of you need all of that now
to go to that next subject which is the actual
subject. This was all review again. So what
I spend so much time to remind you especially
from this cross product which I'm going to
keep it there for time being. And I'm going
to erase that. Remember this determinant always
should be in this format. This is A cross
product of B result. Is that correct or no?
Yes? We all know about that. Now let's get
to the we are going to discuss about a new
idea which you have heard it the name of it
but you don't know exactly what it is moment
and moment vector. [Writing on board] Two
idea which is you are going to deal all the
time with that. Okay, first of all, let's
find out what the definition of a moment is.
Let us say that we have here a force F applied
at point A. A is the point of application
of the force. You have a structure and then
there is a force there, is that correct or
no? Then you want to take the moment of that
force about point B and let's say these are
on the horizontal no, I've tried to find the
moment of force, F about point B. That's the
idea. Not point A. By definition, moment of
force P about point A, about point B I'm sorry,
its magnitude is equal to the magnitude of
force. The more force I have the more moment
I have multiplied a distance D. This is we
call this metaforce distance meter [writing
on board] metaforce distance meter. Now where
is the D? Is this D, DA is D? No. Be careful.
That's not the D, right? The definition of
the D, the definition of D is the shortest
distance, this is very important to write
it, put it in a box. Otherwards you make a
mistake. D is the shortest distance from point
B where you are taking moment about. It's
from point B or where you are taking moment
about, to the line of application of the force.
In other word, not BA. BC. Now BC is perpendicular
to the direction of the force. Is that understood?
Yes? So what's the distant this is? D. Now
if I give you number for example, if I give
you this equal to magnitude of that hundred
and this distance equal to 5 inches, what
is the moment? The moment is 5, I say hundred
pound, hundred pound time 5 inch so it is
500, what? You have the unit there. Pound
inch. That's right. Very good. Sitting there,
force distance so the unit of the moment but
is like unit S pound inch. I can't use it.
This is the length, this is the force. Depend
what system you have. Newton meter. Newton
millimeter. Is that correct or not? So the
moment is about the point depend on two parameter.
The magnitude of the force and distant. You
have noticed this in real life. You want to
raise this table, you cannot do it, you're
afraid from the back, you take a rod, you
make it this rod very long, you put here a
force that goes up. Is that correct or not?
Yes? So you have more force or more distance.
What are, if you get very short wood you have
to put lots of forces there. Everybody understand
that? So the moment depends on the amount
of force and amount of D. The more D you have
the more moment you have. Is that correct?
That's why it's in force distance meter pound
inch, what happen here, what is this moment?
What's the application of this moment? If
this is what you have to see there, look here
if this is a rod you put your force here perpendicular
to this. First of all look this is not D.
This is D. Is that correct or not? This is
an example. If I have a rod here, I am showing
you right now. This rod. Is that correct or
not? And I'm holding it here and I pushing
it this, this is perpendicular. This is hundred
pound, this is I said 5 inch. How much moment
do we have here? 500 pound inch moment, but
look what could happen. This is what I want.
Put it there. Look what happen then. You see
there the result? Rotation about this point.
That's why you say moment about this point.
So we are creating a rotation about this point,
this is your magnitude. The magnitude of that,
so it actually the moment is at rotation,
is that rotation about the point M. And if
this point or whatever is attached to it,
it's going counter clockwise. We are going
to call it positive, if it like angle. If
it goes clockwise we are going to call it
negative. Is that correct or not? So that's
the sign. So this is the unit. Now the sign.
The sign of the moment for 2D problem, this
is a 2D problem, so first of all this is 2D
so we get to the 3D afterward. If the moment
going this way you are going to call it positive.
If the rotation of the item or tendency to
rotate the item in that direction clockwise,
we are going to call it negative. We have
the sign. We have the unit. We have the force
distance but what did we talk about cross
product and not product, nothing no cross
product there, dot product there. Yes or no?
Correct? But now let's take a look more carefully
to see what we have. This is the direction
of the BA is that correct or not? [Inaudible]
you knew that we go like here, this is your
tater. Is that correct or not? Tater would
be this [Inaudible] VA and F. Is that correct
or not? Yep. So I said, this is very interesting,
I said M equal to F times D. Correct? This
is D. This is D, correct? Let's call this
one from D to A equal to F. This is a position
vector started at B goes to A which has nothing
to do with a D, everybody understand that?
Are they some geometrical correlation between
two, is that correct, through that angle.
But if this is tater, this must be also tater,
yes or no? Okay, what is this sign tater?
Sign tater equal to D over L. Yes or no? Correct?
So write it down. Sign tater [writing on board]
equal to D over L, over D must be equal to
L times sign tater. Correct? So this is now
the newer self guys [phonetic] now if I write
M equal to F times D now F is F. But what
is D? D equal to L time sign tater. Eachs
of D I'm using L. Everybody understand it?
So if I'm using multiplied by L multiplied
by sign tater is that something you see on
the board? That's the magnitude of cross product
of A and B. This is your A this is your B
because this is your tater. Is that correct
or not? So therefore, M equal to in magnitude
of A, equal to magnitude of RAB cross product
of F. Did you see how it end up there? Because
that the same. Exactly look. Cross product
of two vector, AB, is AB sign tater. Yes or
no? Forget about the direction yet. Is that,
that's I'm talking about the magnitude of
that. Correct? So magnitude of this vector
M which we were talking about point B so let's
write that point B is you want, but I'm just
giving the definition so leave it at that.
Equal to RAB cross product. So what's RAB?
Look, by definition, go by the definition
in there by the [Inaudible] is the magnitude
of RAB. What's the magnitude of R, RBA, I'm
sorry. So we went. Let's put it this way.
R it started at B minus A. Started at B, finish
at A. Not the other direction. That's very
important. Actually REA cross product of F.
Because if I put AB that's wrong. I am starting
at B going I went this way and that way to
get that tater, is that correct or not? Terrible.
Look what happened. What is the length of
R because magnitude of that vector, magnitude
of that vector equal to L, correct. And then
magnitude of F of course is F. The sign making
these two are this. Is that correct? Or a
sign of tater. As you see we got the same
result. So the moment not even be founded
through this format, be founded through this
one. This format is cross product method.
So we can use either cross product or we can
use it in force distant. However, in force
distant we have to use our right hand rule
to get to the direction of the force, remember?
Because R cross product of F is another vector
perpendicular to the R and F which is coming
toward us, this direction. [Inaudible] in
2D so we get to that in a minute. So this
is its magnitude and if I want to write it
in the vector all I have to do, I have to
do it like that. That's it. That is a vector
representation. That vector is perpendicular
to this and that as you see is coming toward
us. That's the direction of the moment. Notice
the rotation is here like this, or like that.
Everybody understand? This side is positive
but because these two are in the plane of
X and what? This is in the X. This I don't
care. Perpendicular to this plane is K. Is
that correct? Is a 2D problem. Now you are
working 3D problem look what happen. R is
here, F is there. You have to go through that
plane then in space. Is that correct? We get
to that in minute. We have, remember that,
if I have two vector I have to go through
perpendicular to that [Inaudible]. Look at
what I draw here, AB and C. Now one of them
become position vector from B to A. The other
one become force. The third vector will be
perpendicular to these two. As I do the example
you get a better idea of there. Let me explain
that in a minute before we go to treaty because
there is a lot to learn here. It has become
suddenly too complicated to realize what's
going on. Yes or no? Let's go by item by item.
I want you to understand that what you have
to do first. First of all let's analyze that
before we go through the example. Notice if
this is the rod and this is perpendicular
to that as I said earlier, and this is D and
this is F magnitude, this about this point
A or B or O let's put it in general, create
a moment, which a positive moment or negative
moment?
Positive.
Positive. And this one create a
Negative.
Negative moment. Because it turn it that way.
However, I want you to understand this as
well. Now if the force here, now some student
make a mistake. They won't they want to calculate
the moment they go by the sign of the force.
That's a no-no. This force,
Negative.
Is creating negative, this is a positive force
statically. But about point O creating a positive
moment or negative moment?
Negative.
Negative moment. You have to use your hand
or right hand rule to determine the sign.
So what you do you always write the force
distant then with your right hand you should
leave ashwood the first week or so you should
use your hand until you get good at it then
immediately everybody like some of you you
already seeing it, you will see that quickly
whether it is plus or minus. And we have to
do that in this class in the next class continuously
in any structure analysis, structure of beam
column you have to use the moment. To do the
moment you have to have the sign. Without
sign everything is wrong. Everybody understand?
You become good at it, don't worry about it.
But let's go step by step. However, I want
to give you one theory before I do couple
of example and I don't want you to lose that.
Now let's do this. This is a theory in the
book which shows you how. Because we realize
now we have two methods. The summary is this.
I don't want this anymore because I already
explained it as to you. So one method is this
method, yes or no? Force and distant, as I
was explaining that. The second method is
this, one more time. M about a point you just
put point 4A whatever it is R, R comes always
first. R which is start from where you are
taking moment to the point of application
for time being. I may change that in future.
So it is R in this case is the example, BA
cross product on F which gives you both magnitude
and direction, you don't have to worry about
it. Is that, that we call it cross product,
this we call it force distant method. Write
it down. This is basically for 2D problem.
This is basically is for 3D problem because
it would do anything to call it 3D however
later on maybe we'll become good at it, many
of the 3D problem I'm going to turn it into
a 2D problem which people and student usually
appreciate that. But we have to work on it
for a week. Is that correct or not? So basically
for time being, this for 2D this is for 3D
and these both gets you in this one you have
to find the direction yourself. You have to
see which way it will take as you answer meaning
the other problem. Is that correct or no?
Now, a little tilodi is this, let's say I
have here a point because I put the same A,
I put A here. Let's say I have here force
F1 and I want to take the moment of that they
start changing to point O places of B is that
correct or not? What's the moment of that
force about point O in general here? MO equal
to what? MO equal to R, is that RO
OA
A, cross product of it's sitting there RD
become O, this become A I just changed it,
so become ROA cross product of F1, correct?
Let's say at the same point N, not other point,
I have another force F2. What would be the
moment of that about point O? The position
where product didn't change, the position
make sure it still is ROA because these are
at the same point. Plus please write it down
R-O-A cross product of F2. Let's say that
I have another force F3, understand what I'm
saying? I can have more than one force at
point A is that correct or not? Yes? Plus
ROA cross product of F3. This is ROA is coming
in all three of the vector, or three or four
or five, whatever I have. I can write ROA
cross product of F1 plus F2 plus F3, yes or
no? What is F1 plus F2 plus F3 is resultant
force at that point. So what I'm saying that
if you have a force here R this is ROA cross
product of R. You can do it individually or
you can add it together as long as all the
forces are at the same point, is that correct
[Inaudible]. Or wise versa, if I have a force
here and I don't have the right distance,
I can break it now into component and take
your moment. Everybody see where I'm getting
at? Yes? Look at your first question in the
handout. I hope that you put that handout
there as well that everybody can see that.
Is the handout there? The first problem on
the handout as we want to solve this and this
is what you see there. So this is problem
is easy. I'm going to show it for you guys.
So this shows you all the ideas behind what
we just said a minute ago. I'm going to use
every method to get to the same. We have here
a friend here. This friend is this is point
O. There is a force here, this force is 600
newton and this direction is 40 degree. This
length is 2 meter. This is example one in
the handout that I gave you page one, the
top of the page. Actually the solution is
there too. This is the first time, so I'm
showing you the solution to get you understand
the idea of the moment and moment [Inaudible].
So and there the height here this is point
this point this point A, let's say this point
B, that's irrelevant, we change that all the
time. This height is 4 meter, this length
is 2 meter, this is 600 and I want to find
see this is a frame of a structure. You have
a 600 newton node there, obviously this frame
is going to rotate about point A either clockwise
or that's what the moment is. Is that correct
or not? Infusion just become an application
of something. You have a foundation there
and for example I want you to understand what's
the application of this. So you have here
and you have here a plate, you have a column
here, you have both here, you have both there.
Is that correct or not? And if you have a
moment like that, notice this side wants to
come up, this side wants to go in. So these
both want to be pulled out, this one want
to go, everybody see how important this become
in future for you. Is that many you become
a civil engineering or mechanical engineering
you want to design something. Is that correct
or, so moment and direction of it is very
important. How much moment do I have and what
is the direction of it? This is the negative,
the other side would be positive. Their question
is, what is the moment of that force about
point O? Now, do I have the shortest distant
from point O to the line of application of
that? Yes or no?
Not really.
Yes or no? Where is that? Where is that D?
The D, this first of all I didn't say just
OA, I didn't say it is OB everybody understand
that? I said look what the hard you find it
here. You wrote it down. B was the shortest
distant from point we taking moment about
which is O, not to A, to the line of
Action.
So if I draw a line here, hopefully this is
90 degree, this is my D. Do I have that one?
No. So can I use that tilodi? Can I break
my R into component and take that moment,
yes or no? Why I am going to do that because
I have. This is my force 600 is that correct
or not? Notice I have horizontal distant and
I have vertical distant. If I break this two
force into a horizontal and vertical component,
I should be using those distances, yes or
no? As my D, correct? Everybody getting what
I'm saying that so that's exactly what we
are going to do. We are going to take this
force out and break it into two component
parallel to X which I'm going to give you.
One is 600 cosign 40 degree, et cetera, et
cetera. So you have a force going this way
and you have a force going that way, yes or
no? Use reverse of that tilodi. So I use divide
my force into there was an R here I broke
it into two component. Is that correct? That
is 2D problem. And 3D component become 3.
Yes? And then of course I need the magnitude
of that. If you do that we stick on 460 newton,
because 600 time cosign 40 degree and this
become 386 newton. Correct or not? Just you
have to give me couple of minute to finish
this up other word we lose it on everything
that I said. Yes or no? Now, I take a moment
of these two force about point O that I have
the my moment is that of that force. So MO
equal to, write that number. MO now it's magnitude
because I'm using 2D I'm using force distance
method. I'm not using cross product method.
Not yet. I don't, I've run out of time. I'll
do it next time. [Inaudible] do it here too.
MO now, how about this one. Does this one
cause a rotation this way first of all or
that way? Which way does it go? That way.
Is it plus or minus?
Minus.
So 386 times what? What's the shortest distant
from this line to point O?
2.
2.
See the lineup application, please get into
the picture. This is the line of application,
that's exactly. And this is the line perpendicular
to it. Is that correct or not? Therefore the
shortest distance is 2. So times 2 and it's
negative, yes. Of course what's the unit?
It's newton meter. Yes? Okay. And the next,
the next one is that one. Which one that one
goes? That one goes, not the force, the moment.
If I put that moment then put your hand here.
Then put your hand there around the force,
turn it around and you see it is turning that
way, yes or no? With this week practice this,
don't, if you don't get it today it's okay.
On Thursday you're going to practice all the
time, you're going to practice until everybody
be happy with it. Is that correct or not?
So for time being this is just the first exposure
for many of you. So it going to go that way
which is 460 times what? What's the lineup
of application, the lineup application is
there. The shortest distance from O is?
4.
And that's exactly why I use, I turn it into
such a way I can use 2 and 4, is that right?
So 60 times 4 minus and therefore it become
minus 2,612 what? What's the unit?
Newton.
Newton. Not enough. Newton
Meter.
Meter. Correct? Now it comes. We thought by
how do you think in the sky when there is
a airplane flying and they want to find the
shortest distant from this plane to the other
plane, you think they go there and measure,
no, no, we use all the static make. What's
the shortest distant from this point O to
the line of application, can you tell me?
Without calculating? Is this the moment? Yes
or no? Is that the moment equal to F times
D? Yes or no? So D equal to what? D equal
to magnitude of the moment divided by magnitude
of?
4.
4. Please write it down, actually we do it
in reverse. If you want to calculate which
we don't have, then that is equal to magnitude
of the moment is 2,612 newton meter divided
by magnitude from that equation. D equal to
M over F, so therefore let me write it here.
So this is D from now on equal to M divided
by F. Of course D become divided by F. F was
how much? F was actually 600 so 600 newton.
The answer is D meter it become 4 point 35
meter. Which we have to go through lots of
geometry to find it. Is that correct or not?
This is a 2D problem. If we go to 3D problem,
we want to find the shortest distant from
this point that table, that's what we do.
Is that correct or not? Okay, that for time
being that's enough. So this idea is for next
quiz and next homework. Everybody understand?
Your homework and your
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