[ Music ]
I think I's a perpendicular force.
Yes.
So I was just wondering if we could do that
by using the cross product of two of both
vectors.
You don't need that. Can somebody help me
to put this on the board there? OK. That's
the question.
[ Laughter ]
That's how helpful you guys are. OK.
[ Laughter ]
All right. Here we go. All right. Here is
that object. First of all, this obvious is
not [inaudible]. Because there is no friction
in there. Yes or no? So therefore, the first
thing you want to do, actually, many people
ask about [inaudible] office hours, including
someone who came very late, after 12:30. But
still I answered her question. Yes. I know
you are laughing. That means I really appreciate
that you [inaudible] to get all of the questions
answered. . Because I [inaudible] break between
class hours and office hours. Anyhow, so [inaudible]
the object. The object is, because if there
is no friction, it's going to slide. So you
are holding it with two cables. Yes or no?
So therefore, the three-body diagram [inaudible]
any reference to 3-D is an object there. There
is a weight there. The weight was given or
not?
Yes.
The weight was given, some number, I believe
it's 175 pounds. Is that correct or not? Sometimes
that would be a question. Then you have to
draw all of -- it doesn't have to be on a
scale because 3-D is hard to show. Is that
correct or not? But you have to know that
you have here a cable going that way, let's
call it the tension in cable T. [Inaudible].
Still some of you, when you showed me your
homework, I saw there are many mistakes. You
are not distinguishing between the vector
and non-vector. The lambda is all mixed. There
are parts that you have to be a little bit
careful. Many of you doing very good job,
but some of you need to be a little bit more
careful. So this is point A. This cable is
going from A to C. And that's the direction
of the force. Is that correct? We have discussed
that [inaudible]. There is another tension
here. Let's say this call, for example, a
tension AB, going that way. In general I'm
talking about. I'll get to your question in
a minute. Then there is a W. And also, since
this object, [inaudible] any object sitting
here, there is a contact force between the
two. Is that correct or not? That contact
force is also an N. So N has to be a certain
direction, which I cannot show it here, but
I'm going to show it like this for the time
being, just for the sake of it. Is that correct?
Well, that's the vector in a space, yes or
no? Correct? Now, how many forces do we have?
Four forces there. You can use a cross product.
We haven't gotten to that point yet because
we haven't discussed too much of the cross
product yet. But that's an option you have
in future. For time being, you don't need
that. That was your question, asking me how
to calculate N. Of course, everybody knows
how to calculate TAC and TAB because we have
done that. So let me put it down here. TAC
is equal to its magnitude. This is the way
you should write it. Times the lambda. Lambda
is given in the form of position vector divided
by its magnitude. That's one way of doing
it. That was where I was asking a student
in my class, there is another way of doing
it. He couldn't answer. But that's the way
you have to do it for X. So TAC, you all know
how to do it. TAB, you all know how to do
it. Is that correct or not? Yes? That's equal
to TAB times lambda of AB. This is lambda
of AC. So this [inaudible]. W is obvious.
W equal to zero I minus 175j plus zero K pound,
et cetera. These are all in pounds. You have
to use the unit. Now come to N. But I gave
you a hint last time how to find the N. And
you have to use it, that's why I have it here.
Remember that. If there is a line normal to
this force, normal to this surface, that's
the same as normal to that line. Is that correct
or not? That is the key here. So that's what
we are going to use there. So as you look,
I show you that this N can be written in two,
three different formats. One of them is NX,
NY, and NZ. But that is three unknown. I already
have two unknown here, one and two, the blue
ones. So that doesn't work. [Inaudible] if
I have three unknown here, two unknown there,
I have to write three equations of sigma FX
equal FY equal FZ. How can I solve for unknown
with three equations? It cannot work. So it
must be in this format. So it must be in the
format of what I have written here and here.
Is that correct or not? So that's what I asked
you to do. Some of you, of course did it.
Magnitude times lambda of N. So now, the lambda
N, we calculate it here. Position vector divided
by its?
Magnitude.
But do I have a line there?
No.
Do I have any coordinates there? So I cannot
use this technique obviously. What other techniques
did I show you? Cosine theta X, cosine theta
Y, cosine theta Z, or the lambda. All I have
to find, the direction angle of this line
with the X, Y, and Z, which are the positive
[inaudible]. I explain all of that before.
Yes or no? Then you look at the [inaudible],
that's what you have to look at. If this line
is perpendicular to that, all you have to
find, the slope of this line, that would help
you. Is that correct or not? For your problem,
the slope of the dotted line, when you look
at it, this is X. This is Y, not Z. It is
[inaudible] which problem, the one previously,
the one I did in class last Tuesday was Y
and Z of something. Is that correct or not?
This is X and Y. The slope is something like
that. This, I believe, is 60 [inaudible].
And this length is 30 feet or 30 inch. That
doesn't matter. I'm just explaining it. Is
that correct or not? However, while you're
having these numbers, you can find this angle.
Yes? That angle, I believe it comes [inaudible]
about 28 degree. Is that correct or not? For
your problem. Then the line that's in question
is the line, not this line, is the line perpendicular
to this. This is the direction N, that's why
I have this here for you to look at. Is that
correct or not? Yes? OK. So if this angle
is 28 degree, this angle is 90 degree. Obviously
this angle here will be 62 degree. Correct?
But what's that angle by definition?
[Inaudible].
I believe that those of you who did this problem
should have answered me right away. This is
positive. That's correct. This is positive
direction of the N. This is positive direction
of X. So this must be theta X. So theta X
is equal to already 68 degrees. This angle
--
62.
-- positive direction of N. Positive direction
of the Y, which [inaudible] balance of it
up to 90 degree. So that is theta Y. Theta
Y is equal to 20 degree. The only questions
remaining, theta Z, but if you look at it,
well, it goes down here and crosses the line,
line Z. That line and that line are like that
together. It goes like that because it doesn't
matter where line C is. If it goes like that,
it's like that, this angle is 90 degree. Absolutely.
So therefore, cosine B is 90 degree. So in
other words, here we [inaudible]. Write it
down. Now I write it in M -- no, let's write
it in blue. So I write equal to N times cosine
of as you see here, 68 degree I plus cosine
of 20 degree J plus cosine of 90 degree OK.
By the way, if even you don't get this, there
is another formula for it. Is this lambda,
X lambda, Y lambda, Z; yes or no? Lambda X
squared plus lambda Y squared plus lambda
Z squared is equal to one. You have these
two. Can you find the other one from that
equation? If you are not too sure whether
it's 90 degree, which is hard to see, but
you can use the equation, because lambda unit
vector is one. Cosine theta, this is the key
that I gave you long time ago.
A long time ago?
Two weeks ago. Two weeks ago, I gave you the
key that we call this actually theta X, theta
Y, theta Z, what direction angle. Is that
correct? That means that's the direction off
the lambda. Is that correct or not? And cosine
theta X, cosine theta Y, cosine theta Z, which
are called direction cosine, actually are
lambda X, lambda Y, and?
Lambda Z.
So that's the method we use. Is that clear?
Yeah.
Now, using the cross product method, the [inaudible]
product method, the angle between the two
lines, because first of all, you don't have
two lines here, you have three axis and one
line, which is totally different. But there
are some systems to be able to use that, but
that's not the one we are using. Correct?
All right. OK. These are actually the ones
students have the most questions about this
one.
I gave you your homework. Again, these are
the homework for next time. You should be
able to do it. Let's review what I said last
time and go from there. Now, the highlight
of what I said is important. So let's review
that one, and then we're going to immediately
include several problems. Now, the highlight
is this. I said if we want to calculate the
moment of the force about the point, of course
the force will be given. I need to [inaudible],
the moment is force times?
Distance.
But what distance? That's the key.
[Inaudible].
Yeah. Now say it louder that everybody can
hear, including there. OK.
The shortest distance between the point and
the application of the force.
The line of application [inaudible] the line
of the force. It's not point of application.
You see, the force always applies to a point.
From now on, you have a structure here, you
have a force here, you have a force there,
you have [inaudible]. These are point of application.
That's [inaudible]. Absolutely. So if you
have a force here, [inaudible] change all
of those, ABC. And this is point A, and you
want to times the moment of that force at
1.4 [inaudible]. Correct? So all we need is
F, and we need B. These lines are perpendicular
to that. Is that correct or not? That [inaudible]
distance vector. Then through using the concept
of the cross product, we end up two ways.
So M about point O equal to F times D. D [inaudible]
exactly as you -- so if D is given, I'm going
to use it. But many times, it is not given.
Everybody understand that? So [inaudible]
I have to come up with [inaudible] different
scenario for this one. But many times, this
-- let's say [inaudible] is here, so I'm going
to [inaudible] over there. Many times, this
line is given. Is that correct or not? This,
I call it position vector from O to A. So
that's the position that we talked before
about that. This is ROA. But when I went to
explaining the process to you, because this
angle is being called theta, et cetera, et
cetera, we found out that ROA cross product
of F, in that order, R comes first, F comes
second, would also give me moment about point
four. Not even give me the moment, it gives
me as a vector. Because cross product of two
vector must be a vector. Everybody understand
that? This one has both magnitude and the
diagram. Which is more suitable for 3-D problem.
This one seems to be, because 2-D problem,
we have all of the distances and it is not
that different. So FD or force distance method
is more suitable for 2-D problem. Is that
correct or not? Now, this one, of course,
if we are in the plane of XY, what's the perpendicular
plane of XY?
Z.
What's the perpendicular to the plane of XY,
is?
Z.
Z. So this moment, if I'm working in XY plane,
would be in Z [inaudible]. Without knowing
that, this is magnitude. Everybody understand
that? This is not a vector, this is magnitude.
So this magnitude, but we know this M is [inaudible].
Write it down underneath. It is along the
K direction. And I showed you that, you saw
that. So it is become a vector C. Remember,
A cross product, B become vector C. Yes or
no? According to the cross product. This is
vector A, assume it this way. This is vector
B. The next result of the cross product must
be a vector C. So this is the direction of
the moment, and that is the magnitude of the
moment, which we can use the right-hand rule
to do it. So as a result, I have mentioned,
this is the summary again. If it goes this
way, look at my thumb -- everybody should
do that to learn this. This is 2-D problem,
but the moment is in the third dimension,
third line. Everybody understand that? I mentioned
that last time. This was the idea. If I am
drilling here, this way or that way, the moment
goes either this way or that way, but will
move in the direction of the Z. Is that correct
or not? However, if you are drilling down
on the floor, the moment is about the Y. Everybody
[inaudible], and so on and so forth. So here,
this is -- we call that one plus. And this
one minus the unit is force distance or [inaudible].
Now, let's do a couple more examples to see
whether we get the same result. I did the
handout number one. Let's do handout number
two before we forget about this. The only
reason I'm doing that, this is a very simple
problem for you guys to be able to use your
right-hand rule and to get to the sine of
it. When you use force distance, the sine
comes by your right-hand rule, by your hand.
It has nothing to do with the sine of the
force. It has something to do with that, but
you don't look at this force, you look at
the right-hand rule idea. So let's say that
you have -- here is somebody's arm, and this
is the elbow. So this is the arm, and here
on the palm of your hand, you are holding
a -- the picture is clear there -- a bowling
for example. And that ball has a weight of
eight pounds. This is problem number two in
your handout. Is that correct or not? Quickly.
And then assume that this here [inaudible]
your arm has here about eight or five-pound
weight of muscle bone or whatever. Is that
correct or not? Yes? Which is vertical, of
course, all the one. And then this angle is
given 55 degree, the way you are holding it
is not horizontal, it's like that. Is that
correct or not? And then, of course, everybody
has experienced that. If this ball is too
heavy, you give this a try, the hand goes
down. Why? Because you cannot resist it. There
is a moment there, and you want to calculate
the moment, see whether your elbow can tolerate
that moment. Is that correct or not? The question
is, as you see in your handout, how much moment
these two force create about point O. Correct.
Or your elbow. So therefore, look what is
given. This distance is given 13 inches. Yes
or no?
Yes.
Right. Typical of the arm [inaudible]. But
then this distance is given, six inches. Not
the horizontal [inaudible], just to show you
another idea here. And that idea presented
here when we want to solve this problem, of
course it's very simple. I have to add moment
of that and moment of that about point O.
Yes or no? Both moment are positive or negative?
Now, put your hand in this direction. This
is the way you do it. Put your hand in a direction
such a way that you curve it about point O.
Your thumb indicates which way it go, which
is -- of course, you can see it is negative.
Your elbow tried to go that way. Is that correct
or not? So both moment of negative [inaudible]
right-hand. So the answer would be the following:
MO [inaudible] force distance, is equal to
eight pounds times 13 inches, eight pounds.
The answer is negative. Yes. That was the
negative [inaudible]. Now, if I want to calculate
this five pounds, I need the horizontal distance.
Yes or no? These are couple of homework [inaudible].
I don't have a horizontal distance. I have
a distance parallel to the R. Yes or no? So
what I can do, I can break this into two components.
This dash line is the component. Is that correct
or not? Yes? I'm taking advantage of this
six inch. Of course, you can find that [inaudible]
too, but this is another way, because you
have, in the homework, that's been asked.
So of course, if this is 55 degree and you
see this is 55 degree, you can find out what
those are. So this one is becoming -- also,
this is the vertical, yeah. This is 55 degree.
So this is five times cosine of 55 degree,
which we don't care. This one is five times
sine of 55 degree. Is that correct or not?
Yes? So this one goes through the point. If
a force goes through the point, does it create
any moment or not?
No.
No, because the distance is zero. Is that
correct or not? Yes? Everybody agree?
Yes.
OK. But the other one has a moment, and that
is equal to [inaudible] five. Five times sine
of 55. So you just write it down. OK. So it
becomes minus force is five times sine of
55 or cosine of 35, whatever. Both is the
same. Multiply it now by six inch and the
total moment becomes minus 128.6 pound inch
of moment. Yes? Now, can you tell me how much
tension you created into your muscle?
[ Silence ]
This is something for future, not for now.
But there is a muscle here, yes or no? Yes?
Let's put it this way. [Inaudible] correct.
I cannot draw as well as here. Actually, my
drawing is very bad. I'll tell you a story
later. OK.
[ Laughter ]
Yeah. All right. Here is the muscle, and the
muscle represents a tension, yes or no? Yes?
Notice, in future we learn this. This point
must be an equilibrium, yes? This is the action
and the reaction to that. Your muscle reacts
that that. Yes or no? So the moment of that
tension about this point must be equal at
opposite. So that way you can know how much
tension you are creating in your muscle. Is
that correct or not? Therefore, since the
two MO, this one action, and the reaction
there must be equal for future courses. Therefore,
128.6 in magnitude must be equal to tension
T. What is this distance? The distance, yeah,
this is the centroid of your muscle. Distance
of centroid, your muscle to the center of
your bone. Is that correct? It's given how
much in the picture? Two inches. Yes or no?
You must be very muscly phone number. Is that
correct or not? Yeah. Yes? Do you follow what
I'm saying? Correct? So therefore, this distance
is given equal to the center of O. That's
equal to two inches, so obviously it must
be tension times two inches. That's force
times distance or the tension of course ends
up to be equal to 64.3 pounds. If your muscle
can tolerate that much force, you are OK.
If not, you drop the ball or you will sprain
your muscle; one or the other one. Is that
correct or not? Yes? Have you done that before?
Probably. Correct? All right. Now, next problem.
So these are just the basic ideas, we are
not even touching down the real problem. Now,
let's look at this problem which is a little
bit different, but has the same idea. This
is just [inaudible]. This is the bracket.
We have a bracket of six meter here by four
meter there. And then I have here a force
applied here at center of the bracket with
the magnitude of force F equal to 500 meter.
And the slope of the force is 3.4. We did
this problem last time. I want to do it one
more time because this is similar to what
I did last time, a little bit different. This
is not in your handout. This is not in your
handout. Because I want you to show you the
idea of cross product versus the force distance
vector. Yes or no? OK. Notice what we have
to do here. If I want to calculate the moment
of this force, let's say this is point B,
this is point C, let's say this is point A.
The question that I'm raising here: find the
moment of that force about point A, which
is the rotation of this bracket about that
point. Is that correct or not? This is rotating
positively [inaudible] clockwise or counterclockwise.
Remember, clockwise is negative, counterclockwise
is positive, as we said earlier. Is that correct
or not? How do we calculate that? Notice the
magnitude of force given, but the distance
is not given. The distance, if I want to do,
I have to extend this to here, yes or no?
Then I draw a line from here to that line,
and I have to make it perpendicular, and I
have to find that distance. Is that correct
or not? Yes? I don't want to do that. That's
a geometry problem. We are not in geometry
class. Is that correct? So what did I ask
you to do last time? I gave you a theory.
What was the theory? Have we all forgotten?
I said do not forget. Anybody remember? Come
on. That is so obvious that you should remember
that, even if you were not even doing any
problem. Break the force into component. Yes
or no?
Yes.
I said the result -- if there are several
forces at the same point, I can use the resultant
of those forces, yes or no? So reverse of
that, if I have a force here, can I break
it into two components and take the moment
of those two? No. Those two components should
be parallel and perpendicular to the X and
Y, because I have the distances along the
X and Y. Look what I did in the other problem.
Why did I broke it this way? Because I had
this distance. Everybody understand that?
Which makes it perpendicular to each other.
Therefore, taking advantage of the six and
the four, this is what the theory was about.
So break it into two components horizontal
and vertical. In this case, distance become
-- it's actually a little bit shorter. This
was three, the other one was four, so this
becomes 300 meters. And the other one becomes
three-fifth and four-fifth. Remember that?
That was three, four, five ratio. Is that
correct or not? This is just for purpose of
discussion, I made it very simple. Yes or
no? Now, can I find the moment of these two
forces about point A? That would be equivalent
to the moment of original force. Yes or no?
Yes?
Yes.
Come on, shake your head.
Yes.
Yes or no?
Yes.
I'm saying yes, and he's going like that -- or
many of you. Are you following or not?
I wasn't here last week.
Oh, you were not here. That's why he's not
shaking his head.
[ Laughter ]
So the way you smile at me, all of the indications
come from your faces and your eyes, and your
eyebrows [inaudible]. When I talk to you and
your eyebrows go like this --
[ Laughter ]
All right. I said something that has not registered
yet. The whole problem of a static is this,
guys, actually, many students in the past
have told me that it takes them a week or
two or three or four sometimes. Then suddenly,
they're all smile. And so they come to me
[inaudible], "Now everything clicked as if
I know what's going on. Before, I didn't do
it." And the only way to do it is by doing
your homework more and more, and you get to
that point. As soon as you get to that point,
then you will smile all the time. Is that
correct or not? Yes?
[ Laughter ]
OK. All right. So let's see all of those smiles
from now on, OK, guys? So let's go here. What's
the moment about point, what's the moment
of this force 400 about this this point? What
is your guess [inaudible] your right-hand
rule? Is it plus or minus? Is it counterclockwise
or --
[Inaudible].
Counterclockwise. Plus. Therefore, 400 MA
equal, again, magnitude, is equal to 400 times
what? What is the distance? I don't need A
to point C. I need the shortest distance from
A to the line of application which is?
Six.
Six. That's what I'm doing here. Everybody
understand that? Yes?
Yes.
The line of application of 400 is this line,
correct? This line is perpendicular to it,
therefore, that is what I did here. So the
distance is six. So 400 times six --
Is equal to [inaudible].
And of course, it's going that way, therefore,
it is plus. Now again, this is force distance.
This force statically is plus. But notice
at the moment, the moment of this force about
point A is not plus. The moment is going clockwise,
it is negative. So remember, one more time,
the way you do the sine is this. First you
put the magnitude of force, 300. Then you
put the distance of force, that's the [inaudible].
The distance from point A to this not [inaudible]
AC, because that's not the point. You extend
this. This is the perpendicular line. So the
distance is how much?
Four.
Four. That's right. So 300 times four, that's
what the idea I presented to you. If you remember,
what the D is, the answer is correct. It goes
there. Everybody can do it. If you don't know
what D is, then nobody can get it. The sine
is going negative. So as you see, one part
of it is creating positive moment, one part
of it creating negative moment. So that becomes
2400, of course minus 1200. So net result
is positive 1200 -- what was it, meter of
moment. Which direction is the axis of that
moment, one more time? This is Y force, so
it [inaudible] perpendicular. 2-D, this is
something you should remember. 2-D is always
the moment about the k-axis. In other words,
if I want to present this as a vector, then
that becomes a vector. Yes or no? So this
becomes equal to 1200 times K [inaudible]
meters. So we do that in two steps. So this
is how [inaudible]. This gives you the magnitude,
you determine this. As I said, depending on
which plane you are in. Is that correct or
not? If you are in the plane of XY which is
the most common plane, the vector is in the
[inaudible]. However, if you are in the plane
of Z and Y, perpendicular to the Z and Y is
I. Everybody understand that? Yes or no? For
X, correct? And so on and so forth. OK. Now,
let's go to the second method. You see, this
was the first method. This is the second method,
yes or no? I said this is applicable to 3-D
most likely. This is applicable to 2-D. But
can we use that one for this problem? Yes,
we can. So let's find out the moment of that
force about point A using cross product method
-- -- which I said from now on, you are not
supposed to use that, although some of you
will do. In 2-D, definitely you use force
distance, it's not required. I already did
it. Is that correct or not? Just to show you
that we achieve the same results, that's what
I'm doing. So by definition, moment of that
force we have there which was 600 meter. Yes
or no? About point A. Write it down in your
notes. MA as a vector. This time, this is
a vector. That was a magnitude. Becomes equal
to what? R. Now, here is the key. Some people
write this. OK. They write this, they don't
put any subscript there. When they do subscript,
there is two reason for it. Either you are
a little bit lazy and you want to get ahead,
or you don't know it. One or the other one.
I assume you don't know it. Because if you
knew it, you would put it there. Is that correct
or not? Because this R is specific R. It cannot
be any R. Notice the way I explained it before.
Where was it? Started at O, end at A. For
time being, we are going to use that for the
next lecture or so, but later on there is
a modification to that. I don't want you to
-- you understand this is only vector. But
for time being, where we take a moment to
the point of application. Is that correct
or not? That's what I put here. Notice RO.
And in that order. R comes first, make a note
of that. F comes second. And when you write
R, R is started always at the point you are
taking moment. And end up at the point of
application. So for R problem, that R would
be from where to where? You see, A is the
point we are taking moment. C is the point
of application. Everybody understand that?
So my R would be RAC. Is that correct or not?
According to the definition I gave you. OK.
So that's what I'm asking you. Put here or
put it in blue that you see that line is blue.
That is the specific -- define that order.
You change the order of RAC to RCA, the determinant
changes. [Inaudible] determinant of what?
If the determinant, you go RAC or RCA, you
are changing your minus. So the sine of moment
changes. Some people do the homework, they
come to me, "Oh, I got the answer right, but
my sine is not right." Two reasons. One of
them, you are writing RCA instead of RAC.
Another reason that you may get the sine wrong,
you change the F first or second. That changes
also because that, remember, the determinant
I gave you, in the determinant, it's IJK.
Let me write it down for you. In this determinant,
anything changes, the sine changes. Everybody
knows the law of the determinant. Is that
correct? The first rule is IJ and K. The second
rule must be R, the third one was F. If you
shift that, change it F to here or here, the
sine changes. Everybody knows that if you
change the rows of the determinant, the sine
changes. Is that correct or not? So you cannot
do that. So now what are the components of
RAC? It's sitting there. What is X component
of RAC is?
Six.
Y component?
Four.
Z component.
Zero.
Zero. So therefore, if I'm using that technique,
this is six, 4H and zero. Six going that way,
four -- what are the components of the force?
You already seeing that. The component of
the force is -- here, you have to use the
sine. Plus 300. Yes or no? The other one is
plus 400. You are [inaudible]. And then zero.
This is the second method. I'm using my definition
I gave you, approve it. Now this is another
proof to show you that the two answers should
be the same. Is that correct or not? Yes?
Now, MA equal to -- now you have to expand
that. MA equal to -- remember the expansion,
I said they are going to do that. You cross
this column, you cross this line. So you have
this times that minus this times that. OK.
But this is zero, the other one is zero. Of
course I know, I should get zero. Everybody
understand that? There were no I there, no
J there. Is that correct or not? Therefore,
I -- write it down -- four times zero minus
400 times zero is zero. Is that correct or
not? Yes? Then you go to minus J. Why minus
J? Remember the term of expansion. This is
plus, this is minus, this is plus, this is
minus, this is plus. You have to go like that.
Is that correct or not? When it is first row
and second column, when you cut it out, you
have to go to minus. Is that correct or not?
Recall all of your math. So what -- it doesn't
make any difference for this problem. When
I cross this and cross that, I still have
six times zero minus 300 times zero. It is
zero. So finally, this cut, this cut, which
gives me plus OK. And then multiply that.
[Inaudible]. This is gone. This is gone. Six
times 400 minus four times --
[Inaudible].
-- which ends up to be 0I plus 0J plus 1200k.
As a matter of fact, notice, what you have
written here, look. Without thinking about
it, it [inaudible] math. You may never understood
what happened to the frame. It's exactly what
you wrote here. Look at it. Yes or no? You
see that? But the other one, I like force
distance method a lot better than cross product.
But I tell you what, because when you use
force distance method, you [inaudible] the
direction moment. You know what happened to
the object. If in the future you want to design
it, you come to the spring classes or go to
CE classes, some of you are civil engineers.
You know that that is what you need in order
to design a [inaudible] or a shaft, or whatever.
Because you want to know exactly what happened
to that object. Is that correct or not? Here,
you just close your eyes and you use the math,
and the answer comes by itself. Some of you
will love it [inaudible]. Yes? But I don't.
That's what it is. As a matter of fact, I'm
going to turn some of your 3-D problems into
a [inaudible], into all 2-D problem. Because
when you come to the [inaudible] classes or
CE classes at the beginning, we never use
cross product again. Cross product is the
definition, it's used for very specific problem
in 3-D when the problem becomes too complex,
we use it. The forces are vertical and horizontal.
There is a way around it. Is that correct?
To doing it only by force distance. OK. Everybody
clear about that? Now let's go to the other
problem, 3-D problem in your handout three.
So let's start with three and analyze that
problem, see what we get there. This shows
you exactly what you want to know about where
the technique comes from. Is that correct
or not? Now, this problem is difficult problem.
Obviously, I have to use cross product. Is
that correct or not? Because it's a 3-D problem.
Yes? OK. Let's look at it. While I'm erasing
the board, you tell me what's happening with
that problem and what are we trying to do.
[ Silence ]
It's important always to understand what do
we want to achieve and what do we have to
do in order to get to that goal. All right.
I hate that part there. By the way, this is
problem number 21 in your book, but you don't
have to look into your book because I am going
to add to that, because this number could
be different, et cetera, et cetera. Because
there's different versions of the book, has
different number for problem. So here, again,
here is the scenario. This is X, Y, and Z
axis given to you. There is a tree here on
the y-axis shown there and there. And we want
to get rid of this tree. Somehow we want to
dig it out. One way of doing it is first to
dig around it and get it out, which is very
expensive. Yes or no? Right? Everybody has
tried to get rid of a tree at root. You know
how expensive that -- even the tree, they
charge about $1,000 [inaudible]. Yes? Yeah,
I know.
[ Laughter ]
OK. So therefore, another way is this. OK.
I put here a rope, I put here a rope, I put
a wrench here and wrench here. And I create
a tension and pull this out. Is that correct
or not? When you pull it out, how does it
work? It creates a moment about point O and
then [inaudible] ground is not strong enough
to hold that [inaudible], the root and the
soil [inaudible]. Is that correct or not?
Unfortunately, you don't want the tree to
fall on you and your equipment. You want it
to fall in [inaudible] direction. And have
you seen sometimes they break it, I mean,
they put dynamite in the building and the
building goes exactly the way they want it?
Yes or no? This is what [inaudible]. You are
standing here, you don't want this tree to
fall on you, Is that correct or not, when
it comes down? Correct? So if you do it right,
you know exactly that tree is going to fall
down. Let's get to that. Besides more information
than what you -- and then what the problem
says to do. Anyway, let's look at the cable.
There is a cable from point B to point C.
This distance is one meter. And that distance
is 4.25 meter. Correct. On the x-axis, it's
4.25 meter. And this on the z-axis [inaudible].
C is on the horizontal plane. And then there
is another cable going like this to this point.
Again, this distance on the horizontal plane
is .75 meter. And this distance, if I put
a line here like this, this distance is six
meters. This has to be parallel to X. So this
should be [inaudible]. And the height where
I put the point B is given seven meter. Is
that correct or not? Yes? So in other words,
we have to use the position vector. Is that
correct or not? Now, let's look at the data
that we have. In order -- the tree comes out
-- they are not saying that. All they are
saying is that calculate the moment of the
forces acting at B about point O. And you
know what force is acting at B. You have done
enough homework to do that. We have two forces
there, yes? Tension in cable BA and BC, yes
or no? Which typically I asked you to do that;
you have done it in the past. You have to
cut it here, cut it here. Yes or no? Right.
We are not talking about force at this point.
This was point A of course. BA and BC. We
are not talking about that. We are talking
about force here, so you are talking about
these two forces.
[ Silence ]
I don't know whether I'll call it F1 or F2,
whatever [inaudible] call it, it's the same
thing. T, tension BC and tension BA. We are
trying to find those. This should not be any
problem for you. You already have done that.
Yes or no?
Yes.
This was all of your homework [inaudible].
Yes? I told you, you are going to use that
many, many times, didn't I? Yes? How many
times did you use it in those homework?
A lot.
A lot. This is a lot more. Two times more.
Is that correct or not? Or a lot more in every
problem. Anyhow, can you do that? Yes. TBA.
Now, when magnitude of that is given or not,
look at the tension, the magnitude of TBA
is given equal to, so I write it here. TBA
magnitude is given equal to?
[Inaudible].
How much? 500? 565 meter. Correct. At TBC
is given equal to 660 meter. All right. TBA
therefore equal to 555 [inaudible] times lambda,
the lambda is position vector divided by its
magnitude. And those are the following. So
let's see whether this is correct. Minus .75.
Is that what you're reading from your picture?
Yes or no?
Yes.
How much down are we going?
Negative seven I believe.
Minus seven. Because we did that before. Add?
Plus six.
Plus six. Divided by -- you have to calculate
that one, so let's divide it by magnitude
[inaudible]. I don't know [inaudible]. Somebody
calculate the magnitude of that [inaudible].
[Inaudible]. 9.35.
9.35. OK. 9.35. So this is meter over meter.
This is in meter. The answer would be in meter.
So that will be simple enough to calculate
that one. So it is TBA become equal to minus
45i plus 420j plus 360k. We need to put that,
meter. So that's force being vector form.
Yes or no?
[Inaudible] negative?
What?
J?
Minus, minus. This one should be minus. Absolutely.
You are right. Minus, minus. This should be
minus, minus. Correct. Very good. And TBC,
the same thing. TBC, this time is 650. And
the line, you can check it later on if you
want. It is 4.25 [inaudible]. And then minus
seven, that should remain the same, plus one
in meter divided by a number similar to that.
I don't have those numbers because I have
to resolve that. So you would put the magnitude
here [inaudible]. Yes.
It should be 660, right, instead of 650?
660.
Six --
For the magnitude of TBA or TBC.
Let me double check. 660. [Inaudible]. And
if you gave me this number, you [inaudible],
because I have the next [inaudible]. And that
one is equal to 340i plus minus 560j plus
80k meter. Notice now, I can take the moment
of each one of them about point O individually
using cross product. Yes? Or I can use the
[inaudible] like I gave you. I can add it
together and take one moment. Is that correct
or not? Let's do that. So if you want to find
the resultant, the resultant R, which is sitting
here [inaudible] space, which has the same
amount of moment as these two individually
according to that [inaudible] that you wrote
down last time. Correct or not? Yes? So let's
do that. So R equal to TBA plus TBC, which
you already have done it in the previous homework.
[Inaudible] here, you add these two together.
Yes or no? So 340 minus 45, so you get 295i
then minus 560 minus 430 equal to 980 minus
980j. And finally, plus 440k meter. Correct?
All of that look math. These are nothing new
for you except the conclusion. The conclusion
is something that I have to discuss it with
you which is very important. So let's do that.
Now, we go there and we want to do MO. Of
course, this is 3-D and we don't know anything
else yet except the cross product. Yes or
no? Because I don't have distance [inaudible]
that. Is that correct or not? Yes? Definitely
using the cross product. Which cross -- again,
remember, do not write it like that. So this
is position vector, and that's the force.
Is that correct or not? What position vector?
And I already explained it. You have to be
specifically expressing that for yourself
in order not to make any mistakes. [Inaudible].
Start from where and where?
The point of moment to --
Which the point of moment is?
[Inaudible].
Four. The point of application is B which
is very simple position vector from O to B.
Is that correct or not? Yes? So that's exactly
what you gave me, is perfect. So OB. So that
becomes a determinant. Again, remember, that's
all you need. IJK, in that order. ROB is what?
Zero, seven, zero. Very simple. Yes or no?
And then the force. The force, you have to
be careful. Here, you have to put the sine
of the force, because sine of force, you get
in trouble. In FD, actually, we did it. Here,
we have -- OK. What is the force? The force
we end up is 295i plus then minus 980, and
then plus 440. This is your determinant. All
you have to do, expand that into IJ and K.
Notice this time, the vector is in a space.
[Inaudible] the moment vector in a space and
a vector in a space should have three components.
X, Y, and Z. What we are saying in effect
is this. Look at [inaudible] this slope here
to show you. If R is here, for example, F
is here. The cross product of R, cross product
F is a vector perpendicular to this plane
which is here. Remember? [Inaudible] you had
that one. That plane, now you're talking.
That line is perpendicular to the plane of
these two. It is in a space, therefore, add
IJ and K. Similar to that homework that you
did. Is that correct or not? Do we need that?
We don't need to go find N because it comes
by itself after -- actually, it comes afterwards.
But we know the result of this operation is
a vector perpendicular to the plane of R and
plane of OB. Everybody understand what I'm
saying then. Yes. Where that is, we don't
care. Is that correct or not? However, the
answer. The answer would be equal to I, seven
times 440 minus that. So zero times that.
I don't have to worry about it. Then minus
J. Remember that. This and that. Zero. Very
good. Minus J, zero, zero, and then plus eight.
And this time I cross that. Zero minus seven
times 295. Is that correct or not? And the
answer is in meter, meter. That's correct.
So anyhow, MO ends up to 3,085 minus 0J or
plus 0J, [inaudible]. 0J. I put it there for
you to see. And then minus 207 OK meter. That's
it. So if I give you a bunch of 2-D problems
that you can use force distance method, it's
very easy if you understand that. A bunch
of 3-D problems that you can -- don't put
your books [inaudible]. I still have five
minutes discussion.
[ Laughter ]
Maybe a little bit more. And then there are
some which is about axis that I am [inaudible]
now to that, right now. Notice, this vector,
MO, has three components; X, Y, and Z. But
where is that vector, first of all? Show me
the location of that one. Where do you think
it is?
[Inaudible].
It's over here at O. Yes or no? So let's go
redraw that. You should all redraw that. There.
The vector is at O because we are taking moment
at O. Moment is a vector, yes or no? But is
it in the K direction? Absolutely not, because
it has I. If this was zero, this was zero,
that would have been in the K direction. What's
the component of that moment? Wherever that
moment is, we are trying to find out, so the
moment is over here. I don't know yet. I'm
just putting -- if I put it here, it has to
have IX, IJ, and K. But notice what happened.
There is no J. Why there is no J? Can anybody
explain that physical aspect of that?
[Inaudible].
If you put two ropes here, there is no reason
for this tree to rotate about the y-axis.
Everybody understand that. However, it's rotated
about the x-axis and [inaudible] about the
z-axis. That's what the answer is showing
you, but you don't see it if you look at the
number because that show [inaudible]. That
means the rotation is -- now, look at it.
Now, we'll get to that in a minute. So is
there an X component there? This is X [inaudible].
Is there an X component? Where is it?
3,000.
380.
OK. This is -- call it MX. So that M that
we cut out of MO, this has three components.
This is MX, this is MY, and that is MZ. Like
any other vector. Any vector F has FX, FY,
and FZ, which is projection of that vector
over X, Y, and Z. Projection of this vector
here over x-axis is 3,080 and it's positive.
Yes or no? So that -- call it MX. And going
that way in effect is a rotation like that.
So the three coming this way. Is that correct
or not? Not the other way. Is that correct
or not? So if you want to see, you go on the
other side. Yes or no? Not this side. [Inaudible].
Is that correct or not? Yes?
Yeah.
Does it have any component about the y-axis?
No. Does it have any component about the z-axis?
Yes. So let's extend the Z, but apparently
to negative, yes or no? So how much negative?
2,070. So this is the 2,070 going backwards.
So the moment vector, actually this is MO.
Happen to be lying in the XZ plane. This is
your MO. [Inaudible]. Is that correct or not?
This is your MO. Everybody with me? I'm using
that. Is that correct or not? Yes?
Yes.
Now, which way the tree rotate now?
[Inaudible].
Put your hand here around this red line. Perpendicular
to that, that's exactly where the tree is
going to fall down. [Inaudible]. Because the
moment is like that. The rotation is about
that red line. Is that correct or not? The
magnitude of this moment is how much? MO.
Write it down. This is the magnitude of the
moment. The magnitude of the moment is square
of root of 3080 squared plus zero squared,
which we don't care, plus 2070 squared. You
can calculate that. That's how much moment
you are creating about point O. And if that
moment exceeds the moment that you have there,
it's going to -- MO becoming -- I don't know.
Somebody calculate the magnitude.
3710.
How much?
3,710.
3,000?
[Inaudible].
[ Background Conversations ]
I did not hear you.
[Inaudible].
3,000. Oh, ten. 710. Yeah. Something like
that. [Inaudible]. So if the ground is still
not giving up, you may put some water there,
because if you put water, then the ground
becomes weaker and then it falls. Is that
correct? Exactly you know that these are MX,
MY, and MZ. So in a way, not even we calculate
moment about point O, we calculate moment
about x-axis, y-axis, and z-axis. One more
time, please, write down this. So for future
reference, because we are going to talk about
-- because some of your problem is exactly
like that. When we get to the moment about
the line, there are two types of problem.
Let me finish that because by saying that,
you probably can do 80 percent of your homework.
So they are asking you to find the moment
about x-axis. Yes or no? If you want to find
the moment about x-axis, you do exactly what
I did there. First, you find the moment about
point O. Then you get MX, MY, and --
MZ.
You pick up MX, you throw the rest of it out.
[Inaudible]. Is that correct or not? There
is another method that I'll show you which
is much more difficult. So if I can find the
MO here in any vector and then I break it
into -- remember what we did before? This
is MX, this is MY, and that is MZ. In this
format, this is the sense of it, redraw it.
This is MX positive because it's going that
way. This is MY positive, rotation [inaudible].
This is MZ positive. It comes like what we
did here. So it comes either positive or negative.
We leave it at that. Next time I'm going to
show you the [inaudible] find the moment about
a line. Everybody understand that? These lines
are very simple because these lines [inaudible]
X, Y, and Z. But if the force is here, the
line is there, I cannot use that method. Everybody
understand that? That's the only thing left
from that discussion. The rest of the problem,
you can do it then. So you go ahead and do
your homework and we'll stop here. And we'll
continue the discussion later on. But you
have to do this homework in order to understand
the rest of the problem.
