[ Music ]
>> Hi everyone. We have just now worked out
a two-dimensional moment problem. You know
how to calculate moment now. Moment is equal
to force times distance. Moment of a force
is a tendency of that force to rotate an object
about an axis. So in order to calculate moment,
you figure out the distance, which is actually
a perpendicular distance, and you multiply
that distance by your force. All right? So
calculating moment in 2D is easy. But in real
life, you are going to deal with forces in
3D. If you think about many objects that we
see in our everyday life such as a transmission
tower that is secured by many cables, you
are dealing with three-dimensional force.
And if you are dealing with three-dimensional
force, you are also going to deal with three-dimensional
moment. This means you are going to have moment
or components of moment along x, y and z axes.
In theory, though, moment in 3D means you
have to calculate still the perpendicular
distance from an axis to the force. But in
reality, it is a lot harder to visualize,
sketch, and calculate. So we will use vector
algebra. You're all experts in vector algebra,
so we are going to use vectors in calculating
moments. Now I want to recall your attention
to something what we have done several sessions
ago in which we talked about vector addition,
vector subtraction. We also talked about multiplying
a vector by a scalar. Now you remember when
you multiply a vector by a scalar, the length
or the magnitude of the vector will change
but the direction remains the same. At that
time, you might have had a question, can we
multiply a vector by another vector, right?
That's an interesting question. We didn't
talk about it. Well, it turns out you can
and there are two possibilities. You can take
two vectors, cross one vector by another vector.
You could say A, vector A cross vector B.
That's one possibility. Another possibility
is instead of crossing, you can dot. That's
known as the dot product. So multiplying vector
implies two scenarios: a vector product, also
known as the cross product, scalar product,
known as the dot product. Let's talk about
vector product for a minute. What does that
mean? That means you are dealing with two
vectors and you are crossing them. In case
of moment, we are actually dealing with two
vectors. One is the distance. All right? And
I'm going to call it as the position vector.
And then we are also going to deal with force,
which is a force vector. It turns out vector
product that you have been studying in math
classes all along is actually nothing but
moment. Moment is the real application of
vector products. So let's take a look at how
we can use vector product in solving moment
problems, especially three-dimensional moment
problems because it makes your life easier.
So let's take a few minutes, solve an example
problem, review vector product, and complete
our discussion on moments. Hello everyone.
Let's talk about vector product. Many times
this word has come up and you have seen this
in previous math classes. And you know vector
product means you are dealing with two vectors,
for example I have say a vector P and let's
say I have vector Q and vector product is
I am finding the product of these vectors
or more specifically if you want to be saying
it clearly, I would call it a cross product.
So I'm going to say P, vector P cross vector
Q. All right? And from your math classes,
you know when you cross one vector with another
vector, the resultant is going to be a vector
as well and I'm going to call it R. Now what
that means is the resultant is a vector which
has got a magnitude and a direction. The magnitude
of this vector is going to be so P cross Q,
if I want the magnitude of that, from math
class you know it is going to be the magnitude
of vector P times magnitude of vector Q times
sine theta. What is the sine theta? That's
the angle between these two vectors. So if
you want to see this visually, here is vector
P, say here is vector Q. This is the angle
theta between them. And when you cross them,
the resultant vector is going to follow the
right-hand rule, meaning -- Let me put the
vector as a vector here. That's the vector
P. This is vector Q. And when I cross them,
the resultant is going to be defined by the
right-hand rule. That means if I take my palm,
put it along the first vector P and curl it
towards the next vector, the thumb is pointing
in the direction of the resulting vector positive
side, right? So that vector in this case is
going to be R. So this vector R is going to
be perpendicular to the plane containing these
vectors, perpendicular, all right? So that's
the resultant vector. So the direction of
the resulting vector is perpendicular to vector
P and vector Q. And the magnitude is going
to be equal to this. All right? Now I would
like you to explain this in the context of
a unit vector. All of you know, the sine of
unit vector, meaning the magnitude of the
vector is going to be equal to 1 along each
axis x, y, and z. So if I have an axis, x
axis, y axis, and if I have a unit vector
defined along x, which is i, a unit vector
which is defined along j, that means I can
cross now two vectors. Let's cross i and j,
i cross j. And by definition, you will see
it's the magnitude of i, magnitude of j times
the angle sine 90 degrees. Correct? It's 90
degree. And if I did that magnitude of i and
magnitude of j is 1 and 1, so multiply 1,
multiply 1 gives nothing, and sine 90 is equal
to 1. So the magnitude of resulting vector
ij is equal to having a magnitude of 1, right,
the magnitude of this vector is going to be
1, so that means the resultant vector will
have a value magnitude 1, which indicates
it's probably a unit vector. And it is going
to be along the direction perpendicular to
x and y and you know that would be your z.
So the resulting vector is going to be k.
So i cross j is equal to k, a unit vector
along z axis and is going to be equal to 1.
All right, so that's a basic vector product.
All right? Now just to conclude this discussion,
what if I cross i with i? Well, you should
know i cross i, the angle between them is
0 and, therefore, that's going to be equal
to 0. What happens if I cross j with i? Well,
that means I am crossing j with i. That means
the magnitude is going to be still 1. But
if I use the right-hand rule to find the resulting
vector, it is not going to be along positive
z, it is going to be along negative z, meaning
it is going to be negative k. So what is a
vector product? Well, crossing two vectors
and the result is another vector. And the
magnitude of the vector is going to be equal
to the product of vector times the angle between
them, sine angle between them, and the direction
is going to be defined by the right-hand rule.
You did all this in your math class. But in
the context of engineering, P and Q are two
vectors that are more meaningful to us. One
of them is going to be the force vector. Another
one is going to be the position vector. So
how can we apply this for calculating moment
and how can we use this in solving 3D problems
is what we are going to look at. But to complete
the discussion, I'm going to draw or show
the vector product in the context of a moment.
All right? So if I have a force applied here
and I am calculating moment about this point,
this is my force vector. This is my position
vector. Notice how I am showing the position
vector here. This is my position vector. I'm
going to call it r. All right? And that's
my force vector. So I wanted to use the vector
algebra instead of using just perpendicular
distance, I'm going to say I'm going to cross
these two vectors and remember moment vector
is going to be equal to r cross F. So from
now onwards I want you to remember when we
apply vector algebra, moment vector is the
cross product of r, which is a position vector,
all right, please remember that position vector,
and that's your force vector. But you also
realize that position vector r can be returned
in its component form like this. Right? Similarly,
force vector can also be returned in its component
form using Cartesian vector format and I am
right now doing this in 2D. For 3D, you will
have an additional component. So r cross F
means you are going to dot these two expressions.
I'm sorry. You're going to cross these two
expressions. Right? I am not going to spend
time doing all of these calculations because
it's well defined in books. All right? So
you are going to cross these so each of these
components you are going to use your FOIL
to cross it and when i crosses with i, you
are going to get k. When j crosses with i,
you are going to get negative k. And when
j crosses with j -- I'm sorry. I might've
made a mistake here. But you know the vector
algebra. So i cross i, angle is 0. All right?
So it won't give a unit vector. It would be
a 0 vector. But i cross j would give you k
and so on. All right? Now you can write this
expression and calculate, but if you recall
your linear algebra, there is an easier way
to do that. For those who forgot, I'm going
to just give you the expression when you have
to cross two vectors for calculating moment,
you can express this using the determinant
of a matrix form, where you have i, j, k and
then you have rx, ry, rz meaning these are
the components of position vector in 3D and
then the third row will be the components
of force vector. So this is calculating vector
or moment in three-dimensional form and this
form is very useful and very effective because
you know how to calculate the determinant
of this matrix or the determinant of this
matrix, right? And if you can do that, you
can actually calculate the moment very easily.
And this is very effective and very useful
in solving three-dimensional problems. So
any time I recommend you solve 3D problems
dealing with moment, you will likely use this
format. In two-dimensional approach, you could
use this or if you are comfortable doing the
conventional way, the way I did in the first
two examples, you could do that. But 3D problems,
I suggest use this approach because this will
help you in getting the correct answers because
it is rigorous, basic math involving algebra,
vector algebra and you can't go wrong. The
key here is to formulate the problem correctly,
finding the components of your position vector
and finding the components of your force vector.
I hope you understand vector product. And
now we will move on to solving a three-dimensional
problem using vector products. All right,
let's take a look at moment of a force in
3D and use the math behind vector product
to solve this problem. The key for solving
problems like this is to make sure you formulate
the problem correctly because the solution
process itself is very easy. It is basically
solving or finding the determinant of the
matrix. And that's something you can do using
your calculator and you're allowed to keep
those tools. So focus on formulating the problem.
So let's take a look at this one. Here is
a problem from Beer and Johnston. A 200 newton
force is applied, as shown, to the bracket
ABC, a common problem that you encounter in
our everyday life. You use these brackets
for mounting doors, mounting different devices,
and the force is applied. It's in three-dimension,
as you can see. And we want to find the moment
of this force about point A. In this particular
case, they already gave you a coordinate system,
x, y, z with origin being at point B. So the
first thing what you have to do is to make
sure you write your force and the position
vector in Cartesian vector format. So first
identify your force. The force is right here.
That's your force. And you do know the magnitude
of this force, 200 newton, but you want to
write this force in Cartesian vector form.
How do you do that? You need x component,
y component, and z component. And if you write
this, it should look something like this.
Correct? But you need the components. Let's
go and look at the x component. The way this
problem is given, this force is in the plane,
I would say y-z plane. It is not in any of
x and, therefore, there is no x component
and, therefore, it is going to be 0i.You don't
have to write 0 but I wanted to make sure
you understand there is no x components of
0i. Where is the y component? Well, you can
see the y component right here. This is your
y component. And this is 30 degree. So you
can actually visualize a right triangle here.
And if you visualize the right triangle, this
will be your y. This is your y, parallel to
the y and that would be cosine 30. So it is
going to be -- And by the way, the direction
of this force is going to be negative because
it is going in the negative y direction. So
it's going to be negative 200 cosine 30 degree.
I hope you understand that. That's your j.
In other words, I'm taking this force, projecting
it to the y component, and since it is pointing
down, it is negative and this component is
going to be cosine 30. And that leaves now
finally for the z component and the z component,
as you can see, is along here and it is going
this way. So it's positive z. So plus 200
and it is this component I can either say
cosine 60 or sine 30. They are the same but
I am going to use this angle, so it's going
to be cosine 60 k. All right? So what I have
done is writing this force in Cartesian vector
format, 0i, sorry, minus 200 cosine 30 j plus
200 cosine 60 k. Make sure you understand
how to do this. This is a critical step. Once
you have formulated force F, we need the angle.
I'm sorry. We need the position vector and
the position vector is from A to C. Why? Because
they said calculate the moment of a force
about A, so you draw the position vector from
here to here. This may be the easiest way
to do it. All right? So that's my position
vector. Now I need to write this vector in
Cartesian vector form, meaning I want the
x, y, z components. In other words, I want
to do this exactly like what I did with the
force. But I need to find these components
x, y, z components. And the easiest way to
do that would be to find the coordinates of
A, which in this case coordinate of A is 0,
x coordinate is 0, y coordinate is negative
50, and the z coordinate is, z coordinate
is -- x is 0, y is negative 50, and the z
is 0. Right? Similarly, I look at the point
C, x coordinate is 60 millimeter, y coordinate
is going to be up here 25 millimeter, and
the z coordinate is 0. So I know the coordinate
of z. I know the coordinate of A, so I can
actually calculate the position vector r,
which would be equal to C to A 60 i 25 minus
minus 50 plus 75 j plus 0 k. Now this is a
position vector. This is in millimeter and
if you want to convert that to the meter because
we want to calculate the units as newton meters,
I believe this is a newton. So it would be
0.06 i, 0.075 j plus 0 k meter. So I have
my force and position vector returned in Cartesian
vector form. And if I have that, then all
I have to do is to find the cross product.
So if I go back to my problem, all right,
I'm going to take this off but I'm going to
rewrite before I go these units, r is 0.6
i plus 0.075 j plus 0 k is my r. Similarly,
my F, which is the force, equal to 0 i minus
200 cosine 30 degree j plus 200 cosine 60
degree k. So I got all the stuff and all that
is left for me to do is to find r cross F.
All right? I want to make sure you understand
it has to be r cross F, not F cross r. So
we have our force vector. We have our position
vector. And we are crossing them and r cross
F should give a moment vector, which is moment
of this force about point A. All right? Since
it is in 3D, instead of doing all the details,
I am going to put this into the form that
I talked about, i, here comes my r, which
is the x component 0.06, that's my unit vector
i, j 0.075, k I have a 0. So this row belongs
to the components of position vector. The
next row belongs to the components of force
vector, which in this case is 0, negative
200 cosine 30, remember this is just one component
of y, and the last one component, I'm going
to squeeze that in here, 200 cosine 60 degree.
And that's my determinant form. Now I'm not
going to spend time expanding this because
you should know this. If you do not, look
up another lecture where I have explained
this one. And all I have to do is to evaluate.
And if I evaluate r cross F, it is my moment
vector about point B and that happens to be
7.5 i minus 6 j minus 10.4 k. This is the
answer and the answer is not complete without
units and the units in this case is newton
meters. So moment of a force about this point
has been calculated. The first step, define
your position vector and the force vector
in Cartesian vector form and then simply apply
the math or vector algebra and you should
get the answer. All right? That completes
our discussion on three-dimensional moment
calculation as well as on vector product and
now we will move on. Thank you.
