Welcome back. Today we're going to
look at
calculating the moment about a specific
axis,
In this case the z-axis. We want to know
Kind of moment we're creating
about that axis when I apply a force
at the end of that wrench.
I in a previous video, and I’ll put a link up
in the corner if you wanna go see that,
we determine the the moment
about the origin; about this point.
Which is fine - that's good
information and I
repeated the values we came up
with there.
But that's not the moment around that
specific axis – that’s what we want to know - 
what's the torque on the bolt we’re
applying not
what's the torque and direction
perpendicular to our force
and position vector plane. So
this moment, as you can see,it's got
a big z component but
there also is a negative Y
component and a
positive x component. So it kind of tilts 
of a bit
Now, this a particular example is
kinda trivial, but we'll walk through the
process
Anyhow. And will also run it backwards.
Say, if I want a
specific moment,
what does the force need to be? 
We will run it both ways.
um because we're looking at the moment
about an axis the math becomes very
But it’s worth walking it through. Now
if you recall,
In the previous video I calculated my moment
using the cross product between R
and F. And the moment of course is in Newton
meters. I shoulda put
the units up there, because our position
vector is
in meters an our force is in Newtons.
so the moment is Newton-meters. When we
calculated that moment
by evaluating the determinant,
and these are vectors,
that determinant: I J K
and these are all vectors, now we have the R
vector and the M vector, and we evaluate the 
determinate.
so if you don’t recall that, just go
back to last video.
Given that I found this moment
already,
how do I find the moment about a
specified vector? Well
about a specified axes I mean.
Well obviously, we need to project from
this M0; we need to project that onto the
z-axis.
And as we discussed, when we project…
I say project,
you say… yes, dot product. So
my moment about the z-axis, and
because we are
42
00:02:41,769 --> 00:02:46,470
 using the dot product, this will be a
scaler value, is equal to
the M moment about our point
(that vector which we calculated) times
a unit vector in the z-direction - 
- dot, not times, dot; the m0 dot Uz
and then of course Uz is
0 I plus
0 J plus one K
So that's why this becomes,
those are vectors, this one becomes
kind of trivial.
But if I do that dot product,
I do the dot product between this
and this, I get my moment Z
is 0 I dot
times the 9.18 I so that's just
0, and in the J we had zero J, and we have
1.36 well that's still 0,
and we have 1 K times 2.60,
plus 2.60,
and that gives us, of course, 2.60
Newton-meters. That’s kind the two-step process.
Ideally we could do this in one step. If
we want our
moment Z equal to
our unit vector
in the z-direction dot the moment about
the
0 axis… that is equal to
U dot,
U sub Z, dot R cross F,
and these are all vectors.
okay? So I can write that
as a unit vector along the Z axis
U sub z
dot that determinate that we
calculated.
And if I did all the math and worked it all
out, what I get is
is instead of the I J K
up here, I can put my, I can
do the dot product between my unit
vector
and those component vectors,
and what I end up with is just the
magnitude of
the things on the unit vector, so I end
up in the top,
I have, 0 0 1;
which is my unit vector dot I J K.
And then I have the values
of the R vector which is
.2 .02 and negative .06,
00:05:23,240 --> 00:05:26,380
.2 .02 and negative .06, and I have my
force vector, which we calculated as
negative 2.26,
and these columns aren’t lining up very well,
12.8 and 7.5, that's that
determinant.
And let me just clean this up just a
little bit,
not much because
I wouldn't you think that I have
good handwriting. 
this is zero, that's a one.
Yes, I can evaluate that determinate
easily enough,
and that's just, well I'd start out with
0 times the determinant
of this, right? But that's going to be 0
because I don't have, I’m multiplying by 0.
Again a zero and the determinate
of these terms
and these Terms, that's gonna be 0. What I
need,
I can do is just, is just the last
piece
is the that value
times the determinant this 2 by 2,
We have this equal to 1 times this
.2 times 12.8
minus
point 02 times
-2.2 6
and that is equal to
2.56 plus
.452
and that gives me my 2.60
Newton-meters as the moment about
the Z axis. Okay. And that’s same answer we
got
back here; 2.6 there. So that
that works out. Now let's look at
this problem a different way -
kind working backwards. Suppose
I don't know the force,
the magnitude of the force, I know the
direction.
And I want to find the force required to
give me that 2.60
2.60 Newton-meters
moment about that the axis. So I want
to work this backwards, right?
The problem here is find the force F
pointing in the direction specified by
those
projection angles, given
a moment Z of 2.60
2.60 Newton-meters. Well my
problem again, I’ll write it out,is my
moment
in the z-direction is equal to the unit
vector in the z-direction
 dot R, my
position vector, cross F
my known force,
but with unknown magnitude. So here I
know the direction,
but the magnitude is
unknown. Okay?
So I need to sort out
the force F in Cartesian coordinates.
I know that my F sub z is going to be
equal to
the unknown magnitude F times
the sin of
30 degrees, and that's going to be
5.00 F.
And I can project that down
on to the XY plane, or
parallel to the XY plane, and I get
this F prime here.
F prime equals F
cosine(30).
And then given that projection down onto
the,
down parallel to the XY plane, I can
find the X&Y components pretty
straightforward. So
given that projection down parallel to
the XY plane I can find
Fx equals,
and because this is going in the minus
x-direction, negative F cosine(30)
times the sin of this
10 degree angle here, and that’s
equal to .150
.150 F and
the y component equals. and this is
going to be positive, F
cosine(30) from my projection, times the
cosine(10)
cosine(10) degrees. And that’s equal to
.853 F
Now I plug that into my determinant, er,
I N
my equation, 2.60
Newton meters equals the
U dot R cross F, which I can write as
0 0 1
0 0 1 my R vector,
we know from above was .2 
.02 and negative .06
and my F vector is
negative .15 F
-.15 F um,
.853 F,
 and I know that's hard to read my pen
is getting kind of wonkey here,
and ah, .5 F
and I can evaluate that
determinant, and I come up with,
a thing, you know these terms are zero, so
it's just
this term times the determinant of
this here. And if I run through the math, I
just get
2.60 Newton meters
equals .1736,
this would be meters, times
F, which is Newtons, and solving for
F
I find that F equals
14.98
Newtons. And if you go back and double
check,
we originally specified the F as
15 Newtons. So we’ve kinda come full circle,
and got the same result
doing the the problem in both directions.
Right?
Using this determate method is
pretty straightforward.
It simplifies the math, you know trying
to do this with scaler values gets messy.
I'm just remember it’s R cross F
so this… and we do the dot product,
we put our unit vector in
the top row,
put our R vector here, our position
vector,
and our force in that row and
it all works out. So
thanks for watching, and stay tuned for
the next one.
Bye.
