Welcome back to our series of Statics examples,
Jim Kerns, and uh, welcome back to Lawrence
Tech in sunny Southfield MI.
Beautiful day outside, its actually above
zero, Fahrenheit.
Ha-ha.
OK.
First of all, I want to take just a moment
to show how a couple is a free vector.
That seems to confuse students, so we'll look
at that carefully.
And what that means is we can then move a
couple around and basically apply it anywhere
once we've found it.
Then we'll resolve all of the couples shown
on that diagram to the left there and we'll
use different methods.
We'll do the perpendicular distance, we'll
put it into Cartesian notation and find the
distances, and we'll do a cross product just
to see the different methods.
And then sum them up to get the total moment
or the total moment from the couples on that.
Now recall that a couple results when you
have a pair of forces that are parallel, non-collinear,
and opposite direction of equal magnitude.
So by inspection we can see that there is
no net force in either the x or y direction
and obviously not in the z direction but we
do have a moment from each one.
So, lets pick and arbitrary point, we'll call
it point O, we'll put it over here on the
left and let's find the moment due to each
of those forces, well call those F. F1 and
F2.
Now I need the distance from the, our origin
here to those points so ill draw a position
vector from O to F1, I'll call this R1, and
rather than drawing on top of that I'll just
continue on with the second position vector
from F1 to F2 and we'll call that R2.
Okay.
So the sum of my moments it equal to R1 times
F, and that is clockwise so that's a negative,
plus the moment of, we'll just add the vectors
R1 + R2 times F and that is counterclockwise
to that's positive, so if I put those together
and solve for F, I get that the sum of the
moments is equal to (-R1+R1+R2) times F, and
of the course the negative R1 and R1 cancel
each other out so I'm left with R2 times F.
Okay.
And that's my moment.
And you'll notice that no matter where I put
this origin, I can move it up, up and down,
I can move it right and left, I can put the
origin, my point O, anywhere and I still end
up with the moment being simply R2*F, you
know the vector from F1 to F2 is our position
vector times the magnitude of that couple
you want to ask and that's it.
It doesn't matter where I place the origin,
it doesn't matter at what point I calculate
that moment about, so that's why it's a free
vector because I can move, I can move the
point at which that couple is applied anywhere
on the page, up down right left.
So, I hope that clears up any confusion why
we consider a, a couple to be a free vector
that you can apply anywhere.
So let's start working on the vectors.
I going to start with vector B.
We see from the picture above that vector
B has a magnitude of 10 newtons and you know
one end is 5 meters from the end of the block
and the other is 1.5 meters so I've got a
basically 3.5 meters between them, so let's
work out what the moment needs to be.
So we can say that moment B equals our 10
meters, no our 10 newtons times the 3.5 meters
and that gives us a moment of 35 Nm, and that's
going to be counter clockwise, or positive,
or a plus Z so it will be coming out the page
of us, at the page.
Coming out of the page at us in the z-direction.
Moving to the moment for A which is in blue,
one at each of that, that thing.
We can say that the, uh, there's a couple
ways we can do, but were going to use Cartesian
notation for the one rather than trying to
figure out the perpendicular distance which
we could do, we could use cosines and things
to figure out what the distance is and then
work it out but were going to use Cartesian
notation so I'm going to break it down into
the X and Y components.
So I'm going to have a, a Y-component for
A, and I'm going to have an X-component for
A on each end and then we can figure out the
moment of each of those.
We can say that Ax equals -5 because its pointing
to the left.
I'm looking at the right hand version of the
moment.
-5 times the cosine of 15 degrees equals -4.83
Newtons, and Ay is equal to that 5 and that's
positive Y direction.
5 times the Sin of 15 degrees and that is
equal to 1.29 newtons.
So looking at the x component I have the 4.83
Newtons is the force the distance between
the two x components if I were to extend those.
Let me draw those here.
The difference between that line and this
lines is 1 meter.
Given the distance of 1 meter and that force,
my moment in the X-direction from A vector
is equal to that 1 meter times the 4.83.
And actually its negative but that's okay.
And we get a clockwise rotation because the
top force is pushing the right, the bottom
force is pushing the left.
So that's clockwise so that gives a moment
that is negative.
-4.83 newton meters.
Ok.
And M-ay in the y-direction equals the um,
have a distance of 5 meters, one to the other,
5 times the magnitude of 1.9 equals um, 9.15
and that is counterclockwise so that's a positive
moment.
Newton meters.
And that vector will be pointing in the positive
z-direction of the vector form.
So let's, let's look at C now and we'll do
this in the cross product form okay?
So the first thing I want to do is write the
C vector in a Cartesian format.
So my C vector, I'll put the vector signs
on afterwards, equals 12 times the cosine
of 30, um, times i plus 12 times the sin of
30, times j plus, it's two dimensional, so
zero times k.
And that's our, our vector C there.
Our vector in this case is going to be equal
to 1i plus 0k plus, 0k because they're one
meter apart in the x-direction they're aligned
horizontally so there's no y-component and
there no k-component and to calculate the
cross product I'm going to say that my moment
for C is equal to our vector for C times the
um, cross F of C. and give me just a moment.
We want to calculate our moment M as R crossed
F and that's equal to the, we can calculate
that by determining, by evaluating the determinate
of “i”, “j”, “k”.
And our R vector here we'll put in as, um,
the R vector was 1, 0, 0, and our F vector
we determined was, let's see, 12 times the
cosine of 30 degrees “i” which comes to
10.39 and “j” was 12 times the sin of
30 which is 6 and “k” was 0.
So we can evaluate that determinate and what
we end up is that our moment for vector C
is equal to, well if I look at it here, if
I look at the i-component I’m going to do
the determinate of the 0, the 0, minus the
0 times 6, so that’s going to be zero.
The “j” component is 1 times 0, minus
0 times 10.39 those are both zero.
The “k” component is 1 times 6 minus 0
times 10.39 so that’s just going to be 6k.
So we end up with 0i + 0j + 6k as my component
there.
And then I can just add all three for those
couples together and it doesn’t matter that
their applied at different places on that
because we defined that the couple is a free
vector so that I can say that the sum of my
moment total and these are all about the z-axis
because they’re all, this is a two-dimensional
problem here, the moment about the z-axis
is equal to 35 newton-meters from vector B
minus 4.83 newton-meters, from the X-component
of A, plus 9.15 newton-meters, from the Y
component of A and, 6 newton-meters, from
the K-component.
Or, the K-component, yeah from the K-component
from moment C and that’s equal to, if I
did the math all correctly here.
45.82 newton-meters.
So that should uh, I think that covers that.
It’s straight forward once you get the hang
of it and once you get past this concept that
the free vector and the fact you can slide
it around every that makes it reasonably straight
forward.
We covered three different methods for calculating
these.
F is a completely different 3 dimensional
problems obviously the, using the cross product
is the simpler way to do it, but I wanted
to show them all for completeness.
I hope that helped you all some.
Thanks for watching and stay tuned.
