Welcome back to this example.
In our series of examples of doing types of
problems for our statics class.
Here we’re going to look at the equilibrium
of a particle in 3 dimensions.
The physical arrangement of the system is
over here on the left.
I have a boom here going from point A to point
B. I have two cables in green from A to C
and A to D. Those are supporting that load
there, in whatever color that is.
And I have a free-body diagram drawn over
here where I have that Point A right here.
And I have the various forces: the force in
the direction towards C, the force in the
direction towards D, the force in the direction
towards B and the force due to the mass and
gravity pointing straight down.
So looking at this problem, I see three unknowns-
right there- three forces.
So we know that we have, in a three-dimensional
case like this, we have 3 equations: the sum
of the forces in the x-direction, the sum
of the forces in the y-direction, and the
sum of the forces in the z-direction.
So I’ve got 3 equations, 3 unknowns: it’s
fairly straightforward.
So what I need to do is find an expression
for each of these forces.
And looking at the problem and thinking about
how I want to do that, it’s pretty obvious
to me- I want to express those forces in Cartesian
coordinates.
And, even though I don’t know the magnitude
of each of those forces, what I can do is
write them in Cartesian format, or have the
x and y components and z components based
on the direction those vectors take, and the
unknown which we can get from the position
vectors.
If we look at our original diagram, we have,
for example, position vector from point A
to point C. It doesn’t go anywhere in the
z-direction, in the y-direction it goes a
plus (+) 4 meters, in the x-direction it goes
a minus (-) 6 meters, because this here is
6 meters out.
You know, I would draw that…
Well, we will write that out in just a moment.
So, we’ve got a plan of action we’re going
to write out the position vectors.
Given the position vectors, we can find the
unit vectors in the appropriate direction,
multiply that by our unknown force, and get
an expression for each of the three forces.
We can sum the forces in various directions,
making it equal to 0, and we should have an
answer before very long.
Let’s start out by looking at the vector
from A to C- from here to here.
I’ll just write that as C. I’m going to
erase these words here just to give me a little
more room.
So the vector from A to C- we’ll just call
that C for the sake of less writing- we go
a minus (-) 6 in the x-direction, so minus
(-) 6i.
Then we go a plus 4 in the y-direction, so
that’s 4j: that’s our position vector.
We also know that the magnitude of C is equal
to the square root of 6 squared plus 4 squared.
That’s equal to 7.21.
I can convert that into a unit vector in the
C-direction by dividing by that magnitude,
and I get minus (-) 6 over 7.21, which gives
me minus (-) 0.832i plus 0.555j.
These are all, of course, our vectors.
I can write my force in the C-direction, which
I need to solve our problem.
F sub C, add the vector, equals the magnitude
of the force total, times 0.88 minus 0.832
F sub C (the magnitude in the i direction),
plus 0.555 F sub C (the magnitude of the total
force) times the j vector.
So now I have an expression for F sub C. Similarly,
in the D-direction, the position vector is
equal to minus (-) 6i minus 3j plus 1k.
You can stop the recording at any time.
My magnitude is 6.78.
You can pause and go back and see where I
came up with those numbers.
If you want, you can follow along with pencil
and paper.
At some point, we are going to get into a
bunch of Algebra, and I’m going to kind-of
breeze through that.
So, if you’re following along on paper,
you can work it out yourself.
My unit vector in the D delta direction, again,
equals minus (-) 0.884i minus 0.442j plus
0.147k.
My force in the D-direction is equal to minus
(-) 0.884 Fd times i, minus 0.442Fd times
j plus 0.147Fd times k.
I kind of ran out of room: sorry.
It’s the same process for the force in the
B-direction.
So, I come up with my force sub B is equal
to minus (-) 7.68 Fd i-direction minus 0.640
Fd j.
So I have my forces, and I also have my other
force, the force due to the mass, and that’s
just going to be the 200kg times the 9.81,
and that’s going to be in the z-direction.
So, I’ve already written my equilibrium
equations here.
Okay, so in the y-direction, I’ve got to
do it in the other order here, when I write
this- just convenient.
So I’ve got 0.555 times Fc, which we get
from right here.
That’s in our j or y direction.
Minus 0.442 Fd which was from right here.
Excuse me, I’m sorry, right here.
This was unit vector not the force vector.
And those are the only two who have a component
in the y-direction, so I can solve for Fc
equals 0.796 Fd.
If I go down to my sum of the forces in the
x-direction, which components, we can see
that AB has an x-component.
AC and AD all have x-components and they all-
I’ll circle them here- that one.
Again, I’m on a unit vector.
That one and that one.
I’m going to do this in the next slide.
That is equal to minus (-) 0.832 Fc minus
0.884 Fd.
I should make that be consistent with my…
Minus 0.768 Fb.
So all of those unit vectors point in the
negative x-direction.
I can take my value I have here, substitute
that into Fc to get this in terms of Fd, combine
my Fd terms, and, again, if you want to do
this on paper, you can do that and solve.
So I get basically minus (-) 1.54 Fd = 0.768
Fb and that leaves me with Fd equals minus
(-)0.499 Fb, okay.
And I can see that I can substitute that into
here in just one more step.
So this step here, my last equation, the equilibrium
in the z direction, I get 0.442 Fd minus 0.640
Fb minus the force due to the mass, I have
200 kg times 9.81.
That gives me 1962.
I was 10 years old then.
In terms of the force in the z-direction from
that mass.
If I substitute this into here, so I just
have two terms of Fb and a constant, I can
solve for Fb equals minus (-)4570 Newtons.
I can substitute, now that I know Fb, I can
use this equation to find Fd.
Fd equals 2280 Newtons and, going back up
to this equation there, I see that Fc equals
1810 Newtons.
And that’s all there is to it.
Now, if the problem had been stated differently,
and if my objective was not to find the load
in each column and at a given mass.
If I was given the tensile strength of each
element, and compression strength.
By the strength of each element, and if I
was to find the mass, I would’ve gone through
exactly the same process, except, at this
point, instead of having 1962, I would have
left that as an unknown and, essentially,
I could’ve find out the force in terms of
the unknowns.
I could find out which of these was the larger
force and substitute in the actual tensile
strength there and solve for that unknown
mass.
So that would be the same problem, just getting
a different answer and different solution.
Just as a bonus there.
I think that concludes this example.
They really are really straightforward if
you go through the steps one at a time.
Just keep that process in mind.
Get things into Cartesian coordinates.
Do the sum of the forces in each direction
in equilibrium.
It’s just a matter of chugging through the
algebra.
Hopefully, you found this helpful.
I guess if you didn’t you wouldn’t find
this helpful.
Thanks for watching!
