In the previous video we talked about finding moments about a point. But let’s say we need to
find it about a specific axis. Imagine we have a box like this with 3 forces effecting it and
we want to find the moment about the x axis. The scalar process is similar to before, but instead
of thinking of a singular point, we need to think of the whole x axis. Let’s look at just force F1.
When we apply it, we see that the box will turn about the x-axis. It doesn’t matter if
we move this force along the right edge of the box because the moment that’s created about the x-axis
is based on the perpendicular distance from the axis itself. So the moment created by force F1 is
simply the force times the perpendicular distance to the x-axis. If however, we move the force down,
then the moment created would be less since the perpendicular distance is also less. Now if we
look at force F2 and apply it, we see that it would turn the box like this. That moment
is force F2 times the perpendicular distance. Again, if we move this force along the left edge,
the moment will be the same but if we move it along the right edge, the moment created would
change. That’s because moving it along the right edge changes the perpendicular distance to the x
axis. Lastly, we have force F3. Notice that by applying this force, the box won’t turn about
the x-axis. That’s because the line of action of this force is parallel to the x axis. So anytime
a force is parallel to the specified axis, it can’t create a moment about that axis. Now let’s
think about 3D problems and finding moments about a specified axis using vector analysis.
So imagine a pipe system like this and we apply a moment here and we want to find the moment that’s
created along this axis. We can build on what we learned before. Remember we used this equation
when we did a vector analysis. We did the cross product between a position vector and a force. Now
that we are trying to find the moment about this axis, all we need is a unit vector that defines
the direction of this specified axis. We take the dot product between a unit vector and the
result of the cross product, which we can write like this. The rest is the same, so we’d have
a position vector from any point on the AB axis to any point on the line of action of the force.
And as before, we need to make sure our force is expressed in cartesian form.
So really, the only new part is that instead of using I,j,k at the top, we use the x y and zed
components of our unit vector for the specified axis. One thing to keep in mind is that sometimes,
you will get a negative answer when you solve for the moment, all it means is that the moment acts
opposite to the direction of our unit vector. Without further ado, let’s take a look at a few
examples to see how to apply what we learned. Let’s take a look at this problem where we have
a force applied at one end of this contraption and we need to find the moment created about
the x-axis. We are going to use vector analysis since the force is given to us with coordinate
direction angles, so it’ll be easy to express it in cartesian form. So we take the magnitude of the
force and multiply it by the cosine of each angle. Notice that the angle for the x-axis is given to
us with respect to the negative x-axis, so we need to use 120 degrees for the positive angle. If this
is unfamiliar to you, please see the description for the video on cartesian vector forces.
Next, we need a position vector from A to B. Note that we converted mm to m.
Lastly, we need a unit vector for our specified axis. In this case, since we are trying to find
it about the x-axis, our unit vector is simply 1,0,0. For future reference,
the unit vector for the y axis would be 0,1,0 and for the zed axis, it would be 0,0,1.
Now we plug it into our equation. If you are unfamiliar with solving this,
please check the description for solving a scalar triple product. All we are doing is
finding the determinant. Solving gives us the moment created about the x-axis.
Let’s take a look at this problem where we have a concrete slab.
There are 3 forces effecting it and we need to find the moment created about the y axis.
We will use a scalar analysis for this problem. First, we know that force F1 won’t create a
moment about the y axis since it’s parallel to the axis itself, so we can get rid of it.
Now let’s look at force F2. The perpendicular distance to the y axis from force F2
is 1 m. Now we are going to define the positive direction of a moment using the right-hand rule.
So in simple terms, we will assume that if the slab turns counter clockwise about the y axis,
it’s positive, and if it turns clockwise, it’ll be negative. So for force F2, we have force times
the perpendicular distance, and we see that it would make the slab turn clockwise. So it’ll be
a negative moment. Next, we have force F3. The perpendicular distance to the y axis
is 1.5 m. This force would also cause the slab to turn clockwise about the y axis,
so it’ll be negative. The resultant moment is then the addition of the two moments.
Since we got a negative answer, that means the moment vector is facing the negative y direction,
so if we use the right-hand rule, we see that the slab actually turns clockwise about the y-axis.
If you need a refresh about the right-hand rule, please see the video about moment of a force.
Let’s take a look at one last example. In this problem, we need to figure out the
moment created about an axis from points A to B. Since force F is already given to
us in cartesian form, we can set it aside. Next, we need a position vector from A to B
or C to B, both will work since our position vector needs to be from anywhere along the
specified axis to the force. I am going to use A to C since that’s easier to write but I encourage
you to try it with a position vector from A to B as well to see that you will get the same answer.
Lastly, we need a unit vector from A to C. So to get a unit vector, we first need
a position vector from A to C. Next, we find the magnitude of this position vector. Lastly,
we divide each component of our position vector by the magnitude and we get our unit vector.
So now we can plug it into our equation. Solving gives us the moment about the AC axis.
If you used a position vector from A to B, the only part that changes would be the middle row,
which are the components of the position vector.
That should cover the types of problems you will face in this chapter.
I hope this video helped you and thanks so much for watching. Best of luck with your studies!