Let’s talk about moments, maybe you heard of it as torque. In simple terms, it’s when we apply a
force and try to turn it about an axis or a specific point. Imagine a wrench like this,
and we apply a force right here. Now, the moment created about the zed axis is the force we apply
multiplied by the perpendicular distance. So if our force, or the perpendicular distance is large,
then the moment created about the zed axis is larger, and vice versa. If we apply our force
straight along handle, then you can see, that no rotation would occur. Let me draw this line. This
is called the line of action. Because the line of action of our force intersects the zed axis,
it can’t create a moment. Now let’s say, we apply a force at an angle like this. In this case, we
can break this force into x and y components. So the only component that does work in this example,
would be the x component, since the y component is along the line of action. When we apply a force,
the moment vector can be found using the right-hand rule. So here, we are applying
a force counter clockwise, so the moment vector is straight upwards along the zed axis.
In 2 dimensional problems, all we do is multiply the forces by their respective perpendicular
distance to figure out the moment. The units of a moment magnitude is force times distance,
so it’s N meters. If we have multiple forces like this, we can find the resultant moment by
figuring out each individual moment created, and then adding it up. When I go through examples,
this will become much clearer. Now if we need to consider moments in 3 dimensions, then we need
to use the cross product. We find the moment by taking the cross product between the force
and a position vector. The position vector can be from the location where we are finding the moment
about to anywhere along the line of action of the force. The force must be expressed
in cartesian form. That means you need to be able to do that, and if you’re unsure,
or forgot, please check the description for forces along a line, where I explain how to do this step
by step. You also need to know how to take the cross product, again, there is also another video,
where I go through this step by step, so please take a look at those if you forgot. Now,
you will understand these concepts much better with examples, so let’s get started.
Let’s look at a 2D moment problem. So we need to find the moment each of the three forces create
about point A. The first step is to break each of these forces into their x and y components.
Let’s start with force F1. Next force F2. And lastly, we have force F3.
Now, remember, a moment is force times the perpendicular distance. You can see that the
x components of both force F1 and F2 will do no work. That’s because they both go through the line
of action. So if we apply these horizontal forces, the bar won’t spin about point A, which means we
can get rid of them. Now, let’s look at the y component of force F1. The perpendicular distance
from point A to the y component is 2 m. So the moment is then the multiplication of the two.
We also need to show which direction the bar would turn if the moment is applied.
Imagine this bar is free to rotate about point A. If we apply the y component of the force,
notice how it would turn clockwise. So that’s our direction. Now we look at force F2.
The perpendicular distance to the y component is 5 meters. The direction is the same since we see
that it would spin clockwise. Now let’s look at force F3. Now both components of force F3
will create a moment. If we look at just the x component, the perpendicular distance would be
4 m. If we look at the y component, the perpendicular distance is 5 m.
Now, we need to be careful. The x component would cause the bar to turn counterclockwise,
while the y component would cause the bar to turn clockwise. So assuming clockwise to be positive,
the moment created by the x component will be negative, while the moment created by the
y component will be positive. Since our answer is positive, and we assumed clockwise to be positive,
then force F3 would create a clockwise moment about point A. Those are our answers.
Let’s take a look at this problem where we need to find the moment created about point A, and we
also need to find another force applied at point C to create the same moment. So first, we will break
the 70N force into it’s x and y components. Now, let’s assume clockwise to positive. So
the moment at A can be written is the force times the perpendicular distance.
So for the x component, the perpendicular distance is 0.9 m, and for the y component, it’s 0.7 m.
The next part of our question asks us to find another force that would create the same moment.
If we apply a horizontal force at C, we can create a moment about point A.
The perpendicular distance would be 0.9 m. Let’s solve for the force. So this tells us that if we
apply a 82.1 N force at point C horizontally, it’ll create the same moment about point A.
Let’s take a look at a problem involving moments in the x y and zed plane.
We have a bent rod with a force is applied at the end, and we need to figure out the moment
created about point O. So to find the moment about point O, we need to use this equation.
So we will take the cross product between the force and a position vector from O to C.
So that means we first have to express the force given in cartesian form.
Before we do that, let’s write down the locations of all the points of interest.
Now, let’s write our position vector from A to C.
The next step is to find the magnitude of this position vector. Using the magnitude,
we can find the unit vector. So finally, to express our force in cartesian form, we multiply
the magnitude of the force by the unit vector. Next, for our equation, we need a position
vector from point O to point C, which is where the force ends.
Now we can take the cross product. That is our answer.
Let’s take a look at this problem where we need to find the moment created about point B
when a force is applied at point C. Since the force is already given to us in cartesian form,
we can set it aside. The only thing we need is a position vector from B to C.
So let’s write down the locations of points B and C.
Now we can write a position vector. Now we just take the cross product
between the position vector and the force. And that’s our answer.
Let’s take a look at one last example where we have to calculate the resultant moment
created by 2 forces effecting the flag pole. First step is express both forces in cartesian form. I
am going to do this quickly. So we need 2 position vectors, one from A to C, and another from A to B.
Next, we calculate the magnitude of each position vector. Now a unit vector for each cable. Lastly,
we multiply the magnitude of the force by the unit vector and we get our forces in cartesian form.
Now we need 2 position vectors. I am going to pick the 2 vectors from O to B and O to C,
though you can pick a position vector from O to A as well. You will still get the same answer,
since the position vector can be from point O to any location on the line of action of force A.
Now we take the cross product for each force.
Since we are looking for the resultant force, all we need to do is add these to values together.
So remember, when we add, all we do is add I components to I components, j components to j
components and k components to k components. That is our answer. If you did a position vector from
O to A, it’ll be simpler since you don’t need 2 position vectors, but I wanted to show you can
actually can just take the moment from O to where the force is ending which were points B and C.
That should cover the types of problems you will face with finding moments.
I hope this video helped and thanks for watching. Best of luck with your studies.