Let’s talk about simplifying forces and moments. Imagine a metal post like this and we have a few
forces effecting it. Let’s say we want to find an equivalent force and couple moment
acting at point A. So in simple terms, we want a single force and a single moment that represents
all the forces and moments effecting this post. For forces, all we need to do is break each of
them into x and y components and then add it up. For moments, it’s pretty much the same,
we figure out the moment created by each force about point A, and then add it up. Then we can
show a single resultant force and a single couple moment that accounts for all the other forces.
If we have an object in 3D space, the process is similar. We now express our forces in cartesian
form and add it up. To find the moments, we use our trusty equation, and then once
we figure out the moments created by all the forces, we add it up. Now, lets go back to our
metal post. We can actually go one step further than showing a resultant and a couple moment.
We don’t even have to show a moment. Instead of showing the couple moment, what if we can
place our resultant force anywhere on the vertical post to account for the moment as well? In simple
terms, we can show a single resultant force and place it somewhere on this vertical section which
will create the same couple moment we had before, because remember, its force times perpendicular
distance. For example, if we had a couple moment of 20 Nm and our resultant force only
had an x-component of 4N, then all we need is to place the resultant force 5 m away from point O,
and that would create a moment of 20 nm. So now, our resultant force accounts for the moment as
well. We will do this in the very last example to see how its done. Now let’s get to some examples.
Let’s take a look at this problem where we need to replace the forces and moments effecting the beam
by an equivalent resultant force and couple moment at point O. Let’s start with the forces. The 450 N
force can be broken into it’s x and y components. The 200 N force only has a y component so we don’t
need to break it into sperate components. Let’s add up the forces. We will pick down and to the
left to be positive. First the x components. Now the y components. Note that since we picked down
to be positive, the 200 force, which is facing upwards will be negative. Now we can find the
resultant force by finding the magnitude. We also need an angle to show the direction
of the resultant force, so let me draw out the components. Using tan, we can find theta. That’s
the force taken care of, let’s move onto moments. Each component of the forces will create a moment.
Let’s look at the x components, which is only one force. The perpendicular distance is 0.2,
so we multiply the force by the distance. This would create a counter clockwise moment. Next,
we look at y components. We have 2. Looking at the first force, we have a perpendicular distance
of 1.5m. This would create a clockwise moment. Next, we look at the 200N force.
The perpendicular distance is 3.5m. This would create a counter clockwise moment.
Now there is one more moment that we need to keep in mind, which is already given to us.
We will assume clockwise moments to be positive and add it all up. So note that all the counter
clockwise moments will be negative since we picked clockwise to be positive. Those are our answers.
Let’s take a look at a problem involving vector analysis. Same as before,
we need to find an equivalent resultant force and a couple moment at point O.
We will start with the forces first. Force F1 only has a zed component, so we can write
it like this in cartesian form. Force F2 only has a y component so we can write it like this.
Force F3 is already given to us in cartesian form. Now we can find the resultant force by adding it
all up. So remember, all we do is add the I components to I components, j components to j
components and k components to k components. Next, we look at the moments created
about point O by each of the forces. So for force F1, we need a position vector
from O to where force F1 is applied. For F2, we need a position vector from O to where force 2 is
applied. Lastly, for force F3, we need a position vector from point O to where force 3 is applied.
Now, we can find the moment by taking the cross product. Let’s start with force F1.
Next, force F2. Lastly force F3. Now to find the resultant moment,
all we do is add it up, just like when we did the resultant force and that’s our answer.
In the previous 2 questions, we showed a resultant force and a moment about a specific point.
Now we are going to go one step further, and show the resultant force and place it at a location so
that it accounts for the resultant moment as well. So let’s take a look at this problem where we need
to represent the loading on the frame with a single resultant force. The method is similar
to before. Let’s start with the forces, first the x axis forces. We will pick right to be positive,
and there is only one force. Next, y axis forces. Using this, we can find the resultant
force. We also need an angle to show the direction. So let’s draw the components.
Using tan inverse, we get the angle. Now, we will set that aside, and look
at the moments created by the forces. Since the question is asking us to find the line
of action from point A, we will calculate the moments created about point A. The first 200N
won’t create a moment, since it’s line of action intersects point A. The rest create moments,
so let’s figure it out assuming clockwise moments to be positive. This tells us the
resultant moment created by all the forces. Now going back to the resultant force we found,
we need to place it somewhere along this vertical bar so that we can account for the resultant
moment. Since the y component of our resultant force goes through the vertical line of member AB,
we can ignore it. So where can we place our x component to account for the moment?
Well, we will assume this length is x. The moment created would be 600 times x, and we
just need to equal that to the resultant moment. So what we are saying is that if we place our
resultant force at this location, it will account for the resultant moment that’s created as well.
So in simple terms, we can replace all of the forces that’s on this beam
with this single resultant force and it’ll be the same. And that’s our answer.
That should cover the types of problems you will face when it comes to simplification of forces and
moments. I hope this video helped and thanks for watching! Best of luck with your studies.