### Video Transcript

In this video, our topic is moments
in 2D. And even though we do have these
characters drawn in two dimensions, who are enjoying a moment, of course the moments
weโre talking about in this lesson are different. Weโre going to see that they
involve forces, distances, and points or axes of rotation.

To get started, say that we have
these two things, a force ๐น and a point weโll call ๐ over here. Looking at this force, we see that
it acts along a certain line, that is, in a particular direction. The name for this is the line of
action of the force. And we see that this line does not
pass through our point ๐.

So letโs do something to connect
our point ๐ with our force ๐น. Letโs draw in a rod, which weโll
say has negligibly small mass. But one end of this rod is fixed to
point ๐, and the force ๐น acts on the other end. And we can see now what this force
๐น will tend to do to the rod about point ๐. It will tend to make it rotate. The tendency for this force to
cause a rotation about the point is called the moment of force. And we represent it with a capital
๐.

Even if this term is new to us, the
idea of something like this, what weโre calling a moment of force, is actually
familiar. For example, imagine that point ๐
is the hinge of a door and that our orange line represents a top-down view of the
door. In that case, it will be natural
for us to push or pull on this other end of the door โ we could call it โ in order
to open or close it. In doing so, the force we exert
creates a moment.

When it comes to calculating the
moment of force, there are a couple of different ways to do it. To see how this is so, letโs
imagine that our force ๐น is being exerted at an angle of 45 degrees to the axis of
this thin massless rod. To find the moment this force
creates about point ๐, one way we could do that is to break this force up into its
vertical and horizontal components and then multiply the component of the force
thatโs perpendicular to our rod by the overall length of the rod. So if we say that our rod has a
length ๐ฟ and the magnitude of this component of the force perpendicular to it is ๐น
times the cos of 45 degrees, then our moment would be the product of these two
values.

Recalling that the cos of 45
degrees equals the square root of two over two, this moment simplifies to ๐น times
๐ฟ times the square root of two over two. This method of calculating moments,
where we figure out the distance between the point about which weโre considering
rotation and the force thatโs being applied and multiply that distance by the
perpendicular component of the applied force, is the most common way of calculating
a moment of force in general. But as we mentioned, itโs not the
only way.

Another approach we can take is to
find the perpendicular distance between our forceโs line of action and the point
about which weโre considering rotation. If we extend the line of action
weโve drawn in and then draw a perpendicular line from our line of action that
intersects point ๐, another way to calculate the moment of this force is to solve
for this distance โ weโll call it ๐ โ and multiply that by the force ๐น. And notice that if we do this,
weโre once again calculating a moment arm โ originally, that was the length ๐ฟ and
now itโs the distance ๐ โ and multiplying it by a force thatโs perpendicular to
that arm. So then, what is this distance
๐?

We can solve for it by recognizing
that this angle here must be 45 degrees since this one is. And then since this is a right
angle equal to 90 degrees, the third interior angle in our triangle must also be 45
degrees. And this shows us that we can write
the distance ๐ as ๐ฟ times the cosine of that angle. So in this second way of
calculating our moment of force, weโre multiplying the distance ๐ by the component
of ๐น thatโs perpendicular to it. And thatโs the entire force
magnitude. ๐, as weโve seen, is ๐ฟ times the
cos of 45 degrees. And writing out this cosine as an
exact value, we find that the moment of force equals ๐น times ๐ฟ times the square
root of two over two. And we see now that this matches
the result of our earlier method.

So letโs write our equation for the
moment of force ๐ like this. Weโll say that itโs equal to the
perpendicular component of a force ๐น multiplied by the distance between where that
perpendicular component is applied and some point of rotation. And by the way, this quantity that
weโre calling a moment in this context is sometimes called a torque. So if we do hear this word torque
used, we know that itโs referring to a moment of force.

In general, a moment can be either
positive, negative, or zero. We see, for example, that this
force ๐น would tend to create counterclockwise rotation of this rod about the point
weโve picked ๐. By convention, moments in this
direction leading to a counterclockwise rotation are considered positive. This means that we call moments
leading to a clockwise rotation negative. And as we mentioned, itโs also
possible for a moment to be zero.

Say that along with our point ๐ we
pick a second point โ weโll call it point ๐ โ and we locate it along the line of
action of our force. Because ๐ lies along this line,
thereโs no distance between the perpendicular component of our force and this
point. Physically, it would never tend to
create a rotation about point ๐. And thatโs why it canโt give rise
to a nonzero moment. While the moment of force ๐น about
point ๐ is ๐น times ๐ฟ times the square root of two over two, the moment of ๐น
about point ๐ is zero.

One last thing to point out about
moments is the units that we can expect these quantities to have. We see that a moment is given by a
force, with SI base units of newtons, multiplied by a distance, with SI base units
of meters. Typically then, a moment is given
in units of newtons times meters. But regardless of the specific
units of force and distance weโre working with, moments always consist of a unit of
force multiplied by a unit of distance.

Knowing all this, letโs get a bit
of practice now through an example exercise.

If a force having a magnitude
of 498 newtons is eight centimeters away from a point ๐ด, find the norm of the
moment of the force about the point ๐ด, giving your answer in newtons times
meters.

Alright, so letโs say that this
is our point ๐ด and that a distance of eight centimeters away we have a force
weโll call it ๐น with a magnitude of 498 newtons. Knowing all this, we want to
calculate the norm of the moment of this force about point ๐ด.

To start off, we can recall
that the moment of a force about some point โ weโll call this moment capital ๐
โ is equal to the component of an applied force ๐น thatโs perpendicular to a
line from where the force is applied to the point of interest multiplied by that
distance between the point and the applied force. As we look at our particular
scenario, we see that this force ๐น and the distance of eight centimeters are
already perpendicular to one another. Therefore, the moment of this
force about point ๐ด is equal to ๐น times eight centimeters.

If we substitute in the given
value for ๐น, weโre just about ready to multiply to calculate ๐. But before we do, letโs recall
that we want to give our answer in units of newtons times meters. We see that if we go through
our multiplication as is, weโll have units of newtons times centimeters. So before we go any further,
letโs convert this distance from units of centimeters to units of meters. We can do this by recalling
that 100 centimeters equals one meter, which means that eight centimeters is
equal to 0.08 meters. Now, when we multiply these two
values together, we get a result of exactly 39.84 newton meters. This is the norm of the moment
of the force about point ๐ด.

Now letโs look at an example that
involves a more complicated moment arm.

Define the moment of the force
that has a magnitude of 11 newtons about point ๐. Give your answer in newtons
times meters.

Looking at our diagram, we see
this force of 11 newtons directed downward and we also see point ๐ here. To define the moment of this
force about point ๐, we can recall that the moment due to some force about some
point is equal to the component of that force thatโs perpendicular to the line
between where the force is applied and the point of interest multiplied by the
distance between that point and where the force is applied. Practically, what all that
means when it comes to our scenario is that if we sketch in the line along which
our force acts โ this is called the line of action of the force โ then the
moment of this force about point ๐ will equal the force multiplied by this
distance weโve called ๐.

Looking back at this equation
for moment of force, since we already know the force of 11 newtons, our task is
first to solve for this distance ๐. If we begin at point ๐ and
then consider what we could call the three segments of this apparatus, if we
travel from point ๐ along the first segment and then down along the second
segment and finally along the third until weโve reached the line of action of
our force, then the total horizontal displacement of that three-segment journey
will be equal to the distance ๐ we want to calculate.

To determine this distance,
letโs assign motion to the right as motion in a positive direction and motion to
the left in the negative direction. This means that, on the first
leg of our journey, we traveled positive 36 centimeters. And then on the second leg,
traveling along this 29-centimeter segment, since this segment is inclined 60
degrees below the horizontal, then the horizontal displacement we cover moving
along this segment is positive 29 centimeters times the cos of 60 degrees. And then, lastly, along the
third segment, we travel negative 26 centimeters. And thatโs because as weโve
said motion to the left is negative.

When these three values are
combined, they equal ๐, the distance we want to solve for. Recalling that the cos of 60
degrees equals one-half, the total distance we end up with is positive 24.5
centimeters. This then is the value by which
weโll multiply our force ๐น in order to solve for the moment that force creates
about point ๐. Before we do that though, since
we want our answer in the end in newtons times meters, letโs convert this
distance in centimeters to a distance in meters. Recalling that 100 centimeters
equals one meter, we can say that 24.5 centimeters equals 0.245 meters.

So then, the moment that weโll
calculate equals the force ๐น multiplied by this distance in meters. And now, having considered the
positive and negative directions for displacement, letโs do the same thing for
our force. Our force, we can see, is
directed downward. We could give that a negative
value, which would mean that our force ๐น equals negative 11 newtons. Another way to think of this is
to consider the direction of the moment about point ๐ that this force
creates. We can see that, based on the
position of this force and its direction relative to point ๐, it will tend to
create a moment in a clockwise direction. By convention, moments in this
direction are considered negative. This means that the answer we
get for ๐ should be a negative value. And this confirms our choice
for forces in this downward direction to be negative.

All this to say, our force ๐น
is negative 11 newtons, which means the moment that it creates about point ๐ is
negative 2.695 newton meters. This is the moment of that
force about point ๐.

Now letโs look at an example where
we need to combine the moments created by multiple forces.

๐ด๐ต is a rod with length 114
centimeters and negligible weight. Forces of magnitudes 83
newtons, 225 newtons, 163 newtons, and 136 newtons are acting on the rod as
shown in the following figure. ๐ถ and ๐ท are the points of
trisection of ๐ด๐ต, and point ๐ is the midpoint of the rod. Find the algebraic sum of the
moments of these forces about the point ๐.

Okay, so looking at our figure,
we see this rod ๐ด๐ต with point ๐ด on the right end and ๐ต on the left. We also see these four forces
acting either vertically upward or vertically downward on this rod, and as well
as this, the points of trisection of the rod, ๐ท and ๐ถ โ these are points that
divide the rod into three even lengths โ and the midpoint of the rod point
๐.

Putting our focus on this
midpoint, we want to calculate the total moment about point ๐ due to these four
forces. As we begin to do that, letโs
recall the fact that the length of our rod, the length of line segment ๐ด๐ต, is
114 centimeters. And weโll also say that we want
to calculate the moment about point ๐. Weโll call this ๐ sub ๐.

We can now clear away our
problem statement. And letโs recall the fact that
when it comes to moments of force, the moment ๐ created by some force around a
given point is equal to the component of that force perpendicular to the line
between where the force is applied and the point of interest times the distance
from that point of interest to the applied force. To calculate ๐ sub ๐ then,
the net moment about point ๐ due to these four forces, weโll add together the
moment created by each one individually to get that total.

And notice that we have a sign
convention for the moments that are generated by these forces. If a force tends to cause
rotation in a counterclockwise direction about point ๐, then that moment is
considered positive. So, for example, letโs consider
the force applied at point ๐ด. This force will create a moment
of 83 newtons multiplied by this distance ๐ด๐. And we see, because this will
tend to create a counterclockwise rotation about point ๐, this individual
moment will be positive.

To keep track of the moments
created by these four forces, letโs solve for them individually. The moment created by this
force โ weโll call it ๐ sub 83 โ is equal to the force of 83 newtons times the
length of this line segment ๐ด๐. Weโre told that ๐ is the
midpoint of line segment ๐ด๐ต, which has a length of 114 centimeters. 114 divided by two is 57, and
so this is ๐ 83, the moment due to the 83-newton force about point ๐.

Next, weโll consider ๐ 225 the
moment due to the 225-newton force. This is equal to that force
times the distance of the line segment ๐ถ๐. If we recall that point ๐ถ is
at one of the trisection points of our line segment ๐ด๐ต, then that indicates
that this distance here is 114 centimeters divided by three. Thatโs 38 centimeters. And if we subtract that
distance from the distance from point ๐ด to point ๐, half the length of the
rod, then we find that line segment ๐ถ๐ equals half the length of the rod, 57
centimeters, minus 38 centimeters. That equals 19 centimeters.

Weโre just about done with this
moment. But notice that this force of
225 newtons would tend to create a clockwise rotation about point ๐. Therefore, weโll add a negative
sign to this moment. So while the 83-newton force
created a positive moment, the 225-newton force creates a negative moment about
point ๐.

Moving on to our next force,
163 newtons, this force creates a moment which equals that force magnitude times
the length of the line segment ๐๐ท. And we can note that the length
of this line segment is equal to that of ๐ถ๐. Thatโs because point ๐ท, just
like point ๐ถ, is a trisection point of the line segment ๐ด๐ต. So line segment ๐๐ท is 19
centimeters as well. And we see that this force
tends to create a counterclockwise moment about point ๐ and therefore is
positive. And then, lastly, we consider
the moment created by our 136-newton force. This equals 136 newtons
multiplied by the length of our line segment ๐๐ต. And that distance, we can
recall, is equal to the length of ๐ด๐, 57 centimeters. Lastly, we consider what
direction of rotation this force would tend to create around point ๐. We see that its moment would
rotate in a clockwise direction and therefore is negative by our sign
convention.

Alright, now that weโve
calculated each moment individually, letโs combine them to solve for the net
moment about point ๐. ๐ sub ๐ is equal to ๐ 83
plus ๐ 225 plus ๐ 163 plus ๐ 136. And if we multiply and then
combine all of these terms on our calculator, the answer we get is negative 4199
newton centimeters. This is the cumulative or net
moment around point ๐.

Letโs finish up now by summarizing
a few key points from this lesson. In this video, weโve seen that a
moment of force measures a forceโs tendency to cause a body to rotate about a
specific point. We learned further that if a
forceโs line of action passes through a point, then the force creates no moment
about that point. And lastly, considering the case of
a force which generates a moment about some point, we said that that moment can be
represented by a capital ๐ and that itโs equal to the component of that force
perpendicular to the line between where the force is applied and the point of
interest multiplied by the distance between those two locations.