Video: Moment of a Force about a Point in 2D: Scalar

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 we’re talking about in this lesson are different. We’re 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 𝐹 and a point we’ll call 𝑃 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 𝑃.

So let’s do something to connect our point 𝑃 with our force 𝐹. Let’s draw in a rod, which we’ll say has negligibly small mass. But one end of this rod is fixed to point 𝑃, and the force 𝐹 acts on the other end. And we can see now what this force 𝐹 will tend to do to the rod about point 𝑃. 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 𝑀.

Even if this term is new to us, the idea of something like this, what we’re calling a moment of force, is actually familiar. For example, imagine that point 𝑃 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 — we could call it — 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, let’s imagine that our force 𝐹 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 𝑃, 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 that’s perpendicular to our rod by the overall length of the rod. So if we say that our rod has a length 𝐿 and the magnitude of this component of the force perpendicular to it is 𝐹 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 𝐹 times 𝐿 times the square root of two over two. This method of calculating moments, where we figure out the distance between the point about which we’re considering rotation and the force that’s 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, it’s not the only way.

Another approach we can take is to find the perpendicular distance between our force’s line of action and the point about which we’re considering rotation. If we extend the line of action we’ve drawn in and then draw a perpendicular line from our line of action that intersects point 𝑃, another way to calculate the moment of this force is to solve for this distance — we’ll call it 𝑑 — and multiply that by the force 𝐹. And notice that if we do this, we’re once again calculating a moment arm — originally, that was the length 𝐿 and now it’s the distance 𝑑 — and multiplying it by a force that’s perpendicular to that arm. So then, what is this distance 𝑑?

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 𝑑 as 𝐿 times the cosine of that angle. So in this second way of calculating our moment of force, we’re multiplying the distance 𝑑 by the component of 𝐹 that’s perpendicular to it. And that’s the entire force magnitude. 𝑑, as we’ve seen, is 𝐿 times the cos of 45 degrees. And writing out this cosine as an exact value, we find that the moment of force equals 𝐹 times 𝐿 times the square root of two over two. And we see now that this matches the result of our earlier method.

So let’s write our equation for the moment of force 𝑀 like this. We’ll say that it’s equal to the perpendicular component of a force 𝐹 multiplied by the distance between where that perpendicular component is applied and some point of rotation. And by the way, this quantity that we’re calling a moment in this context is sometimes called a torque. So if we do hear this word torque used, we know that it’s referring to a moment of force.

In general, a moment can be either positive, negative, or zero. We see, for example, that this force 𝐹 would tend to create counterclockwise rotation of this rod about the point we’ve picked 𝑃. 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, it’s also possible for a moment to be zero.

Say that along with our point 𝑃 we pick a second point — we’ll call it point 𝑄 — and we locate it along the line of action of our force. Because 𝑄 lies along this line, there’s no distance between the perpendicular component of our force and this point. Physically, it would never tend to create a rotation about point 𝑄. And that’s why it can’t give rise to a nonzero moment. While the moment of force 𝐹 about point 𝑃 is 𝐹 times 𝐿 times the square root of two over two, the moment of 𝐹 about point 𝑄 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 we’re working with, moments always consist of a unit of force multiplied by a unit of distance.

Knowing all this, let’s 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 𝐴, find the norm of the moment of the force about the point 𝐴, giving your answer in newtons times meters.

Alright, so let’s say that this is our point 𝐴 and that a distance of eight centimeters away we have a force we’ll call it 𝐹 with a magnitude of 498 newtons. Knowing all this, we want to calculate the norm of the moment of this force about point 𝐴.

To start off, we can recall that the moment of a force about some point — we’ll call this moment capital 𝑀 — is equal to the component of an applied force 𝐹 that’s 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 𝐹 and the distance of eight centimeters are already perpendicular to one another. Therefore, the moment of this force about point 𝐴 is equal to 𝐹 times eight centimeters.

If we substitute in the given value for 𝐹, we’re just about ready to multiply to calculate 𝑀. But before we do, let’s 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, we’ll have units of newtons times centimeters. So before we go any further, let’s 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 𝐴.

Now let’s 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 𝑂. 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 𝑂 here. To define the moment of this force about point 𝑂, we can recall that the moment due to some force about some point is equal to the component of that force that’s 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 — this is called the line of action of the force — then the moment of this force about point 𝑂 will equal the force multiplied by this distance we’ve called 𝑑.

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 𝑑. If we begin at point 𝑂 and then consider what we could call the three segments of this apparatus, if we travel from point 𝑂 along the first segment and then down along the second segment and finally along the third until we’ve reached the line of action of our force, then the total horizontal displacement of that three-segment journey will be equal to the distance 𝑑 we want to calculate.

To determine this distance, let’s 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 that’s because as we’ve said motion to the left is negative.

When these three values are combined, they equal 𝑑, 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 we’ll multiply our force 𝐹 in order to solve for the moment that force creates about point 𝑂. Before we do that though, since we want our answer in the end in newtons times meters, let’s 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 we’ll calculate equals the force 𝐹 multiplied by this distance in meters. And now, having considered the positive and negative directions for displacement, let’s 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 𝐹 equals negative 11 newtons. Another way to think of this is to consider the direction of the moment about point 𝑂 that this force creates. We can see that, based on the position of this force and its direction relative to point 𝑂, 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 𝑀 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 𝐹 is negative 11 newtons, which means the moment that it creates about point 𝑂 is negative 2.695 newton meters. This is the moment of that force about point 𝑂.

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

𝐴𝐵 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. 𝐶 and 𝐷 are the points of trisection of 𝐴𝐵, and point 𝑂 is the midpoint of the rod. Find the algebraic sum of the moments of these forces about the point 𝑂.

Okay, so looking at our figure, we see this rod 𝐴𝐵 with point 𝐴 on the right end and 𝐵 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, 𝐷 and 𝐶 — these are points that divide the rod into three even lengths — and the midpoint of the rod point 𝑂.

Putting our focus on this midpoint, we want to calculate the total moment about point 𝑂 due to these four forces. As we begin to do that, let’s recall the fact that the length of our rod, the length of line segment 𝐴𝐵, is 114 centimeters. And we’ll also say that we want to calculate the moment about point 𝑂. We’ll call this 𝑀 sub 𝑂.

We can now clear away our problem statement. And let’s recall the fact that when it comes to moments of force, the moment 𝑀 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 𝑀 sub 𝑂 then, the net moment about point 𝑂 due to these four forces, we’ll 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 𝑂, then that moment is considered positive. So, for example, let’s consider the force applied at point 𝐴. This force will create a moment of 83 newtons multiplied by this distance 𝐴𝑂. And we see, because this will tend to create a counterclockwise rotation about point 𝑂, this individual moment will be positive.

To keep track of the moments created by these four forces, let’s solve for them individually. The moment created by this force — we’ll call it 𝑀 sub 83 — is equal to the force of 83 newtons times the length of this line segment 𝐴𝑂. We’re told that 𝑂 is the midpoint of line segment 𝐴𝐵, which has a length of 114 centimeters. 114 divided by two is 57, and so this is 𝑀 83, the moment due to the 83-newton force about point 𝑂.

Next, we’ll consider 𝑀 225 the moment due to the 225-newton force. This is equal to that force times the distance of the line segment 𝐶𝑂. If we recall that point 𝐶 is at one of the trisection points of our line segment 𝐴𝐵, then that indicates that this distance here is 114 centimeters divided by three. That’s 38 centimeters. And if we subtract that distance from the distance from point 𝐴 to point 𝑂, half the length of the rod, then we find that line segment 𝐶𝑂 equals half the length of the rod, 57 centimeters, minus 38 centimeters. That equals 19 centimeters.

We’re just about done with this moment. But notice that this force of 225 newtons would tend to create a clockwise rotation about point 𝑂. Therefore, we’ll 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 𝑂.

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 𝑂𝐷. And we can note that the length of this line segment is equal to that of 𝐶𝑂. That’s because point 𝐷, just like point 𝐶, is a trisection point of the line segment 𝐴𝐵. So line segment 𝑂𝐷 is 19 centimeters as well. And we see that this force tends to create a counterclockwise moment about point 𝑂 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 𝑂𝐵. And that distance, we can recall, is equal to the length of 𝐴𝑂, 57 centimeters. Lastly, we consider what direction of rotation this force would tend to create around point 𝑂. We see that its moment would rotate in a clockwise direction and therefore is negative by our sign convention.

Alright, now that we’ve calculated each moment individually, let’s combine them to solve for the net moment about point 𝑂. 𝑀 sub 𝑂 is equal to 𝑀 83 plus 𝑀 225 plus 𝑀 163 plus 𝑀 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 𝑂.

Let’s finish up now by summarizing a few key points from this lesson. In this video, we’ve seen that a moment of force measures a force’s tendency to cause a body to rotate about a specific point. We learned further that if a force’s 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 𝑀 and that it’s 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.