Lesson Video: Moment of a Force about a Point in 2D: Vectors Mathematics

In this video, we will learn how to find the moment of a planar system of forces acting on a body about a point as a vector.

16:08

Video Transcript

In this video, we’ll learn how to find the moment of a planar system of forces acting on a body about a point as a vector.

We know that a force or a system of forces can have a rotational effect on a body. And we call that the moment of the force or the system of forces about the point. In planar motion, the moment of some force 𝐅 about a point is a scalar whose magnitude is given by the product of the magnitude of the force with the perpendicular distance 𝑑 between the point and the line of action of that force. Then, we determine the sign by considering whether the rotational effect is clockwise or counterclockwise.

It’s conventional to define the moment with a counterclockwise effect to be positive, meaning that the moment with a clockwise rotational effect is said to be negative. Now, whilst this works well for planar motion, we want to extend the definition of the moment to three-dimensional motion from the scalar moment defined for planar motion. And in order to preserve the notion of the orientation, the direction of the rotation, we define the moment to be a vector as follows.

The moment of the vector force 𝐅 about 𝑂 is given by the cross product of the vector 𝐫 with the vector 𝐅, where the vector 𝐫 is the position vector of the point of application of that force. Now, of course, in this definition, the coordinate system has been chosen such that the origin coincides with the point about which we take the moment. Of course, this might not necessarily be the case. Suppose instead we wanted to work out the moment of a force 𝐅 about some point 𝑃 that’s not the origin. If we define 𝐴 to be the point of application of the force, then we replace the vector 𝐫 with the vector 𝐏𝐀.

So the moment of the force about 𝑃 is the cross product of the vector 𝐏𝐀 with the vector 𝐅. So with all this in mind, let’s demonstrate how to use this formula to calculate the vector moment of a force about some point.

If the force vector 𝐅 equals negative five 𝐒 plus π‘šπ£ is acting at the point 𝐴 seven, three, determine the moment of the force 𝐅 about the point 𝐡 seven, negative two.

In order to calculate the moment of a planar force about a point, let’s recall two of the formulae we can use. If we’re considering the moment of some force 𝐅 taken about the origin, then we calculate the cross product of the vector 𝐫 with the vector 𝐅, where 𝐫 is the position vector of the point of application of the force. In this case though, we want to determine the moment of the force acting at 𝐴 about point 𝐡.

And so we need to reorient ourselves within the coordinate plane. And to do so, we replace the vector 𝐫 with the vector 𝐁𝐀. This now is the vector moment of the force 𝐅 acting at point 𝐴 about point 𝐡, the cross product of 𝐁𝐀 with vector 𝐅.

So let’s begin by finding vector 𝐁𝐀. The vector 𝐁𝐀 is given by subtracting the vector 𝐎𝐁 from the vector πŽπ€. Now, of course, the point 𝐴 has coordinates seven, three. So in the three-dimensional plane, it has the vector seven, three, zero. Similarly, the vector 𝐎𝐁 is seven, negative two, zero. Then, we simply subtract the individual components. And we find the vector 𝐁𝐀 is the vector zero, five, zero. Then, inspecting that vector force 𝐅, we see we can alternatively represent it as the vector negative five, π‘š, zero.

By representing each vector in this way, we can then find the cross product of the vector 𝐁𝐀 with the vector 𝐅. Remember, the cross product can be expressed as a determinant. If we think about the vector 𝐚 with elements π‘Ž sub one, π‘Ž sub two, π‘Ž sub three and the vector 𝐛 with elements 𝑏 sub one, 𝑏 sub two, and 𝑏 sub three, then the cross product is the determinant of the three-by-three matrix with elements 𝐒, 𝐣, 𝐀, π‘Ž sub one, π‘Ž sub two, π‘Ž sub three, 𝑏 sub one, 𝑏 sub two, 𝑏 sub three.

So in our case, the cross product of 𝐁𝐀 and 𝐅 is the determinant of the matrix 𝐒, 𝐣, 𝐀, zero, five, zero, negative five, π‘š, zero. Then, to find the determinant of this three-by-three matrix, we multiply 𝐒 by the determinant of the two-by-two matrix that remains if we eliminate the first row and the first column. Then, we multiply 𝐣 by the determinant of the matrix with elements zero, zero, negative five, zero. And then we add 𝐀 times the determinant of the final two-by-two matrix. So it’s 𝐒 times five times zero minus zero times π‘š minus 𝐣 times zero times zero minus zero times negative five plus 𝐀 times zero times π‘š minus five times negative five. And this simplifies really nicely to get 25𝐀.

Now, of course, that unknown π‘š canceled out when we completed our cross product. And so we calculated the moment of the force 𝐅 about our point 𝐡. It’s simply 25𝐀.

In this example, we calculated the vector moment of a planar force about a point. And interestingly, the resulting vector only contained a 𝐀-component; the 𝐒- and 𝐣-components vanished. Now, this isn’t really surprising if we think about the geometric property of a cross product. The vector that results from the cross product of two vectors is perpendicular to those two vectors. So since the moment is defined to be the cross product of two vectors, it must be perpendicular to those. And since in this case the vector 𝐁𝐀 and the vector 𝐅 lay on the π‘₯𝑦-plane, π‘š should be perpendicular to this. In other words, it’s parallel to the unit vector 𝐀 in the three-dimensional system.

Since this will always be the case, we can simplify the calculation of the cross product by using the two-dimensional cross product. Suppose we have two two-dimensional vectors π‘Ž, 𝑏 and 𝑐, 𝑑. Their two-dimensional cross product is given by π‘Žπ‘‘ minus 𝑏𝑐 times the unit vector 𝐀. Since this is much quicker to calculate, we’ll use this formula to work out the cross product between two vectors for the rest of this lesson.

In our previous example though, we calculated the vector moment of the force. But what about the magnitude? Well, since the moment is given by the cross product of the vector 𝐫 and the vector 𝐅, then its magnitude is the magnitude of this cross product. This can, however, be alternatively represented by multiplying the magnitude of the vector force with 𝑑, the perpendicular distance between the point and the line of action of that force. And that’s really useful because we can rearrange to obtain a formula that allows us to calculate that perpendicular distance. It will be the magnitude of the moment divided by the magnitude of the force.

In our next example, we’ll use this formula to find the perpendicular distance between a point and the line of action of a force.

Given that the force 𝐅 equals four 𝐒 minus three 𝐣 acts through the point 𝐴 three, six, determine the moment 𝐌 about the origin 𝑂 of the force 𝐅. Also, calculate the perpendicular distance 𝐿 between 𝑂 and the line of action of the force.

Remember, we can calculate the moment about 𝑂 of some force 𝐅 by finding the cross product of the vector πŽπ€, where 𝐴 is the point at which the force acts, with the vector 𝐅. And we can simplify this by using the two-dimensional definition of a cross product.

So since the point 𝐴 has coordinates three, six, the vector πŽπ€ is the vector three, six. Then, the force 𝐅 is four 𝐒 minus three 𝐣. So in component form, that’s the vector four, negative three. The cross product of two-dimensional vectors is defined as shown. π‘Ž, 𝑏 crossed with 𝑐, 𝑑 gives us π‘Žπ‘‘ minus 𝑏𝑐 times the unit vector 𝐀. In this case then, we’re going to multiply three by negative three and then subtract six times four. And we’ll multiply that by the vector 𝐀. This simplifies to negative 33𝐀. So the moment 𝐌 about the origin of our force is negative 33𝐀.

Next, we need to find the perpendicular distance 𝐿 between the origin and the line of action of the force. And the formula we can use to calculate that is to divide the magnitude of the moment by the magnitude of the force. Well, since the moment is vector negative 33𝐀, its magnitude is simply 33. But of course the magnitude of force 𝐅 is the square root of the sum of the squares of its components. That’s the square root of four squared plus negative three squared, which is equal to five.

This means the perpendicular distance 𝐿 is the quotient of these. It’s 33 divided by five, which is equal to 6.6. So the moment 𝐌 is negative 33𝐀, and that distance 𝐿 is 6.6 length units.

Now, in our last two examples, we noted that the moment of a force about a point results in a vector that’s parallel to the unit vector 𝐀. In other words, there is some scalar 𝑐 such that the vector 𝐌 is equal to 𝑐 times 𝐀. We also observed that the magnitude of the moment is equal to the magnitude of the scalar moment. If we define that to be equal to π‘š, then that means 𝑐 is either equal to π‘š or negative π‘š.

To determine which is true, we examine whether or not the sign of 𝑐 matches the sign of the scalar moment. In particular, if 𝑐 is positive, the moment vector comes out of the plane; it points up. And that corresponds to the counterclockwise rotation. And if 𝑐 is negative, the moment vector goes into the plane, down, and we have that clockwise rotation.

Let’s formalize this a little. Suppose we have π‘š and vector 𝐌 as the scalar and vector moment of a force or system of forces, respectively, on a plane about a point. Then, the vector 𝐌 is equal to the scalar π‘š times the unit vector 𝐀. This property allows us to establish why the vector moment is a reasonable extension of the scalar moment for a planar force. Furthermore, we can generalize it to represent a moment of a general 3D force about a point since it’s obtained using the cross product.

Now, there’s another property that links the moment of a force with the point about which the force is acting. It’s outside of the scope of this video to demonstrate where this comes from. But we must be aware that the vector moment 𝐌 of a force about a point is independent of the point at which the force acts, as long as the point lies in the same line of action of the force. Let’s demonstrate the application of this.

End 𝐴 of line segment 𝐴𝐡 is at negative six, seven and 𝐴𝐡 has midpoint 𝐷 negative seven, one. If the line of action of the force 𝐅 equals negative two 𝐒 minus six 𝐣 bisects 𝐴𝐡, determine the moment of the force 𝐅 about point 𝐡.

Remember, we can calculate the moment of a force 𝐅 taken about some point 𝑂 by finding the cross product of 𝐫 with vector 𝐅, where 𝐫 is the position vector of 𝐴, which is the point of application of the force. But in this case we’re not given the point at which the force is acting. We are, however, told that the line of action of the force 𝐅 bisects 𝐴𝐡. This tells us then that the line of action of the force passes through the midpoint 𝐷 negative seven, one. Now, we know that as long as the point lies in the same line of action of the force, then the vector moment 𝐌 is independent of the initial point. And so we can calculate the moment by considering that initial point to be at 𝐷 negative seven, one.

Since we’re not working with the origin, we’re going to replace the vector 𝐫 with the vector 𝐁𝐃. We’re finding the moment of 𝐅 about 𝐡, assuming that the force is acting at 𝐷. So let’s begin by finding the vector 𝐁𝐃. Now, we know that the midpoint of line segment 𝐴𝐡 is point 𝐷. So if we plot point 𝐴 and point 𝐷 on a coordinate plane, this then allows us to show that the magnitude, the length, of the vector 𝐀𝐃 must be equal to the length of the vector 𝐁𝐃. But of course these are acting in opposite directions, so we can actually say that the negative vector 𝐀𝐃 is equal to the vector 𝐁𝐃.

Then, we can find the vector 𝐀𝐃 by subtracting the vector πŽπ€ from the vector πŽπƒ. So that’s the vector negative seven, one minus the vector negative six, seven. Subtracting the individual components, and we get negative one, negative six. Then, the vector 𝐁𝐃 is the negative of this vector. So we multiply through by negative one, and we find vector 𝐁𝐃 is simply one, six.

And we now have enough to calculate the moment of our force. Since the force 𝐅 is negative two 𝐒 minus six 𝐣, the moment is the cross product of the vector one, six with the vector negative two, negative six. And using the 2D definition of the cross product, we multiply one by negative six and then subtract six times negative two. And that’s all multiplied by the unit vector 𝐀. Well, one times negative six minus six times negative two is positive six. And so we found the moment of our force 𝐅 about point 𝐡 to be six 𝐀.

We’ve now demonstrated how to find the vector and magnitude of a moment of a force 𝐅 about some point. We’ve shown how to use the two-dimensional definition of the cross product to simplify this process and how we can use the formula for the magnitude of a vector moment to find the distance 𝑑 between the moment and the line of action of the force.

It’s worth noting at this stage that we can actually work with a system of forces by considering their sum. In particular, the moment of a system of forces is equal to the sum of the individual moments of each force in that system about the same point.

Let’s now recap the key points from this lesson. The vector moment of a force 𝐅 acting at a point 𝐴 about the origin is given by the cross product of 𝐫 with 𝐅, where 𝐫 is the vector from point 𝑂 to point 𝐴, the point at which the force is acting. The magnitude of the vector moment is given by the magnitude of the force times 𝑑, the perpendicular distance between the point and the line of action of that force.

We saw that the vector moment 𝐌 of a force about a point is independent of the initial point, as long as the point lies in the same line of action. And we also saw that if π‘š and vector 𝐌 are the scalar and vector moments of a force, respectively, about some point, then the vector 𝐌 is equal to the scalar π‘š times the unit vector 𝐀. We learned that we can simplify our calculations by using a two-dimensional cross product and that the moment of a system of forces is equal to the sum of the individual moments of each force in that system about the same point.

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