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 𝐢 plus 𝑚𝐣 is acting at the point 𝐴 seven, three, determine the
moment of the force 𝐅 about the point 𝐵 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 𝐵.
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 𝐵, 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 𝐎𝐀. 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, 𝑚,
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 𝑎 sub one, 𝑎 sub two, 𝑎 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 𝐢, 𝐣, 𝐤, 𝑎 sub one, 𝑎 sub two, 𝑎
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 𝐢, 𝐣, 𝐤, zero, five, zero,
negative five, 𝑚, zero. Then, to find the determinant of
this three-by-three matrix, we multiply 𝐢 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 𝐤 times the
determinant of the final two-by-two matrix. So it’s 𝐢 times five times zero
minus zero times 𝑚 minus 𝐣 times zero times zero minus zero times negative five
plus 𝐤 times zero times 𝑚 minus five times negative five. And this simplifies really nicely
to get 25𝐤.
Now, of course, that unknown 𝑚
canceled out when we completed our cross product. And so we calculated the moment of
the force 𝐅 about our point 𝐵. It’s simply 25𝐤.
In this example, we calculated the
vector moment of a planar force about a point. And interestingly, the resulting
vector only contained a 𝐤-component; the 𝐢- 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 𝑥𝑦-plane, 𝑚 should be perpendicular to
this. In other words, it’s parallel to
the unit vector 𝐤 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 𝑎, 𝑏 and 𝑐, 𝑑. Their two-dimensional cross product
is given by 𝑎𝑑 minus 𝑏𝑐 times the unit vector 𝐤. 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
𝐢 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 𝐎𝐀,
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 𝐎𝐀 is the vector three, six. Then, the force 𝐅 is four 𝐢 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. 𝑎, 𝑏 crossed with 𝑐, 𝑑 gives us
𝑎𝑑 minus 𝑏𝑐 times the unit vector 𝐤. 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 𝐤. This simplifies to negative
33𝐤. So the moment 𝐌 about the origin
of our force is negative 33𝐤.
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 33𝐤, 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 33𝐤,
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 𝐤. In other words, there is some
scalar 𝑐 such that the vector 𝐌 is equal to 𝑐 times 𝐤. 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
𝑚, then that means 𝑐 is either equal to 𝑚 or negative 𝑚.
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 𝑚 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 𝑚 times the unit vector 𝐤. 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 𝐴𝐵 is at
negative six, seven and 𝐴𝐵 has midpoint 𝐷 negative seven, one. If the line of action of the force
𝐅 equals negative two 𝐢 minus six 𝐣 bisects 𝐴𝐵, determine the moment of the
force 𝐅 about point 𝐵.
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 𝐴𝐵. 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 𝐵, 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 𝐴𝐵 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 𝐎𝐀 from the vector 𝐎𝐃. 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
𝐢 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 𝐤. 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 𝐵 to be six 𝐤.
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 𝑚 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 𝑚 times the unit vector 𝐤. 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.
