In this explainer, we will learn how to find the moment of vector forces acting on a body about a point in 3D.
A force, or a system of forces, acting on a body may have a turning effect on this body. This turning effect is described by the moment of the force (or of the system of forces). You may already be familiar with the moment of a force defined as a scalar given by the product of the magnitude of the force with the perpendicular distance between the line of action of the force and the point about which the moment is taken:
In this explainer, we will learn that the moment of a force is properly defined as a vector.
The moment of the force depends on the point of the body about which it is taken. The direction of the moment vector gives the direction of rotation about that point that the force would generate, while its magnitude indicates the strength of the force’s turning effect.
Let us first see how the moment vector is defined.
Definition: Moment of a Force
The moment of force acting on a body, taken about point , is given by where is the position vector of , the point of application of force .
In this definition, we see that the coordinate system has been chosen such that its origin coincides with the point about which we take the moment. If we wanted to work out the moment of force about point that is not the origin, then we would simply replace with :
The letter has been added as a subscript to to indicate that the moment is taken about point .
The properties of the cross product allow us to conclude first that is a vector perpendicular to the plane defined by and . The direction of is given by the right-hand rule. This rule is sometimes explained by referring to the rotation of a screw: the direction of the vector corresponds to the direction of the movement (up or down) of a bottle lid or a nut that one would turn in the same sense of rotation as when going from to , as illustrated in the following diagram.
This reference to rotation makes a lot of sense when talking about the moment of a force. Let us take the diagram used in the definition box above and imagine the movement of a rod fixed at and subjected to a single force at . It is worth noting that, for the sake of simplicity, the coordinate system is chosen here such that and are in the -plane, so that is parallel to the -axis.
We see that the rod would turn clockwise until it is aligned with , that is, when has no turning effect anymore on the rod, which is indeed described by the fact that its moment is zero (which is consistent with the cross product of two collinear vectors being zero). This clockwise rotation corresponds to the rotation of vector toward vector .
Using the right-hand rule as explained above, it means that the moment of about points down. In other words, its only component, which is along the -axis, is negative since the positive -axis points up. Remember that, in a three-dimensional Cartesian coordinate system, the right set of unit vectors is such that .
The magnitude of the moment is given by where is the angle between and . We have used absolute value bars around here because if one uses directed angles, can be negative, as it is the case for the clockwise rotation discussed above, where is then a negative angle. If one uses geometric angles, then , and thus .
Let us see now why this way of defining the magnitude of is equivalent to what you may have learned before; namely, , where is the perpendicular distance between the line of action of and the point about which the moment is taken.
Taking our previous example of the rod, we can draw a circle with center and radius . We can think of it as a unit circle scaled by and rotated in such a way that the -axis associated with the circle is in the same direction as . The scalar projection of vector on the -axis associated with the unit circle is the segment , whose length is then . Considering now the segment (observe that triangles and are congruent), we see that this length is also the perpendicular distance between the line of action of and .
To show that the perpendicular distance between the line of action of and is given by , we can also use the property that two supplementary angles have the same sine value . As (here is a geometric angle), we have . In triangle , we have
This result means that, to work out the magnitude of the moment, we can take whatever angle between the line of action of and the line ; the result will be the same.
As said above, the magnitude of the moment of a force indicates the strength of the force’s turning effect. We can link the equation with something we all have experienced in our daily life, namely, that the turning effect of a given force will increase if the force is applied further away from the pivot, the point about which the object turns. It is the working principle of a lever.
The method to calculate the cross product of two vectors from its components can be applied to calculate the moment of a force. With the force, , and the position vector of the point of application of force defined in the coordinate system as and , the moment of about point is given by
Let us look at an example where a three-dimensional moment vector is determined.
Example 1: Determining the Moment of a Force about a Point in Three Dimensions
A force of magnitude 6 N is acting on and is represented by a vector in a plane perpendicular to the -axis as shown in the figure. Determine its moment vector about in newton-centimeters.
To answer this question, we will first determine the components of vector , which is the vector from the point about which the moment is taken and the point of application of the force, and of force acting on .
From the figure, we find and , with 1 cm being the unit length of the coordinate system. Hence,
The magnitude of force is 6 N. Let us draw the direction of in a plane perpendicular to the -axis, that is, in a plane parallel to the -plane. Beware of the orientation of the -axis when working in three dimensions! We have represented here the plane by looking at it from left to right in the given diagram, that is, such that the unit vector is pointing downward with respect to our screen.
lies in a plane perpendicular to the -axis; hence, . From the figure, we see that
Since , we find that
We can now work out the moment of about as
Let us now look at an example where multiple forces act at a point to produce a moment.
Example 2: Finding the Resultant Moment Vector of Two Forces about the Origin in Three Dimensions
In the figure, if the forces and are acting on the point , where and are measured in newtons, determine the moment vector of the resultant about the point in newton-centimetres.
Two forces act at , and . Since the two forces act at the same point, the sum of their moments is equal to the moment of their resultant. For this reason, we are asked to find the moment of their resultant (i.e., their sum).
Let us start by finding the resultant, , of and :
We want to calculate the moment of about point , the origin, so we need to find the position vector of , . From the figure, we find that
The moment of about point is given by
Let us now look at an example in which unknown components of a force vector are determined from the moment about a point due to the force.
Example 3: Finding the Unknown Components of a Force given Its Position Vector and the Moment Components about an Axis in Three Dimensions
If the force is acting at a point whose position vector is and the - and -components of the moment of the force about the origin point are 73 and 242 units of moment, respectively, find the values of and .
As we are given here the - and -components of the moment of the force about the origin, let us first work out the moment using the position and force vectors. It will give us the three components of the moment in terms of and :
Equating the - and -components of with the values given in the question gives
Let us solve equation (1):
And equation (2) gives
The answer is and .
Let us now look at another example in which we are asked to determine unknown components of the position vector of the point at which the force acts using the moment of the force.
Example 4: Finding the Unknown Coordinates of a Point Lying on the Line of Action of a Force given Its Vector and the Moment of the force about a Point in Three Dimensions
The moment of the force about the origin is , where and . Given that the force passes through a point whose -coordinate is 4, find the - and -coordinates of the point.
The moment of a force acting at point about the origin is given by the cross product of the position vector of , , and the force, :
In this question, we are not given the point of application of the force, but we are asked to find the -coordinate of a point that lies on the line of action of the force.
We know, however, that where is the angle between and and is a unit vector perpendicular to the plane defined by and (provided and are noncollinear). As is equal to the perpendicular distance between the line of action of and the point about which the moment is taken (here the origin), , we can write
We see that taking as the position vector of any point on the line of action of will give the same moment, .
Therefore, we can now calculate by working out with being the position vector of the point on the line of action of with coordinates :
We are told in the question that
Equating both expressions for gives an equation for each component:
The first equation can be solved in this way:
And the third equation gives the value of :
We can now check that these solutions verify the second equation to make sure our system of equations is consistent:
The force passes through the point of coordinates .
Let us look at an example in which the perpendicular distance between the line of action of the force and a point is determined.
Example 5: Finding the Moment Vector of a Force Acting on a Given Point and the Length of the Perpendicular between the Origin Point and the Line of Action of the Force
Find the moment of the force about the origin point, given that and is acting at a point whose position vector is with respect to the origin point, then determine the length of the perpendicular segment drawn from the origin point to the line of action of the force .
We are asked to find first the moment of force about the origin. It is given by
Then, we are asked to determine the length of the perpendicular segment drawn from the origin point to the line of action of the force . is what is commonly referred to as the perpendicular distance between the origin and the line of action of , denoted .
Knowing that we see that, as , is given by
We have and
Let us now summarize what we have learned in this explainer.
- The moment of a force about a point describes the turning effect about this point of the force.
- The moment of force acting on a body,
taken about point , the origin of the coordinate system,
is given by
where is the position vector of ,
the point of application of force .
When the moment of force is taken about a point that is not the origin, in the above equation is replaced by .
- Given the properties of the cross product, the magnitude of the moment is given by where is the angle between and .
- Since the perpendicular distance, , between the line of action of force and the point about which the moment is taken is equal to (or ), we have that is,