Question Video: Using the Cross Product to Determine the Moment of a Force Mathematics • Third Year of Secondary School

Video Transcript

True or False: If the force 𝐅 acts at point 𝐴, then its moment in 3D space about point 𝐵 is given by 𝐴 to 𝐵 cross 𝐅.

Let’s consider the moment of a force 𝐹 acting from a point 𝐴 about a pivot point 𝐵 from first principles. The magnitude of the moment 𝑀 is given by the magnitude of the force 𝐹 multiplied by the perpendicular distance of the line of action of 𝐹 from the pivot point 𝐵. The line of action of the force is simply the line that extends to ∞ in both directions, passing through the point of action to the force and parallel to the force. The perpendicular distance between the line of action and the pivot point 𝐵 is therefore given by the length of this line.

Since this is a perpendicular distance, by definition, we have formed a right angle between this line and the line of action. We can therefore form a right triangle whose hypotenuse is the magnitude of the vector 𝐴 to 𝐵 and whose opposite side is the perpendicular distance. If we take this angle here to be 𝜃, then 𝑑 is given by the magnitude of the vector 𝐴 to 𝐵 multiplied by sin 𝜃. The magnitude of the moment 𝑀 is therefore given by the magnitude of the force 𝐹 multiplied by the magnitude of the vector 𝐴 to 𝐵 multiplied by sin 𝜃.

Recall that the cross product of two vectors 𝐮 cross 𝐯 is equal to the magnitude of 𝐮 multiplied by the magnitude of 𝐯 multiplied by the sin of the angle between them 𝜃 multiplied by the unit vector 𝐧. If we take both vectors to be acting from the same point, then the angle between them 𝜃 is the smallest angle between the two vectors. The unit vector 𝐧 is the vector of length one that is perpendicular to both 𝐮 and 𝐯 and therefore perpendicular to the plane containing both 𝐮 and 𝐯 and in the direction according to the right-hand screwing rule.

In this example, both 𝐮 and 𝐯 are in the plane at the screen. Therefore, the unit vector 𝐧 will be either straight into or straight out of the screen. The right-hand screwing rule tells us that if we imagine the screw turning from the vector 𝐮 to the vector 𝐯 about their point of origin, then the direction in which the screw moves gives us the direction of the unit vector. In this case, therefore, the unit vector 𝐧 goes straight into the screen.

Now notice that this formula for the cross product is exactly the same as the formula for the magnitude of the moment of the force apart from the unit vector 𝐧. We have the magnitude of the first vector 𝐅 multiplied by the magnitude of the second vector 𝐴 to 𝐵 multiplied by the sin of the angle between them 𝜃. As an aside, the angle between 𝐹 and the vector 𝐴 to 𝐵 is in fact 𝜋 minus 𝜃. But we can use the property of the sine function that sin of 𝜋 minus 𝜃 is equal to sin of 𝜃, which therefore gives us the same answer. The magnitude of the moment 𝑀 is therefore equal to the magnitude of the cross product 𝐴 to 𝐵 cross 𝐹.

But now, we need to consider the direction of the moment. Looking at the diagram, it’s clear that the force 𝐹 will have a turning force about the pivot point 𝐵 in the anticlockwise direction. If we consider the cross product 𝐴 to 𝐵 cross 𝐹, then by the right-hand screwing rule, we imagine turning a screw about the point 𝐴 from the vector 𝐴 to 𝐵 to the vector 𝐅. This turning force is in the clockwise direction. So this will give us a vector with the same magnitude but in the opposite direction to the moment of the force.

The true moment of this force will therefore be given by the negative of this vector 𝐵 to 𝐴 cross 𝐹. The answer is therefore false. The vector from the point of action of the force to the pivot point 𝐴 to 𝐵 cross 𝐹 will give us a moment of the correct magnitude but the wrong direction. The true moment of the force is given by the vector from the pivot point to the point of action 𝐵 to 𝐴 cross 𝐹.