Question Video: Using the Cross Product to Find the Line of Action and Moment of a Force about Two Points Mathematics • Third Year of Secondary School

Video Transcript

The force 𝐅 equals three 𝐢 minus four 𝐣 is acting on the point 𝐴 zero, two. Given that 𝐵 equals two, three and 𝐶 equals negative two, one, determine the moment of this force about both points 𝐵 and 𝐶. Which of the following would you conclude about the line of action of the force 𝐅? (A) The line of action of 𝐅 passes through the point 𝐶. (B) The line of action of 𝐅 bisects the line segment 𝐵𝐶. (C) The line of action of 𝐅 is parallel to the line 𝐵𝐶. Or (D) none of the answers are correct.

Let’s begin with a rough diagram of the scenario. We have the three points 𝐴 zero, two; 𝐵 two, three; and 𝐶 negative two, one. And we have the force 𝐅 equals three 𝐢 minus four 𝐣 acting on the point 𝐴. Recall that the moment 𝐌 of a force 𝐅 acting from a point 𝑃 about a pivot point 𝑂 is given by 𝐫 cross 𝐅, where 𝐫 is the vector 𝑂 to 𝑃.

In this case, we are asked to determine the moment of 𝐅 about the points 𝐵 and 𝐶. We therefore need to find the vectors 𝐫 𝐵, which is equal to 𝐵 to 𝐴, and 𝐫 𝐶, which is equal to the vector 𝐶 to 𝐴. So 𝐫 𝐵 is equal to the position vector of 𝐴 zero, two minus the position vector of 𝐵 two, three, which comes to negative two, negative one. Similarly, 𝐫 𝐶 is equal to the position vector of 𝐴 minus the position vector of 𝐶, which is zero, two minus negative two, one, which comes to two, one.

Notice that this is exactly the same vector but with the opposite sign to 𝐫 𝐵. So, when we take the cross product, the moment will be the same but with the opposite sign. This allows us to immediately conclude that the line of action of 𝐅 bisects the line 𝐵𝐶. Since the moment is the same magnitude about both points, the perpendicular distance of the line of action of the force must be the same for both points. And since the sign of the moment is opposite for the two points, the line of action must pass between the two points rather than on the same side if it were parallel to the line between the two points.

Now we need to determine the moment of the force about the points. The moment 𝐌 𝐵 about the point 𝐵 is equal to 𝐫 𝐵 cross 𝐅, which is given by the determinant of the three-by-three matrix 𝐢, 𝐣, 𝐤, negative two, negative one, zero, three, negative four, zero. Both of these vectors are in the 𝑥𝑦-plane and have a 𝐤-component of zero. Therefore, only the 𝐤-component of their cross product will be nonzero.

Taking the determinant of this matrix by expanding along the top row gives us 11𝐤. Since 𝐫 𝐶 is equal to negative 𝐫 𝐵, the moment about the point 𝐶 will just be equal to the negative of the moment about the point 𝐵. So this gives us our final answer. The moment about the point 𝐵 is equal to 11𝐤, and the moment about the point 𝐶 is equal to negative 11𝐤.