# Question Video: Finding the Moment of a Force Vector Acting At a Point When Given Information about the Line of Action of the Force Mathematics

End 𝐴 of line segment 𝐴𝐵 is at (−6, 7) and 𝐴𝐵 has midpoint 𝐷(−7, 1). If the line of action of the force 𝐅 = −2𝐢 − 6𝐣 bisects 𝐴𝐵, determine the moment of 𝐅 about point 𝐵.

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### Video Transcript

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.