Question Video: Finding the Moment of a Force Vector about a Point Mathematics

If the force 𝐅 = βˆ’5𝐒 + π‘šπ£ is acting at the point 𝐴(7, 3), determine the moment of the force 𝐅 about the point 𝐡(7, βˆ’2).

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Video Transcript

If the force vector 𝐅 equals negative five 𝐒 plus π‘šπ£ is acting at the point 𝐴 seven, three, determine the moment of the force 𝐅 about the point 𝐡 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 𝐡.

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 𝐡, 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 πŽπ€. 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, π‘š, 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 π‘Ž sub one, π‘Ž sub two, π‘Ž 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 𝐒, 𝐣, 𝐀, π‘Ž sub one, π‘Ž sub two, π‘Ž 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 𝐒, 𝐣, 𝐀, zero, five, zero, negative five, π‘š, zero. Then, to find the determinant of this three-by-three matrix, we multiply 𝐒 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 𝐀 times the determinant of the final two-by-two matrix. So it’s 𝐒 times five times zero minus zero times π‘š minus 𝐣 times zero times zero minus zero times negative five plus 𝐀 times zero times π‘š minus five times negative five. And this simplifies really nicely to get 25𝐀.

Now, of course, that unknown π‘š canceled out when we completed our cross product. And so we calculated the moment of the force 𝐅 about our point 𝐡. It’s simply 25𝐀.

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