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

If the force π = β5π’ + ππ£ is acting at the point π΄(7, 3), determine the moment of the force π about the point π΅(7, β2).

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Video Transcript

If the force vector π equals negative five π’ plus ππ£ is acting at the point π΄ seven, three, determine the moment of the force π about the point π΅ seven, negative two.

In order to calculate the moment of a planar force about a point, letβs recall two of the formulae we can use. If weβre considering the moment of some force π taken about the origin, then we calculate the cross product of the vector π« with the vector π, where π« 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 π΄ about point π΅.

And so we need to reorient ourselves within the coordinate plane. And to do so, we replace the vector π« with the vector ππ. This now is the vector moment of the force π acting at point π΄ about point π΅, the cross product of ππ with vector π.

So letβs begin by finding vector ππ. The vector ππ is given by subtracting the vector ππ from the vector ππ. Now, of course, the point π΄ has coordinates seven, three. So in the three-dimensional plane, it has the vector seven, three, zero. Similarly, the vector ππ is seven, negative two, zero. Then, we simply subtract the individual components. And we find the vector ππ is the vector zero, five, zero. Then, inspecting that vector force π, we see we can alternatively represent it as the vector negative five, π, zero.

By representing each vector in this way, we can then find the cross product of the vector ππ with the vector π. Remember, the cross product can be expressed as a determinant. If we think about the vector π with elements π sub one, π sub two, π sub three and the vector π with elements π sub one, π sub two, and π sub three, then the cross product is the determinant of the three-by-three matrix with elements π’, π£, π€, π sub one, π sub two, π sub three, π sub one, π sub two, π sub three.

So in our case, the cross product of ππ and π is the determinant of the matrix π’, π£, π€, zero, five, zero, negative five, π, zero. Then, to find the determinant of this three-by-three matrix, we multiply π’ by the determinant of the two-by-two matrix that remains if we eliminate the first row and the first column. Then, we multiply π£ by the determinant of the matrix with elements zero, zero, negative five, zero. And then we add π€ times the determinant of the final two-by-two matrix. So itβs π’ times five times zero minus zero times π minus π£ times zero times zero minus zero times negative five plus π€ times zero times π minus five times negative five. And this simplifies really nicely to get 25π€.

Now, of course, that unknown π canceled out when we completed our cross product. And so we calculated the moment of the force π about our point π΅. Itβs simply 25π€.