Question

A Constant force $F = 3i + 2j - 4$ is applied at the point $(1, - 2,2)$. Find the vector moment of F about the point $(2, - 1,3)$
$(1){\text{ }}6i - 7j + k$
$(2){\text{ }} - 2i + 6j - 10k$
$(3){\text{ }} - 4i - 2j - 5k$
$(4){\text{ }}$ None of the above.

Answer 1

Hint: Generally, we know that the moment of force is called torque.
We know that the torque is the cross-product of the position vector and force vector.
The torque measures the force that makes an object rotate around the axis. We called this a pivot. The sum of the anticlockwise moments about the axis should be equal to the sum of the clockwise moments about the same axis. Hence, the resultant moment is zero.

Complete answer:
The moment of force generally causes a rotation. The force’s turning effect is known as the Moment and it is the product of the force multiplied by the object’s turning point.
The unit of moment of force is the newton meter.it is a vector quantity. The moment of force is a measure of rotation about the axis
The question can be solved as follows,
Let’s consider the point \[A\left( {1, - 2,2} \right){\text{ }}and{\text{ }}B\left( {2, - 1,3} \right)\]
The resultant vector \[r\] becomes,
\[r{\text{ }} = {\text{ }}\vec B.\vec A{\text{ }} = {\text{ }}\left( { - i - j - k} \right)\]
In this context, the vector momentum of force is about the point “\[B\] ” where it can be written as follows,
\[\vec B{\text{ }} = {\text{ }}\vec r{\text{ }} \times {\text{ }}\vec F\]
Now by substituting the value in the above equations
\[ = {\text{ }}\left( { - i - j - k} \right) \times {\text{ }}\left( {3i{\text{ }} + {\text{ }}2j{\text{ }}-{\text{ }}4} \right)\]
After solving the above equation we get the vector momentum of \[F\]
\[ = {\text{ }}6i{\text{ }}-{\text{ }}7j{\text{ }} + {\text{ }}k\]
Hence the option (C) is the correct answer.

Note: The force is the vector quantity because it has both magnitude and direction.
The force can create motion in the object.
The direction of the torque is always perpendicular to \[r\] and \[F\].