A force \(\vec F = 3\hat i + 2\hat j - 4\hat k\) is applied at the point (1, -1, 2). What is the moment of the force about the point (2, -1, 3)?

Concept:

• Moment of a force: It is equal to the magnitude of force multiplied by the perpendicular distance between its line of action and the axis of rotation. Or cross product of distance between line of action and axis of rotation and force acting.⇔ \({\bf{\vec T}} = {\bf{\vec r}} \times {\bf{\vec F}}\)
• Where, \(\vec T\) is torque or moment of force, \(\vec r\) is the position vector, measured from the moment center to any point along the line of action of the force and  is the force acting.

Formulas used:

• Position vector, from \(\left( {{x_1},{y_1},{z_1}} \right)\) to \(\left( {{x_2},{y_2},{z_2}} \right)\):

\(\vec r = \left( {{x_2} - {x_1}} \right)\hat i + \left( {{y_2} - {y_1}} \right)\hat j + \left( {{z_2} - {z_1}} \right)\hat k\)

• Cross product:

\(\vec r \times \vec F = \left[ {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ {{r_1}}&{{r_2}}&{{r_3}}\\ {{F_1}}&{{F_2}}&{{F_3}} \end{array}} \right]\)

Calculation:

Given:

\(\vec F = 3\hat i + 2\hat j - 4\hat k\)

\(\left( {{x_1},{y_1},{z_1}} \right) = \left( {2,\;-1,\;3} \right)\)

\(\left( {{x_2},{y_2},{z_2}} \right) =\left( {1,\;-1,\;2} \right)\)

Find: Moment of force.

Position vector,

\({\rm{\vec r}} = \left( {1 - 2} \right){\rm{\hat i}} + \left( {\left( -1 \right) - \left(- 1 \right)} \right){\rm{\hat j}} + \left( {2 - 3} \right){\rm{\hat k}}\)

⇒ \(\vec r = - \hat i - \hat k\)

Now, Moment of force,

\(\vec T = \vec r \times \vec F\)

⇒ \(\vec T = \left[ {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ -1&0&-1\\ 3&2&-4 \end{array}} \right]\)

\(\Rightarrow \vec T = \hat i\left\{ {0 - \left( -2 \right)} \right\} - \hat j\left\{ {4 - \left(- 3 \right)} \right\} + \hat k\left\{ {\left( -2 \right) - 0} \right\}\)

\( \Rightarrow \vec T = 2\hat i - 7\hat j - 2\hat k\)