The moment of P about a line joining points O and D is [tex]\overrightarrow M_{OD} = (33.909, 16.955, 11.303)\,[lb\cdot in][/tex].
In this case we need to determine the expression of the force momentum around line OD ([tex]\overrightarrow{M}_{OD}[/tex]), in pound-force-inches, whose vectorial expression is described below:
[tex]\overrightarrow{M}_{OD} = \left[\frac{(\vec r_{EO}\,\times\,\vec P)\,\bullet\,\vec r_{OD}}{\|\vec r_{OD}\|^{2}} \right]\cdot \vec r_{OD}[/tex] (1)
Where:
- [tex]\vec r_{EO}[/tex] - Distance between force and origin, in inches.
- [tex]\vec P[/tex] - External forces, in pounds.
- [tex]\vec r_{OD}[/tex] - Distance from point O to point D, in inches.
- [tex]\|\vec r_{OD}\|[/tex] - Norm of the distance from point O to point D, in inches.
By direct inspection we get the following vectors:
[tex]\vec r_{EO} = (0, 7.5, 10)\,[in][/tex], [tex]\vec r_{OD} = (30,15,10)\,[in][/tex], [tex]\|r_{OD}\| = 35\,in[/tex] and [tex]\vec P = (480,-120,160)\,[lb][/tex]
By cross product we get the following outcome, which can be obtained by the Sarrus rule for 3 x 3 determinants:
[tex]\vec r_{EO}\,\times\,\vec P = (138.463, 46.154, -346.155) [lb\cdot in][/tex]
And lastly we get the expression for the force moment around the line joining points O and D:
[tex]\overrightarrow M_{OD} = \left[\frac{(138.463, 46.154, -346.155)\,\bullet \,(30,15,10)}{35^{2}}\right]\cdot (30, 15,10)\,[lb\cdot in][/tex]
[tex]\overrightarrow M_{OD} = (33.909, 16.955, 11.303)\,[lb\cdot in][/tex]
The moment of P about a line joining points O and D is [tex]\overrightarrow M_{OD} = (33.909, 16.955, 11.303)\,[lb\cdot in][/tex].
We kindly invite to check this question on force moments: https://brainly.com/question/6278006
