Answer:
The force-moment system felt at O =
[tex]F = -T = 8.28i + 7.52 j + 12.79 k[/tex]
[tex]\\\\M = -296.36i +234.06j - 54.24k[/tex]
Explanation:
The diagram shown below completes the other half of the question.As such, we will make sure to work in conjunction to that :
Given that;
the tension in the cable, T, in kN, is 10 plus the last digit of your SID,
T = (10 + 7)kN
T = 17 kN
Also; from the diagram ; Co-ordinates at points are
O (0,0,0)
A(6,12,9)
B(17,22,2)
However;
AB = OA - OB
AB = (6i + 12j + 19k) -(17i +22j + 2k)
AB = -11i -10j + 17 k
[tex]|AB| = \sqrt{(-11)^2+(-10^2)+(17)^2} \\\\|AB| = 22.58[/tex]
For the unit vector along AB is expressed as:
[tex]\lambda_{AB} = \frac{AB}{22.58}\\\\\lambda = \frac{-11i -10j+17k}{22.58}[/tex]
Component of Tension [tex]T = \lambda_AB *T[/tex]
[tex]T = \frac{-11i-10j+17k}{22.58}*17\\\\T = -8.28i -7.52j +12.79 k[/tex]
From the diagram , at point O; the reactive species generated are opposite in nature.
Therefore; [tex]F = -T = 8.28i + 7.52 j + 12.79 k[/tex]
Also, to determine the moment of T about point O ; we have:
[tex]\\\\M = AO * T = ( -6i - 12j - 19k ) * (-8.28i - 7.52j +12.79k)[/tex]
[tex]\\M \left[\begin{array}{ccc}1&j&k\\-6&-12&-19\\-8.28&-7.52&12.79\end{array}\right][/tex]
[tex]\\\\\\M = [ (-12*12.79)-(-19*7.52)i - (-6)(12.79)-(-19)(-8.28)j+(-6)(-7.52)-(-12)(-8.28)][/tex]
[tex]\\\\M = (-296.36)i - (-234.06)j + (-54.24)k[/tex]
[tex]\\\\M = -296.36i +234.06j - 54.24k[/tex]
