Question

A force of 140 kN passes through point C (-6,2,2) and goes to point B (6,6,8). Calculate moment of force about origin.

Answer 1

Given : C (-6,2,2)
B (6,6,8)
To find : Moment of force about origin
Assume 𝑏̅ and 𝑐̅ be the position vectors of points B and C respectively w.r.t O (0,0,0)
`overline(OB)=overlineb=6overlineI+6overlineJ+8overlineK`
`overline(OC)=6overlineb+2overlinej+2overlinek`
`overline(CB)=(6overlineI+6overlineJ+8overlineK)-(-6overlineI+2overlineJ+overlineK)`
`=2(6overlineI+2overlineJ+3overlineK)`
`overline(|CB|)=2xsqrt(6^2+2^2+3^2)`
`=14`
Unit vector along `overline(CB)=(CB)/(|CB|)=(6i+2j+3k)/7`
Force along `overline(CB)=overlineF=140x(6i+2j+3k)/7`
`=120overlineI+40 overlineJ+60overlineK`
Moment of 𝐹̅ about O =`overline(OB)xoverlineF`

𝑖	𝑗	𝑘
6	6	8
120	40	60

`=40overlineI+600overlineJ-480K`
Moment of F about C is `40overlineI+600overlineJ-480overlinek kNm`