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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. - Engineering Mechanics

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Answer in Brief

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.

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Solution

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 π‘˜
6 6 8
120 40 60


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

Concept: Moment of force about a point
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